Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

<strong>Contents</strong><strong>1.</strong> <strong>Introduction</strong> 2<strong>2.</strong> <strong>Preliminaries</strong> 4<strong>2.</strong><strong>1.</strong> <strong>Some</strong> <strong>results</strong> on equalizers and coequalizers 4<strong>2.</strong><strong>2.</strong> Contractible Equalizers and Coequalizers 11<strong>2.</strong>3. Adjunction 133. Monads 153.<strong>1.</strong> Liftings of module functors 233.<strong>2.</strong> The category of balanced bimodule functors 363.3. The comparison functor for monads 404. Comonads 444.<strong>1.</strong> Lifting of comodule functors 514.<strong>2.</strong> The comparison functor for comonads 625. Liftings and distributive laws 815.<strong>1.</strong> Distributive laws 815.<strong>2.</strong> Liftings of monads and comonads 846. (Co)Pretorsors and (co)herds 896.<strong>1.</strong> Pretorsors 896.<strong>2.</strong> Herds 926.3. Herds and comonads 936.4. Herds and distributive laws 946.5. Herds and Galois functors 946.6. The tame case 986.7. Copretorsors 1026.8. Coherds 1126.9. Coherds and Monads 1136.10. Coherds and distributive laws 1166.1<strong>1.</strong> Coherds and coGalois functors 1226.1<strong>2.</strong> The cotame case 1267. Herds and Coherds 1287.<strong>1.</strong> Constructing the functor Q 1287.<strong>2.</strong> From herds to coherds 1287.3. Constructing the functor ̂Q 1327.4. From coherds to herds 1417.5. Herd - Coherd - Herd 1458. Equivalence for (co)module categories 1548.<strong>1.</strong> Equivalence for module categories coming from copretorsor 1548.<strong>2.</strong> Equivalence for comodule categories coming from pretorsor 1719. EXAMPLES 1729.<strong>1.</strong> H-Galois extension 1819.<strong>2.</strong> H-Galois coextension 1879.3. Galois comodules 20310. Bicategories 2191<strong>1.</strong> Construction of BIM (C) 2201<strong>2.</strong> Entwined modules and comodules 235Appendix A. Gabriel Popescu Theorem 241References 2651

2<strong>1.</strong> <strong>Introduction</strong>The starting point of this work was the notion of a torsor which comes from principlebundles in classical geometry. A torsor is a principal homogeneous space overa group, i.e. a G-set X where G is a group acting freely and transitively over X. Anidea due to Baer which goes back to the 1920’s allows to reformulate the definitionof a torsor without specifying the group G: a torsor is a set, sometimes called herd,X together with a structure X ×X ×X → X satisfying some parallelogram relations(see [Ba, p. 202] or [Pr, p. 170]).A noncommutative analogue is the notion of a Hopf-Galois object as introducedby Kreimer and Takeuchi [KT]. Let H be a Hopf algebra, flat over the basering k, a (right) H-Galois object A is a (right) H-comodule algebra such that theGalois map β : A ⊗ A → A ⊗ H given by β (x ⊗ y) = xy (0) ⊗ y (1) is bijective(where δ : A → A ⊗ H : x ↦→ x (0) ⊗ x (1) is the H-comodule structure of A) andA co(H) = {x ∈ A | δ (x) = x ⊗ 1 H } = k.A similar concept for the noncommutative case was introduced by Grunspan in [G]as the notion of a quantum torsor. Together with the definition, Grunspan givesthe proof that every quantum torsor gives rise to two Hopf algebras over which itis a bi-Galois extension of the base field. Conversely, Schauenburg in [Sch4] provesthat every Hopf-Galois extension of the base field is a quantum torsor in the senseof Grunspan.The axioms defining a quantum torsor were simplified allowing to prove anywaya correspondence between faithfully flat torsors and Hopf-(bi)Galois object (see[Sch1]). Moreover, Schauenburg in [Sch4] could prove that the two Hopf algebrascoming from a torsor are Morita-Takeuchi equivalent, i.e. their categories of comodulesare equivalent. Another equivalence between module categories has beenstudied in [BMV] and it is related with Morita contexts defined in the pure categoricalsetting. This gave the hint to investigate a special class of herds at this level ofgenerality.The simplified version of the torsor axioms admits a generalization to arbitraryGalois extensions (not only of the base ring or field) and gave rise to different <strong>results</strong>which we try to summarize here. Hopf Galois extensions of an arbitrary algebra Bby introducing the notion of a B-torsor in [Sch1], Galois extensions by bialgebroidsby means of A-B-torsors in [Ho, Chapter 5] and [BB], Galois extensions by coringsusing the notion of a pretorsor in [BB] and Galois comodules of corings arising fromentwining structures using the notion of a bimodule herd in [BV]. Generalizing thenotion of pretorsor given in [BB], pretorsors over two adjunctions are introduced in[BM, Section 4]. Such categorical setting is the one we choose for this work tryingto develop the notion of pretorsor and herd at this pure general level.The first aim of this thesis is to give a unified and self-contained treatment of anumber of known <strong>results</strong> related to the theory of herds. This gives us the technicaltools to deal with the second aim of our work which is to obtain some new <strong>results</strong>about herds and coherds in the pure categorical setting. A herd at this level is apretorsor with respect to a formal dual structure M = (A, B, P, Q, σ A , σ B ). This isgiven by two monads A and B over two different categories, two bimodule functors Pand Q with respect to the monads and functorial morphisms σ A and σ B satisfying

linearity and compatibility conditions. In particular we refer to the study of thespecial class of tame herds, which yields a correspondence with Galois functors(generalizing the historical <strong>results</strong> we started with). Moreover, under the regularityassumption on the formal dual structure related to a herd, one can construct acoherd. Conversely, beginning with a coherd over a regular formal codual structure,a herd can be obtained. By applying twice this process, one can start with a formaldual structure and a herd, construct a formal codual structure and a coherd andthen compute also a new formal dual structure. Under the extra assumption that thestarting formal dual structure is also a Morita context, the final formal dual structurecomputed from a tame herd comes out to be closely related to the starting one. Weconsider a few cases in which the monads are in fact isomorphic. As an application,we develop some examples. The first is given by a right Galois comodule from whichwe derive the herd. Then we simplify the setting and we study the Schauenburg caseof A/k a faithfully flat Hopf-Galois extension with respect to H. In this example wecan compute the comonads associated to the herd and the coherd. The last exampleis a non trivial example of a coherd. It allows to compute the two monads associatedand the equivalence between the module categories with respect to these monads.Finally we investigate the bicategory of balanced bimodule functors which are oneof the most useful tools in this work and is inspired by the balanced bimodules inthe classical sense.We developed the portions of the theory of herds, resp. coherds, we found moresuitable for our purposes.In the first part we collect some well-known <strong>results</strong> including proofs. It is aboutequalizers and coequalizers, contractible equalizers and coequalizers and notationfor adjunctions.Then we concentrate, in the second section, the needed materials for the sequelabout monads. Similarly we do for comonads. At this point we also includethe Beck’s Tripleability Theorem and the generalized version which introduce theEilenberg-Moore comparison functor and the categories of modules and comodules.We reserve a short section to the notion of distributive laws and above all thecorrespondence between distributive laws and liftings of monads and comonads.The next section introduces the notion of pretorsor and herd bringing all thedetails and the <strong>results</strong> needed to prove the equivalence between herds and Galoisfunctors in the tame case. The same has been done for the dual case of copretorsorsand coherds.Later on a section dedicated to a new fundamental functor built from a herdand a coherd respectively and then the theorem relating the starting formal dualstructure and the one obtained after applying the two processes from a herd andfrom a coherd.One section is dedicated to the equivalence between the module categories obtainedfrom a copretorsor and to the equivalence between the comodule categoriesobtained from a pretorsor.The following section is a collection of the examples we provide in this work, aboutherds and coherds first and then applications of Beck’s Theorem and of its generalization.In particular, the example of a coherd was produced during some usefuldiscussions with T. Brzeziński. In the subsection dedicated to Galois comodules we3

4need some material related to Gabriel-Popescu theorem which is contained in theappendix.The last part is the first outcome of a joint work with J. Gomez-Torrecillas.It is devoted to the introduction of the bicategory of balanced bimodule functorsBIM (C). First we fix some notation and terminology about 2-categories and bicategories.Then we define the bicategory of balanced bimodule functors and finallywe study how it can be related to entwined modules and comodules.<strong>2.</strong> <strong>Preliminaries</strong><strong>2.</strong><strong>1.</strong> <strong>Some</strong> <strong>results</strong> on equalizers and coequalizers. In the following, most ofthe computations are justified. We denote by the name of a functorial morphism,its naturality property.Definitions <strong>2.</strong><strong>1.</strong> Let α : B → C be a functorial morphism. We say that α is• a functorial monomorphism, or simply a monomorphism, if for every β, γ :A → B such that α ◦ β = α ◦ γ we have β = γ.• a functorial regular monomorphism, or simply a regular monomorphism, ifα is the equalizer of two functorial morphisms.• a functorial epimorphism, or simply an epimorphism, if for every β, γ : C →D such that β ◦ α = γ ◦ α we have β = γ.• a regular epimorphism, or simply a regular epimorphism, if α is the coequalizerof two functorial morphisms.Definition <strong>2.</strong><strong>2.</strong> A parallel pair α, β : F → F ′ is said to be reflexive if the twoarrows have a common right inverse δ : F ′ → F.Definition <strong>2.</strong>3. A reflexive equalizer is an equalizer of a reflexive parallel pair.Definition <strong>2.</strong>4. A reflexive coequalizer is a coequalizer of a reflexive parallel pair.Lemma <strong>2.</strong>5. Let F, G, H be functors and let f : F → G, g : G → H and h : F → Hbe functorial morphisms such that h = g ◦ f. Assume that f is a functorial isomorphism.Then h is a regular epimorphism if and only if g is a regular epimorphism.Proof. First, let us assume that g is a regular epimorphism, i.e.(H, g) = Coequ Fun (α, β). Then we haveh ◦ f −1 ◦ α = g ◦ f ◦ f −1 ◦ α = g ◦ f ◦ f −1 ◦ β = h ◦ f −1 ◦ β.Now, let χ : F → X be a functorial morphism such that χ ◦ f −1 ◦ α = χ ◦ f −1 ◦ β.By the universal property of the coequalizer (H, g) = Coequ Fun (α, β), there existsa unique functorial morphism χ : H → X such that χ ◦ g = χ. Then, by composingto the right with f we getχ ◦ h = χ ◦ g ◦ f = χ ◦ f −1 ◦ f = χ.Moreover, let χ ′ be another functorial morphism such that χ ′ ◦ h = χ. Since we alsohave χ ◦ h = χ we haveχ ′ ◦ g ◦ f = χ ′ ◦ h = χ = χ ◦ h = χ ◦ g ◦ f

5and since g ◦ f is an epimorphism, we deduce that χ ′ = χ so that(H, h) = Coequ Fun(f −1 ◦ α, f −1 ◦ β ) .Conversely, let us assume that h is a regular epimorphism, i.e.(H, h) = Coequ Fun (α, b). Then we haveg ◦ f ◦ a = h ◦ a = h ◦ b = g ◦ f ◦ b.Now, let ξ : G → X be a functorial morphism such that ξ ◦ f ◦ a = ξ ◦ f ◦ b. Bythe universal property of (H, h) = Coequ Fun (α, b), there exists a unique functorialmorphism ξ : H → X such that ξ ◦ h = ξ ◦ f, i.e. ξ ◦ g ◦ f = ξ ◦ f and since f isan isomorphism we deduce that ξ ◦ g = ξ. Let us assume that there exists anotherfunctorial morphism ξ ′ : H → X such that ξ ′ ◦ g = ξ. Since we also have ξ ◦ g = ξwe get thatξ ′ ◦ h = ξ ′ ◦ g ◦ f = ξ ◦ f = ξ ◦ g ◦ f = ξ ◦ hand since h is an epimorphism, we deduce that ξ ′ = ξ. Therefore, (H, g) =Coequ Fun (f ◦ a, f ◦ b).□Lemma <strong>2.</strong>6. Let F, G, H be functors and let f : F → G, g : G → H and h : F → Hbe functorial morphisms such that h = g ◦ f. Assume that g is a functorial isomorphism.Then h is a regular epimorphism if and only if f is a regular epimorphism.Proof. Assume first that f is a regular epimorphism, i.e. (G, f) = Coequ Fun (α, β).Then we haveh ◦ α = g ◦ f ◦ α = g ◦ f ◦ β = h ◦ β.Let ξ : F → X be a functorial morphism such that ξ ◦ α = ξ ◦ β. By the universalproperty of the coequalizer (G, f) = Coequ Fun (α, β), there exists a unique functorialmorphism ξ such that ξ ◦ f = ξ. Then we haveξ ◦ g −1 ◦ h = ξ ◦ g −1 ◦ g ◦ f = ξ ◦ f = ξso that ξ factorizes through h via ξ ◦g −1 . Moreover, if there exists another functorialmorphism ξ ′ : F → X such that ξ ′ ◦ h = ξ, since we also have ξ = ξ ◦ g −1 ◦ h wehaveξ ′ ◦ g ◦ f = ξ ′ ◦ h = ξ = ξ ◦ g −1 ◦ h = ξ ◦ g −1 ◦ g ◦ f = ξ ◦ fand since f is epi we getfrom which we deduce thatTherefore we obtainedξ ′ ◦ g = ξξ ′ = ξ ◦ g −1 .(H, h) = Coequ Fun (α, β) .Conversely, let now assume that h is a regular epimorphism, i.e.(H, h) = Coequ Fun (a, b). Then we haveand since g is mono we get thatg ◦ f ◦ a = h ◦ a = h ◦ b = g ◦ f ◦ bf ◦ a = f ◦ b.

6Let now χ : F → X be a functorial morphism such that χ ◦ a = χ ◦ b. by theuniversal property of the coequalizer (H, h) = Coequ Fun (a, b) there exists a uniquefunctorial morphism χ : H → X such that χ ◦ h = χ and hence χ ◦ g ◦ f = χ so thatχ factorizes through f via χ ◦ g. Moreover, let χ ′ : F → X be another functorialmorphism such that χ ′ ◦ f = χ. Since we also have χ ◦ h = χ we haveχ ′ ◦ g −1 ◦ h = χ ′ ◦ g −1 ◦ g ◦ f = χ ′ ◦ f = χ = χ ◦ hand since h is epi we get that χ ′ ◦ g −1 = χ from which we deduce thatTherefore (G, f) = Coequ Fun (a, b) .χ ′ = χ ◦ g.Lemma <strong>2.</strong>7. Let A and B be categories, let F, F ′ : A → B be functors and α, β : F →F ′ be functorial morphisms. If, for every X ∈ A, there exists Coequ B (αX, βX),then there exists the coequalizer (C, c) = Coequ Fun (α, β) in the category of functors.Moreover, for any object X in A, we have (CX, cX) = Coequ B (αX, βX).Proof. Define a functor C : A → B with object map (CX, cX) = Coequ B (αX, βX)for every X ∈ A. For a morphism f : X → X ′ in A, naturality of α and β impliesthat(F ′ f) ◦ (αX) = (αX ′ ) ◦ (F f) and (F ′ f) ◦ (βX) = (βX ′ ) ◦ (F f)and hence(cX ′ ) ◦ (F ′ f) ◦ (αX) = (cX ′ ) ◦ (αX ′ ) ◦ (F f) ccoequ= (cX ′ ) ◦ (βX ′ ) ◦ (F f)= (cX ′ ) ◦ (F ′ f) ◦ (βX)i.e. (cX ′ ) ◦ (F ′ f) coequalizes the parallel morphisms βX and αX. In light of thisfact, by the universal property of the coequalizer (CX, cX) , Cf : CX → CX ′ isdefined as the unique morphism in B such that (Cf) ◦ (cX) = (cX ′ ) ◦ (F ′ f). Byconstruction, c is a functorial morphism F ′ → C such that c ◦ α = c ◦ β. It remainsto prove universality of c. Let H : A → B be a functor and let χ : F ′ → H bea functorial morphism such that χ ◦ α = χ ◦ β. Then, for any object X in A,(χX) ◦ (αX) = (χX) ◦ (βX). Since ⊂ (CX, cX) = Coequ B (αX, βX), there is aunique morphism ξX : CX → HX such that (ξX) ◦ (cX) = χX. The proof iscompleted by proving naturality of ξX in X. Take a morphism f : X → X ′ in A.Since c and χ functorial morphisms,(Hf) ◦ (ξX) ◦ (cX) = (Hf) ◦ (χX) χ = (χX ′ ) ◦ (F ′ f)= (ξX ′ ) ◦ (cX ′ ) ◦ (F ′ f) = (ξX ′ ) ◦ (Cf) ◦ (cX).Since cX is a epimorphism, we get that ξ is a functorial morphism.Lemma <strong>2.</strong>8 ([BM, Lemma <strong>2.</strong>1]). Let C and K be categories, let G,G ′ : C → K befunctors and γ, θ : G → G ′ be functorial morphisms. If, for every X ∈ C, thereexists Equ K (γX, θX), then there exists the equalizer (E, i) = Equ Fun (γ, θ) in thecategory of functors. Moreover, for any object X in C, (EX, iX) = Equ K (γX, θX).Lemma <strong>2.</strong>9. Let A and B be categories, let F ,F ′ : A → B be functors, and let α, β :F → F ′ be functorial morphisms. Assume that, for every X ∈ A, B has coequalizersof αX and βX and let (Q, q) = Coequ Fun (α, β). Under these assumptions, for anyfunctor P : D → A, Coequ Fun (αP, βP ) = (QP, qP ) .□□

Proof. Clearly (qP )◦(αP ) = (qP )◦(βP ). Let y : F ′ P → Y be a functorial morphismsuch that y ◦ (αP ) = y ◦ (βP ). Then,(yD) ◦ (αP D) = (yD) ◦ (βP D)and hence, since by Lemma <strong>2.</strong>7 (QP D, qP D) = Coequ B (αP D, βP D), there existsa unique d D : QP D → Y D such thatd D ◦ (qP D) = yD.Let us prove that the assignment D ↦→ d D yields a functorial morphism d : QP → Y .Let h : D → D ′ be a morphism in D. We compute(Y h) ◦ d D ◦ (qP D) = (Y h) ◦ (yD) y = (yD ′ ) ◦ (F ′ P h)Since qP D is an epimorphism, we conclude.= d D ′ ◦ (qP D ′ ) ◦ (F ′ P h) q = d D ′ ◦ (QP h) ◦ (qP D) .Lemma <strong>2.</strong>10 ([BM, Lemma <strong>2.</strong>2]). Let G, G ′ : C → K be functors, and let γ, θ :G → G ′ be functorial morphisms. Assume that every pair of parallel morphisms inK has an equalizer and let (E, i) = Equ Fun (γ, θ). Under these assumptions, for anyfunctor P : D → C, Equ Fun (γP, θP ) = (EP, iP ).Lemma <strong>2.</strong>1<strong>1.</strong> Consider the following serially commutative diagram in an arbitrarycategory KAfgBn mn ′m ′n ′′ m ′′A ′ f ′ B ′ i ′ C ′g ′ee ′ e ′′A ′′ f ′′ g ′′ B ′′ i ′′ C ′′Assume that all columns are coequalizers and also the first and second rows arecoequalizers. Then also the third row is a coequalizer.Proof. In order to see that the third row is a fork, note that, by commutativity ofthe diagram and fork property of the second row,i ′′ ◦ f ′′ ◦ e = i ′′ ◦ e ′ ◦ f ′ = e ′′ ◦ i ′ ◦ f ′ = e ′′ ◦ i ′ ◦ g ′ = i ′′ ◦ e ′ ◦ g ′= i ′′ ◦ g ′′ ◦ e.Since e is an epimorphism, this proves that the third row is a fork that is i ′′ ◦ f ′′ =i ′′ ◦ g ′′ .To conclude we want to prove the universality of i ′′ . To do so, let us take anymorphism x : B ′′ → X such that x ◦ f ′′ = x ◦ g ′′ . Then we want to prove that thereexist a unique functorial morphism z : C ′′ → X such that z ◦ i ′′ = x.We observex ◦ e ′ ◦ f ′ = x ◦ f ′′ ◦ e = x ◦ g ′′ ◦ e = x ◦ e ′ ◦ g ′ .so we get thatx ◦ e ′ ◦ f ′ = x ◦ e ′ ◦ g ′ .iC7□

8Since the second row is a coequalizer by assumption, there is a unique morphismy : C ′ → X such that(1) y ◦ i ′ = x ◦ e ′ .We calculatey ◦ n ′′ ◦ i = y ◦ i ′ ◦ n ′ (1)= x ◦ e ′ ◦ n ′ = x ◦ e ′ ◦ m ′ (1)= y ◦ i ′ ◦ m ′= y ◦ m ′′ ◦ iand since i is an epimorphism we get thaty ◦ n ′′ = y ◦ m ′′ .Since the third column is a coequalizer, there exists a unique morphism z : C ′′ → Xsuch thatThenz ◦ e ′′ = y.z ◦ i ′′ ◦ e ′ = z ◦ e ′′ ◦ i ′ = y ◦ i ′ = x ◦ e ′so we get that z ◦ i ′′ = x. Since e ′′ ◦ i ′ = i ′′ ◦ e ′ and e ′′ , e ′ , i ′ are epimorphism, wededuce that i ′′ is epimorphism and hence z is unique with respect to z ◦ i ′′ = x. □Corollary <strong>2.</strong>1<strong>2.</strong> Let F , F ′ : A → B be functors and α, β : F → F ′ be functorialmorphisms. Assume that, for every X ∈ A, B has coequalizers of αX and βXhence there exists (Q, q) = Coequ Fun (α, β), cf. Lemma <strong>2.</strong>7. Assume that (P, p) =Coequ A (f, g) of morphisms f, g : X → Y in A and that both F and F ′ preserveCoequ A (f, g). Then also Q preserves Coequ A (f, g).Proof. The following diagram (in B) is serially commutative by naturalityF XαXβXF ′ XqXQXF fF gF ′ fF ′ gQfQgF YαYβYF ′ YqY QYF pF ′ pQpF PαPβPF ′ PqP QPThe columns are coequalizers by Lemma <strong>2.</strong>7. The first and second rows are coequalizersby the assumption that F and F ′ preserve coequalizers. Thus the third row isa coequalizer by Lemma <strong>2.</strong>1<strong>1.</strong>□

Lemma <strong>2.</strong>13 ([BM, Lemma <strong>2.</strong>5]). Consider the following serially commutative diagramin an arbitrary category KAiBee ′ e ′′A ′ i ′ B ′ f ′ C ′g ′n mn ′m ′n ′′ m ′′A ′′ i ′′ B ′′ f ′′ g ′′ C ′′Assume that all columns are equalizers and also the second and third rows are equalizers.Then also the first row is an equalizer.Proof. Dual to Lemma <strong>2.</strong>1<strong>1.</strong>Corollary <strong>2.</strong>14. Let G,G ′ : C → K be functors and γ, θ : G → G ′ be functorialmorphisms. Assume that, for every X ∈ C, K has equalizers of γX and θX hencethere exists (E, e) = Equ Fun (γ, θ), cf. Lemma <strong>2.</strong>8. Assume that (I, i) = Equ C (f, g)of morphisms f, g : X → Y in C and that both G and G ′ preserve Equ C (f, g). Thenalso E preserves Equ C (f, g).Proof. Dual to Corollary <strong>2.</strong>1<strong>2.</strong>Lemma <strong>2.</strong>15. Let Z, Z ′ , W, W ′ : A → B be functors, let a, b : Z → W and a ′ , b ′ :Z ′ → W ′ be functorial morphisms, let ϕ : Z → Z ′ and ψ : W → W ′ be functorialisomorphisms such thatψ ◦ a = a ′ ◦ ϕ and ψ ◦ b = b ′ ◦ ϕ.Assume that there exist (E, i) = Equ Fun (a, b) and (E ′ , i ′ ) = Equ Fun (a ′ , b ′ ). Then ϕinduces an isomorphism ̂ϕ : E → E ′ such that ϕ ◦ i = i ′ ◦ ̂ϕ.fgC9□□EˆϕiZϕE ′ i ′Z ′ b ′aWProof. Let us define ̂ϕ. Let us computebψa ′ W ′a ′ ◦ ϕ ◦ i = ψ ◦ a ◦ i defi= ψ ◦ b ◦ i = b ′ ◦ ϕ ◦ iand since (E ′ , i ′ ) = Equ Fun (a ′ , b ′ ) there exists a unique functorial morphism ̂ϕ : E →E ′ such thati ′ ◦ ̂ϕ = ϕ ◦ i.Note that ̂ϕ is mono since so are i and i ′ and ϕ is an isomorphism. Considerϕ −1 : Z ′ → Z and ψ −1 : W ′ → W . Then we havea ◦ ϕ −1 = ψ −1 ◦ a ′ and b ◦ ϕ −1 = ψ −1 ◦ b ′ .

10Let us computea ◦ ϕ −1 ◦ i ′ = ψ −1 ◦ a ′ ◦ i ′ def i ′= ψ −1 ◦ b ′ ◦ i ′ = b ◦ ϕ −1 ◦ i ′and since (E, i) = Equ Fun (a, b) there exists a unique functorial morphism ̂ϕ ′ : E ′ →E such thati ◦ ̂ϕ ′ = ϕ −1 ◦ i ′ .Then we havei ◦ ̂ϕ ′ ◦ ̂ϕ = ϕ −1 ◦ i ′ ◦ ̂ϕ = ϕ −1 ◦ ϕ ◦ i = iand since i is a monomorphism we deduce thatSimilarlŷϕ ′ ◦ ̂ϕ = Id E .i ′ ◦ ̂ϕ ◦ ̂ϕ ′ = ϕ ◦ i ◦ ̂ϕ ′ = ϕ ◦ ϕ −1 ◦ i ′ = i ′and since i ′ is a monomorphism we obtain that̂ϕ ◦ ̂ϕ ′ = Id E ′.Lemma <strong>2.</strong>16. Let K : B → A be a full and faithful functor and let f, g : X → Y bemorphisms in B. If (KE, Ke) = Equ A (Kf, Kg) then (E, e) = Equ B (f, g).Proof. Since K is faithful, from (Kf) ◦ (Ke) = (Kg) ◦ (Ke) we get that f ◦ e = g ◦ e.Let h : Z → X be a morphism in B such that f ◦ h = g ◦ h. Then in A we get(Kf) ◦ (Kh) = (Kg) ◦ (Kh) and hence there exists a unique morphism ξ : KZ →KE such that (Ke) ◦ ξ = (Kh). Since ξ ∈ Hom A (KZ, KE) and K is full, thereexists a morphism ζ ∈ Hom B (Z, E) such that ξ = Kζ. Since K is faithful, from(Ke) ◦ (Kζ) = Kh we get e ◦ ζ = h. From the uniqueness of ξ, the one of ζ easilyfollows.□Lemma <strong>2.</strong>17. Let α, γ : F → G be functorial morphisms where F, G : A → Bare functors. Assume that, for every X ∈ A there exists Equ B (αX, γX). Let(E, i) = Equ Fun (α, γ) , where i : E → F . Then, for every X ∈ A and Y ∈ B wehave that(Hom B (Y, EX) , Hom B (Y, iX)) = Equ Sets (Hom B (Y, αX) , Hom B(Y, γX))which means thatwhere(Hom B (−, E) , Hom B (−, i)) = Equ Fun (Hom B (−, α) , Hom B(−, γ))Hom B (−, E) and Equ Fun (Hom B (−, α) , Hom B(−, γ)) : B op × A → Sets.Proof. We have thatHom B (Y, αX) ◦ Hom B (Y, iX) = Hom B (Y, (αX) ◦ (iX))= Hom B (Y, (γX) ◦ (iX)) = Hom B (Y, γX) ◦ Hom B (Y, iX)□

i.e. Hom B (Y, iX) equalizes Hom B (Y, αX) and Hom B (Y, γX) , for every X ∈ A andY ∈ B. Let now ζ : Z → Hom B (Y, F X) be a map such that Hom B (Y, αX) ◦ ζ =Hom B (Y, γX) ◦ ζ. Then, for every X ∈ A, Y ∈ B and for every z ∈ Z we have(αX) ◦ ζ (z) = Hom B (Y, αX) (ζ (z)) = Hom B (Y, γX) ◦ (ζ (z))= (γX) ◦ ζ (z) .Then, for every X ∈ A and Y ∈ B there exists a unique morphism θ z : Y → EX inB such that(iX) ◦ θ z = ζ (z)i.e.Hom B (Y, iX) (θ z ) = ζ (z) .The assignment z ↦→ θ z defines a map θ : Z → Hom B (Y, EX) such that Hom B (Y, iX)◦θ = ζ.□<strong>2.</strong><strong>2.</strong> Contractible Equalizers and Coequalizers.11Definition <strong>2.</strong>18. Let C be a category.eightuple (Z, X, Y, d, d 0 , d 1 , s, t) whereA contractible ( or split) equalizer is asuch thatZ d Xsd 0td 1 Yt ◦ d 0 = Id Xs ◦ d = Id Zt ◦ d 1 = d ◦ sd 0 ◦ d = d 1 ◦ d.Proposition <strong>2.</strong>19. Let C be a category and let (Z, X, Y, d, d 0 , d 1 , s, t) be a contractibleequalizer. Then (Z, d) = Equ C (d 0 , d 1 ) .Proof. Let ξ : L → X be such thatthenLetso thatd 0 ◦ ξ = d 1 ◦ ξξ = Id X ◦ ξ = t ◦ d 0 ◦ ξ = t ◦ d 1 ◦ ξ = d ◦ (s ◦ ξ) .ξ ′ = s ◦ ξ : L → Zξ = d ◦ ξ ′ .Let now ξ ′′ : L → Z be such that d ◦ ξ ′′ = ξ. Thenξ ′′ = Id Z ◦ ξ ′′ = s ◦ d ◦ ξ ′′ = s ◦ ξ = ξ ′ .□

12Proposition <strong>2.</strong>20. Let C be a category, let (Z, X, Y, d, d 0 , d 1 , s, t) be a contractibleequalizer and let F : C → D be a functor. Thenis a contractible equalizer in D.F Z F d F XF sF d 0F tF d 1 F YProof. Since functors preserve composition, the statement is proved.Proposition <strong>2.</strong>2<strong>1.</strong> Assume thatZ d Xis an equalizer and there exists t : Y → X such thatd 0d 1 Y□t ◦ d 0 = Id Xd 1 ◦ t ◦ d 1 = d 0 ◦ t ◦ d 1Then there exists s : X → Z such that (Z, X, Y, d, d 0 , d 1 , s, t) is a contractible equalizer.Proof. Since d 1 ◦ t ◦ d 1 = d 0 ◦ t ◦ d 1 and (Z, d) = Equ (d 0 , d 1 ), there exists s : X → Zsuch thatt ◦ d 1 = d ◦ s.Let us computed ◦ s ◦ d = t ◦ d 1 ◦ d = t ◦ d 0 ◦ d = dand since d is mono we gets ◦ d = Id Z .□Definition <strong>2.</strong>2<strong>2.</strong> Let F : C → D be a functor. An F -contractible equalizer pair isa parallel pairin C such that there exists a contractible equalizerin D.XD d F Xsd 0d 1All the previous <strong>results</strong> can be considered in the opposite category so that theygive the dual notion, namely contractible coequalizers.Definition <strong>2.</strong>23. Let C be a category.(C, X, Y, c, d 0 , d 1 , u, v) whereF d 0tF d 1 Y F YA contractible coequalizer is a eightupleXd 0vd 1 YcuC

13such thatd 0 ◦ v = Id Yd 1 ◦ v = u ◦ cc ◦ u = Id Cc ◦ d 0 = c ◦ d 1 .Proposition <strong>2.</strong>24 ([BW, Proposition 2 (a)]). Let C be a category and let(C, X, Y, c, d 0 , d 1 , u, v) be a contractible coequalizer. Then (C, c) = Coequ C (d 0 , d 1 ) .Proof. Dual to Proposition <strong>2.</strong>24.Definition <strong>2.</strong>25. Let F : C → D be a functor. An F -contractible coequalizer pairis a parallel pairin C such that there exists a contractible coequalizerin D.<strong>2.</strong>3. Adjunction.F XXF d 0vF d 1d 0d 1<strong>2.</strong>26. Let L : B → A and R : A → B be functors. Recall that L is called a leftadjoint of R, or R is called a right adjoint of L if there exists functorial morphisms F Y Yη : Id B → RL and ɛ : LR → Id Asuch that(ɛL) ◦ (Lη) = L and (Rɛ) ◦ (ηR) = R.In this case we also say that (L, R) is an adjunction and η is called the unit of theadjunction while ɛ is called the counit of the adjunction. Leta X,Y : Hom A (LY, X) → Hom B (Y, RX)be the isomorphism of the adjunction (L, R). Then, for every ξ ∈ Hom A (LY, X)and for every ζ ∈ Hom B (Y, RX) we also havea X,Y (ξ) = (Rξ) ◦ (ηY ) and a −1X,Y(ζ) = (ɛX) ◦ (Lζ) .Moreover, for every X ∈ A, Y ∈ B, unit and counit of the adjunction are given byηY = a LY,Y (Id LY ) and ɛX = a −1X,RX (Id RX) .<strong>2.</strong>27. Let (L, R) be an adjunction. Then L preserves colimits and thus coequalizersand R preserves limits and thus equalizers. We also say that L is right exact andthat R is left exact.Lemma <strong>2.</strong>28. Let (L, R) be an adjunction with unit η and counit ɛ, where L : B → Aand R : A → B. For every Y ′ ∈ B the following conditions are equivalent:(1) L −,Y ′ = a −1LY ′ ,− ◦ Hom B (−, ηY ′ ) is a functorial isomorphismcuC□

14(2) Hom B (−, ηY ′ ) is a functorial isomorphism(3) ηY ′ is an isomorphism (η is a functorial isomorphism).Proof. Since (L, R) is an adjunction, a X,Y : Hom A (LY, X) → Hom B (Y, RX) is anisomorphism for every X ∈ A and for every Y ∈ B, so that (1) is equivalent to (2).(3) ⇒ (2) Let η −1 Y ′ be the two-sided inverse of ηY ′ . Then Hom B (−, η −1 Y ′ ) is theinverse of the functor Hom B (−, ηY ′ ) . In fact, let f ∈ Hom B (Y, Y ′ ) and compute[ (HomB −, η −1 Y ′) ◦ Hom B (−, ηY ′ ) ] ((f) = Hom B −, η −1 Y ′) (ηY ′ ◦ f)= ( η −1 Y ′) ◦ (ηY ′ ) ◦ f = fand[HomB(−, ηY ′ ) ◦ Hom B(−, η −1 Y ′)] (f) = Hom B (−, ηY ′ ) (( η −1 Y ′) ◦ f )= (ηY ′ ) ◦ ( η −1 Y ′) ◦ f = f.Thus Hom B (−, ηY ′ ) is a functorial isomorphism.(2) ⇒ (3) Since Hom B (−, ηY ′ ) is a functorial isomorphism, in particularHom B (RLY ′ , ηY ′ ) : Hom B (RLY ′ , Y ′ ) → Hom B (RLY ′ , RLY ′ ) is an isomorphism.Thus, there exists f ∈ Hom B (RLY ′ , Y ′ ) such that (ηY ′ ) ◦ f = Id RLY ′ which impliesthat ηY ′ is an epimorphism. Moreover we also have Hom B (Y ′ , ηY ′ ) (Id Y ′) = ηY ′ =(ηY ′ ) ◦ f ◦ (ηY ′ ) = Hom B (Y ′ , ηY ′ ) (f ◦ (ηY ′ )) . Since Hom B (−, ηY ′ ) is a functorialisomorphism, also Hom B (Y ′ , ηY ′ ) is an isomorphism. Thus we deduce that Id Y ′ =f ◦ (ηY ′ ) which implies that ηY ′ is also a monomorphism and moreover ηY ′ has atwo-sided inverse f : RLY ′ → Y ′ .□Remark <strong>2.</strong>29. Note that, for every f ∈ Hom B (Y, Y ′ ) we haveL Y,Y ′ (f) = [ a −1LY ′ ,Y ◦ Hom B (Y, ηY ′ ) ] (f) = a −1LY ′ ,Y (ηY ′ ◦ f)= (ɛLY ′ ) ◦ (LηY ′ ) ◦ (Lf) (L,R)adj= Lf.Lemma <strong>2.</strong>30. Let (L, R) be an adjunction with unit η and counit ɛ, where L : B → Aand R : A → B. For every X ∈ A the following conditions are equivalent:(1) R X,− = a −,RX ◦ Hom A (ɛX, −) is a functorial isomorphism(2) Hom A (ɛX, −) is a functorial isomorphism(3) ɛX is an isomorphism (ɛ is a functorial isomorphism).Proof. Dual to proof of Lemma <strong>2.</strong>28.Remark <strong>2.</strong>3<strong>1.</strong> Note that, for every f ∈ Hom A (X, X ′ ) we haveR X,X ′ (f) = [a X ′ ,RX ◦ Hom A (ɛX, X ′ )] (f) = a X ′ ,RX (f ◦ ɛX) = R (f ◦ ɛX) ◦ (ηRX)= (Rf) ◦ (RɛX) ◦ (ηRX) (L,R)adj= Rf.Proposition <strong>2.</strong>3<strong>2.</strong> Let (L, R) be an adjunction with unit η and counit ɛ, whereL : B → A and R : A → B. Then R is full and faithful if and only if ɛ is afunctorial isomorphism.Proof. To be full and faithful for R means that the mapφ : Hom A (X, X ′ ) −→ Hom B (RX, RX ′ )□

f ↦→ Rfis bijective for every X, X ′ ∈ A. Since this φ (f) = R (f) = R X,X ′ (f) , φ is anisomorphism if and only if R X,X ′ is an isomorphism for every X, X ′ ∈ A and, byLemma <strong>2.</strong>30, if and only if ɛX is an isomorphism for every X ∈ A.□Lemma <strong>2.</strong>33. Let (L, R) be an adjunction where L : B → A and R : A → B suchthat L is an equivalence of categories. Then R is also an equivalence of categories.Proof. By assumption L : B → A is an equivalence of categories with R ′ : A → B.Then it is well-known that (L, R ′ ) is an adjunction. By the uniqueness of theadjoint, we have that R ≃ R ′ which is an equivalence. Thus R is also an equivalenceof categories.□3. MonadsDefinition 3.<strong>1.</strong> A monad on a category A is a triple A = (A, m A , u A ) , whereA : A → A is a functor, m A : AA → A and u A : A → A are functorial morphismssatisfying the associativity and the unitality conditions:m A ◦ (m A A) = m A ◦ (Am A ) and m A ◦ (Au A ) = A = m A ◦ (u A A) .Definition 3.<strong>2.</strong> A morphism between two monads A = (A, m A , u A ) and B =(B, m B , u B ) on a category A is a functorial morphism ϕ : A → B such thatϕ ◦ m A = m B ◦ (ϕϕ) and ϕ ◦ u A = u B .Example 3.3. Let (A, m A , u A ) be an R-ring where R is an algebra. Then• A is an R-R-bimodule• m A : A ⊗ R A → A is a morphism of R-R-bimodules• u A : R → A is a morphism of R-R-bimodules satisfying the followingm A ◦(m A ⊗ R A) = m A ◦(A ⊗ R m A ) , m A ◦(A ⊗ R u A ) = r A and m A ◦(u A ⊗ R A) = l Awhere r A : A ⊗ R R → A and l A : R ⊗ R A → A are the right and leftconstraints. LetA = − ⊗ R A : Mod-R → Mod-Rm A = − ⊗ R m A : − ⊗ R A ⊗ R A → − ⊗ R Au A = (− ⊗ R u A ) ◦ r−−1 : − → − ⊗ R R → − ⊗ R AWe prove that A = (A, m A , u A ) is a monad on the category Mod-R. Forevery M ∈ Mod-R we compute[m A ◦ (m A A)] (M) = (M ⊗ R m A ) ◦ (M ⊗ R A ⊗ R m A ) = M ⊗ R [m A ◦ (A ⊗ R m A )]= M ⊗ R [m A ◦ (m A ⊗ R A)] = (M ⊗ R m A ) ◦ (M ⊗ R m A ⊗ R A)= [m A ◦ (Am A )] (M)[m A ◦ (Au A )] (M) = (M ⊗ R m A ) ◦ [ ](M ⊗ R u A ) ◦ r −1M ⊗R A= (M ⊗ R m A ) ◦ (M ⊗ R u A ⊗ R A) ◦ ( r −1M ⊗ R A )= (M ⊗ R [m A ◦ (u A ⊗ R A)]) ◦ ( r −1M ⊗ R A )= (M ⊗ R l A ) ◦ ( r −1M⊗ R A ) = M ⊗ R A = AM15

16and[m A ◦ (u A A)] (M) = (M ⊗ R m A ) ◦ (M ⊗ R A ⊗ R u A ) ◦ r −1M⊗ R A= (M ⊗ R [m A ◦ (A ⊗ R u A )]) ◦ r −1M⊗ R A= (M ⊗ R r A ) ◦ r −1M⊗ R A = M ⊗ R A = AM.Proposition 3.4 ([H]). Let (L, R) be an adjunction with unit η and counit ɛ whereL : B → A and R : A → B. Then RL = (RL, RɛL, η) is a monad on the category B.Proof. We have to prove that(RɛL)◦(RLRɛL) = (RɛL)◦(RɛLRL) and (RɛL)◦RLη = RL = (RɛL)◦(ηRL) .In fact we haveand(RɛL) ◦ (RLRɛL) ɛ = (RɛL) ◦ (RɛLRL)(RɛL) ◦ RLη (L,R)= RL (L,R)= (RɛL) ◦ (ηRL) .Definition 3.5. A left module functor for a monad A = (A, m A , u A ) on a categoryA is a pair ( Q, A µ Q)where Q : B → A is a functor and A µ Q : AQ → Q is a functorialmorphism satisfying:A µ Q ◦ ( A A µ Q)= A µ Q ◦ (m A Q) and Q = A µ Q ◦ (u A Q) .Definition 3.6. A right module functor for a monad A = (A, m A , u A ) on a categoryA is a pair ( P, µ A P)where P : A → B, is a functor and µAP : P A → P is a functorialmorphism such thatµ A P ◦ ( µ A P A ) = µ A P ◦ (P m A ) and P = µ A P ◦ (P u A ) .Remark 3.7. Let A = (A, m A , u A ) be a monad on a category A and let ( )Q, A µ Q bea left A-module functor and ( )P, µ A P be a right A-module functor. By the unitalityproperty of A µ Q and µ A P we deduce that they are both epimorphism.Definition 3.8. For two monads A = (A, m A , u A ) on a category A and B =(B, m B , u B ) on a category B, a A-B-bimodule functor is a triple ( Q, A µ Q , µ Q) B , whereQ : B → A is a functor and ( ) ( )Q, A µ Q is a left A-module functor, Q, µBQ is a rightB-module functor such that in additionA µ Q ◦ ( )Aµ B Q = µBQ ◦ (A µ Q B ) .Definition 3.9. A module for a monad A = (A, m A , u A ) on a category A is a pair(X, A µ X)where X ∈ A and A µ X : AX → X is a morphism in A such thatA µ X ◦ ( A A µ X)= A µ X ◦ (m A X) and X = A µ X ◦ (u A X) .A morphism between two A-modules ( X, A µ X)and(X ′ , A µ X ′)is a morphism f :X → X ′ in A such thatA µ X ′ ◦ (Af) = f ◦ A µ X .We will denote by A A the category of A-modules and their morphisms. This is theso-called Eilenberg-Moore category which is sometimes also denoted by A A .□

Remark 3.10. Let A = (A, m A , u A ) be a monad on a category A and let ( X, A µ X)∈AA. From the unitality property of A µ X we deduce that A µ X is epi for every(X, A µ X)∈ A A and that u A X is mono for every ( X, A µ X)∈ A A, i.e. u A is amonomorphism.Definition 3.1<strong>1.</strong> Corresponding to a monad A = (A, m A , u A ) on A, there is anadjunction ( A F, A U) where A U is the forgetful functor and A F is the free functorAU : AA → A AF : A → AA(X, A µ X)→ X X → (AX, mA X)f → f f → Af.Note that A U A F = A. The unit of this adjunction is given by the unit u A of themonad A:u A : A → A U A F = A.The counit λ A : A F A U → A A of this adjunction is defined by settingTherefore we haveAU ( λ A(X, A µ X))= A µ X for every ( X, A µ X)∈ A A.(λ AA F ) ◦ ( A F u A ) = A F and ( A Uλ A ) ◦ (u AA U) = A U.Proposition 3.1<strong>2.</strong> Let A = (A, m A , u A ) be a monad on a category A. Then A U isa faithful functor. Moreover, given Z, W ∈ A A we have thatZ = W if and only if A U (Z) = A U (W ) and A U (λ A Z) = A U (λ A W ) .In particular, if F, G : X → A A are functors, we haveF = G if and only if A UF = A UG and A U (λ A F ) = A U (λ A G)Proposition 3.13. Let A = (A, m A , u A ) be a monad on a category A. Then( A U, ( A Uλ A )) is a left A-module functor.Proof. We have to prove that( A U A λ) ◦ (A A U A λ) = ( A U A λ) ◦ (m AA U) and( A U A λ) ◦ (u AA U) = A U.Let us consider ( X, A µ X)∈ A A. We have to show that(AU A λ ( X, A µ X))◦(AA U A λ ( X, A µ X))=(AU A λ ( X, A µ X))◦(mAA U ( X, A µ X))17and that(AU A λ ( X, A µ X))◦(uAA U ( X, A µ X))= A U ( X, A µ X)i.e. thatA µ X ◦ ( A A µ X)= A µ X ◦ (m A X)andA µ X ◦ (u A X) = Xwhich hold since ( X, A µ X)is an A-module.□

18Proposition 3.14. Let A = (A, m A , u A ) be a monad on a category A and let(X, A µ X)be a module for A. Then we have(X, A µ X)= CoequA(A A µ X , m A X ) .In particular if ( Q, A µ Q)is a left A-module functor, then we have(Q, A µ Q)= CoequFun(A A µ Q , m A Q ) .Proof. Note thatAAXm A Xu A AXA µ X AXis a contractible coequalizer and thus, by Proposition <strong>2.</strong>24,(X, A µ X)= CoequA(A A µ X , m A X ) . Let now ( Q, A µ Q)be a left A-module functorwhere Q : B → A. Then, by the foregoing, for every Y ∈ B we have that(QY, A µ Q Y ) = ( QY, A µ QY)= CoequA(A A µ QY ,m A QY ) = Coequ A(A A µ Q Y,m A QY ) .A µ XThen, by Lemma <strong>2.</strong>7, we have that ( Q, A µ Q)= CoequFun(A A µ Q , m A Q ) .Corollary 3.15. Let A = (A, m A , u A ) be a monad on a category A and let ( A F, A U)be the associated adjunction. Then ( A U, ( A Uλ A )) is a left A-module functor andu A X( A U, ( A Uλ A )) = Coequ Fun (A A Uλ A , m AA U) .Proof. By Proposition 3.13 ( A U, ( A Uλ A )) is a left A-module functor. By Proposition3.14 we get that ( A U, ( A Uλ A )) = Coequ Fun (A A Uλ A , m AA U) . □Proposition 3.16. Let A = (A, m A , u A ) be a monad on a category A and let(P, µAP)be a right A-module functor, then we have(2)Proof. Note thatX(P, µAP)= CoequFun(µAP A, P m A).P AAP m AP Au Aµ A P A P Ais a contractible coequalizer and thus, by Proposition <strong>2.</strong>24,(P, µAP)= CoequFun(µAP A, P m A). □Lemma 3.17. Let A = (A, m A , u A ) be a monad on a category A and let ( Q, A µ Q)bea left and ( P, µ A P)be a right A-module functors where Q : Q → A and P : A → P.Let F : X → Q and G : P → B be functors. Then(1) ( QF, A µ Q F ) is a left A-module functor and(2) ( GP, Gµ A P)is a right A-module functor.Proof. Fromwe deduce thatµ A PP u AA µ Q ◦ ( A A µ Q)= A µ Q ◦ (m A Q) and Q = A µ Q ◦ (u A Q)A µ Q F ◦ ( A A µ Q F ) = A µ Q F ◦ (m A QF ) and QF = A µ Q F ◦ (u A QF )P□

19and fromµ A P ◦ ( µ A P A ) = µ A P ◦ (P m A ) and P = µ A P ◦ (P A)we deduce thatGµ A P ◦ ( Gµ A P A ) = Gµ A P ◦ (GP m A ) and GP = Gµ A P ◦ (GP A) .□Proposition 3.18. Let A = (A, m A , u A ) be a monad on a category A and let( A F, A U) be the adjunction associated. Then A U reflects isomorphisms.Proof. Let f : ( X, A µ X)→(Y, A µ Y)be a morphism in A A such that A Uf has atwo-sided inverse f −1 in A. SinceA µ X ′ ◦ (Af) = f ◦ A µ Xwe get thatf −1 ◦ A µ X ′ = A µ X ◦ ( Af −1) .□Lemma 3.19 ([BMV, Lemma 4.1]). Let A = (A, m A , u A ) be a monad on a categoryA, let ( P, µ A P)be a right A-module functor and let(Q, A µ Q)be a left A-modulefunctor where P : A → B, Q : B → A. Then any coequalizer preserved by P A isalso preserved by P and any coequalizer preserved by AQ is also preserved by Q.Proof. Consider the following coequalizerXfg Yz Zin the category A and assume that P A preserves it. By applying to it the functorsP A and P we get the following diagram in BP u A XP AXP Xµ A P XP AfP AgP fP gP u A YP AYP Yµ A P YP AzP zP u A ZP AZP Z.µ A P ZBy assumption, the first row is a coequalizer. Assume that there exists a morphismh : P Y → H such thath ◦ (P f) = h ◦ (P g) .Then, by composing with µ A P X we geth ◦ (P f) ◦ ( µ A P X ) = h ◦ (P g) ◦ ( µ A P X )and since µ A Pis a functorial morphism we obtainh ◦ ( µ A P Y ) ◦ (P Af) = h ◦ ( µ A P Y ) ◦ (P Ag) .Since (P AZ, P Az) = Coequ B (P Af, P Ag), there exists a unique morphism k :P AZ → H such that(3) k ◦ (P Az) = h ◦ ( µ A P Y ) .

20By composing with P u A Y we getk ◦ (P Az) ◦ (P u A Y ) = h ◦ ( µ A P Y ) ◦ (P u A Y )and thusk ◦ (P u A Z) ◦ (P z) = h.Let l := k ◦ (P u A Z) : P Z → H. Then we havel ◦ (P z) = k ◦ (P u A Z) ◦ (P z) u A= k ◦ (P Az) ◦ (P u A Y )(3)= h ◦ ( µ A P Y ) ◦ (P u A Y ) = h.Let l ′ : P Z → H be another morphism such thatThen we havel ′ ◦ (P z) = h.l ◦ ( µ A P Z ) ◦ (P Az) = l ◦ (P z) ◦ ( µ A P Y ) = h ◦ ( µ A P Y )= l ′ ◦ (P z) ◦ ( µ A P Y ) = l ′ ◦ ( µ A P Z ) ◦ (P Az) .Since P A preserves coequalizers, we have that P Az is an epimorphism. Since µ A P Zis also an epimorphism, we deduce that l = l ′ . Therefore we obtain that (P Z, P z) =Coequ B (P f, P g) . The second statement can be proved similarly. We consider theabove coequalizerXfg Yin the category B and assume that AQ preserves it. By applying to it the functorsAQ and Q we get the following diagram in Au A QXAQXQXA µ Q XAQfAQgQfQgu A QYAQYQYA µ Q YzAQzQz Zu A QZAQZQZA µ Q ZBy assumption, the first row is a coequalizer. Assume that there exists a morphismh : QY → H such thath ◦ (Qf) = h ◦ (Qg) .Then, by composing with A µ Q X we geth ◦ (Qf) ◦ (A µ Q X ) = h ◦ (Qg) ◦ (A µ Q X )and since A µ Q is a functorial morphism we obtainh ◦ (A µ Q Y ) ◦ (AQf) = h ◦ (A µ Q Y ) ◦ (AQg) .Since (AQZ, AQz) = Coequ B (AQf, AQg), there exists a unique morphism k :AQZ → H such that(4) k ◦ (AQz) = h ◦ (A µ Q Y ) .By composing with u A QY we getk ◦ (AQz) ◦ (u A QY ) = h ◦ (A µ Q Y ) ◦ (u A QY )

and thusk ◦ (u A QZ) ◦ (Qz) = h.Let l := k ◦ (u A QZ) : QZ → H. Then we havel ◦ (Qz) = k ◦ (u A QZ) ◦ (Qz) u A= k ◦ (AQz) ◦ (u A QY )(4)= h ◦ (A µ Q Y ) ◦ (u A QY ) = h.Let l ′ : QZ → H be another morphism such thatThen we havel ′ ◦ (Qz) = h.l ◦ (A µ Q Z ) ◦ (AQz) = l ◦ (Qz) ◦ (A µ Q Y ) = h ◦ (A µ Q Y )= l ′ ◦ (Qz) ◦ (A µ Q Y ) = l ′ ◦ (A µ Q Z ) ◦ (AQz) .Since AQ preserves coequalizers, we have that AQz is an epimorphism. Since A µ Q Zis also an epimorphism, we deduce that l = l ′ . Therefore we obtain that (QZ, Qz) =Coequ B (Qf, Qg) .□Lemma 3.20 ([BMV, Lemma 4.2]). Let A = (A, m A , u A ) be a monad on a categoryA and let f, g : ( X, A µ X)→(Y, A µ Y)be morphisms in A A. Assume that there exists(C, c) = Coequ A ( A Uf, A Ug) and assume that AA preserves coequalizers. Then thereexists (Γ, γ) = Coequ A A (f, g) and A U (Γ, γ) = (C, c) .Proof. Since AA preserves coequalizers and (A, m A ) is a right A-module functor, alsoA preserves coequalizers by Lemma 3.19, in particular, A preserves (C, c) . Sincec ◦ A µ Y ◦ (A A Uf) f∈ AA= c ◦ ( A Uf) ◦ A µ Xccoequ= c ◦ ( A Ug) ◦ A µ Xg∈ A A= c ◦ A µ Y ◦ (A A Ug)by the universal property of the coequalizer (AC, Ac) there exists a unique morphismA µ C : AC → C such thatc ◦ A µ Y = A µ C ◦ (Ac) .Moreover, by composing with u A Y this identity we getSince c is an epimorphism we obtainc = A µ C ◦ (Ac) ◦ (u A Y ) u A= A µ C ◦ (u A C) ◦ c.C = A µ C ◦ (u A C) .Now, consider the following serially commutative diagram21m A XAAX A A µ XAA A Uf AA A Ug m A YAAYAAcAACAA µ Ym A CAA µ CA A UfAXAY ACA A UgAcA µ XA µ YXAUfAUgYcA µ C C.

22Since we already observed that the columns are coequalizers and also the first andthe second row are coequalizers by Proposition 3.14, in view of Lemma <strong>2.</strong>11 also thethird row is a coequalizer, so that (C, c) has a left A-module structure, i.e. thereexists (Γ, γ) ∈ A A such that (Γ, γ) = Coequ A A (f, g) and A U (Γ, γ) = (C, c) . □Lemma 3.21 ([BMV, Lemma 4.3]). Let A = (A, m A , u A ) be a monad on a categoryA with coequalizers and let ( A F, A U) be the adjunction associated. The followingstatements are equivalent:(i) A : A → A preserves coequalizers(ii) AA : A → A preserves coequalizers(iii) A A has coequalizers and they are preserved by A U : A A → A(iv) A U : A A → A preserves coequalizers.Proof. (i) ⇒ (ii) and (iii) ⇒ (iv) are clear.(ii) ⇒ (iii) follows by Lemma 3.20.(iv) ⇒ (i) Note that A F is a left adjoint, so that in particular it preserves coequalizers.Then A U A F = A also preserves coequalizers.□Lemma 3.2<strong>2.</strong> Let A = (A, m A , u A ) be a monad over a category A and assume that Apreserves equalizers. Then A F preserves equalizers where ( A F, A U) is the adjunctionassociated to the monad.Proof. LetE e Xbe an equalizer in A. Let us consider the fork obtained by applying the functor A Fto the equalizeri.e.(AE, m A E)AF E A F e AF XAe (AX, m A X)fgAF fAF gAfAg Y AF Y (AY, m A Y )Now, let ( Z, A µ Z)∈ A A and z : ( Z, A µ Z)→ (AX, mA X) be a morphism in A A suchthat (Af) ◦ z = (Ag) ◦ z. Since A preserves equalizers, we know that (AE, Ae) =Equ A (Af, Ag) . By the universal property of the equalizer (AE, Ae) in A, thereexists a unique morphism z ′ : Z → AE in A such that (Ae) ◦ z ′ = z. We now wantto prove that z ′ is a morphism in A A, i.e. that (m A E) ◦ (Az ′ ) = z ′ ◦ A µ Z . Since z isa morphism in A A we have that(m A X) ◦ (Az) = z ◦ A µ Zand since also Ae is a morphism in A A we have thatThen we have(m A X) ◦ (AAe) = (Ae) ◦ (m A E) .(Ae) ◦ (m A E) ◦ (Az ′ ) Ae∈ AA= (m A X) ◦ (AAe) ◦ (Az ′ )propz= (m A X) ◦ (Az) z∈ AA= z ◦ A µ Zpropz= (Ae) ◦ z ′ ◦ A µ Z

and since A preserves equalizers, Ae is a monomorphism, so that we get(m A E) ◦ (Az ′ ) = z ′ ◦ A µ Z .Lemma 3.23. Let A = (A, m A , u A ) be a monad over a category A, let L, M : B → Abe functors and let µ : AL → L be an associative and unital functorial morphism,that is (L, µ) is a left A-module functor. Let h : L → M and let ϕ : AM → M befunctorial morphisms such that(5) h ◦ µ = ϕ ◦ (Ah) .If AAh and h are epimorphisms, then ϕ is associative and unital, that is (M, ϕ) isa left A-module functor.Proof. We calculateϕ ◦ (Aϕ) ◦ (AAh) (5)= ϕ ◦ (Ah) ◦ (Aµ) (5)= h ◦ µ ◦ (Aµ)µis ass.= h ◦ µ ◦ (m A L) (5)= ϕ ◦ (Ah) ◦ (m A L) m A= ϕ ◦ (m A M) ◦ (AAh) .Since AAh is an epimorphism, we deduce that ϕ is associative. Moreover we haveϕ ◦ (u A M) ◦ h u A= ϕ ◦ (Ah) ◦ (u A L) (5)= h ◦ µ ◦ (u A L) = h.Since h is an epimorphism, we get that ϕ is unital.3.<strong>1.</strong> Liftings of module functors.Proposition 3.24 ([Ap] and [J]). Let A = (A, m A , u A ) be a monad on a categoryA and let B = (B, m B , u B ) be a monad on a category B and let Q : A → B be afunctor. Then there is a bijection between the following collections of dataF functors ˜Q : A A → B B that are liftings of Q (i.e. B U ˜Q = Q A U)M functorial morphisms Φ : BQ → QA such thatΦ ◦ (m B Q) = (Qm A ) ◦ (ΦA) ◦ (BΦ) and Φ ◦ (u B Q) = Qu Agiven by( )a : F → M where a ˜Q =( )BUλ B ˜QA F ◦ ( B U B F Qu A )b : M → F where B Ub (Φ) = Q A U and B Uλ B b (Φ) = (Q A Uλ A ) ◦ Φ i.e.b : M → F where b (Φ) (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX))and b (Φ) (f) = Q (f) .Proof. Let ˜Q : A A → B B be a lifting of the functor Q : A → B (i.e. B U ˜Q = Q A U).Define a functorial morphism φ : B F Q → ˜Q A F as the composite( )φ := λ B ˜QA F ◦ ( B F Qu A )where u A : A → A U A F = A is also the unit of the adjunction ( A F, A U) and λ B :BF B U → B B is the counit of the adjunction. Let now defineΦ def= B Uφ : B U B F Q = BQ → B U ˜Q A F = Q A U A F = QA.23□□

24We have to prove that such a Φ satisfies Φ ◦ (m B Q) = (Qm A ) ◦ (ΦA) ◦ (BΦ) andΦ ◦ (u B Q) = Qu A . First, let us compute()(Qm A ) ◦ (ΦA) ◦ (BΦ) = (Qm A ) ◦ BUλ B ˜QA F A ◦ ( B U B F Qu A A)()◦ B B Uλ B ˜QA F ◦ (B B U B F Qu A )()AUλ AA F= (Q A Uλ AA F ) ◦ BUλ B ˜QA F A()◦ ( B U B F Qu A A) ◦ B B Uλ B ˜QA F ◦ (B B U B F Qu A )(eQlifting= BU ˜Qλ) ()AA F ◦ BUλ B ˜QA F A()◦ ( B U B F Qu A A) ◦ B B Uλ B ˜QA F ◦ (B B U B F Qu A )[( ) ( )]= B U ˜QλAA F ◦ λ B ˜QA F A ◦ ( B F Qu A A)()◦ B B Uλ B ˜QA F ◦ (BBQu A )[( ) (λ=BB U λ B ˜QA F ◦ BF B U ˜Qλ)]AA F ◦ ( B F Qu A A)()◦ B B Uλ B ˜QA F ◦ (BBQu A )[( )]eQlifting= B U λ B ˜QA F ◦ ( B F Q A Uλ AA F ) ◦ ( B F Qu A A)()◦ B B Uλ B ˜QA F ◦ (BBQu A )[( )]AUλ AA F= B U λ B ˜QA F ◦ ( B F Qm A ) ◦ ( B F Qu A A)()◦ B B Uλ B ˜QA F ◦ (BBQu A )( ) ()Amonad= BUλ B ˜QA F ◦ B B Uλ B ˜QA F ◦ (BBQu A )( ) (eQisBmod= BUλ B ˜QA F ◦ m BB U ˜Q)A F ◦ (BBQu A )( )m=BBUλ B ˜QA F ◦ (BQu A ) ◦ (m B Q)( )= BUλ B ˜QA F ◦ ( B U B F Qu A ) ◦ (m B Q)Moreover we have= ( B Uφ) ◦ (m B Q) = Φ ◦ (m B Q) .( )BUλ B ˜QA F ◦ ( B U B F Qu A ) ◦ (u B Q)Φ ◦ (u B Q) = ( B Uφ) ◦ (u B Q) =( )u=BBUλ B ˜QA F ◦ (u B Q A U A F ) ◦ (Qu A )( ) (eQlifting= BUλ B ˜QA F ◦ u BB U ˜Q)A F◦ (Qu A ) ( BF, B U)adj= Qu A .

Conversely, let Φ be a functorial morphism satisfying Φ ◦ (m B Q) = (Qm A ) ◦ (ΦA) ◦(BΦ) ( and ) Φ ◦ (u B Q) = Qu A . We define ˜Q : AA → B B by setting, for everyX, A µ X ∈ A A,˜Q (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX)).We have to check that ( Q (X) , ( Q A µ X)◦ (ΦX))∈ B B, that is)B µe QX◦(B B µe QX= B µe QX◦ (m B QX) and B µe QX◦ (u B QX) = QX.We compute)B µe QX◦(B B µe QXpropertyofΦ=Moreover we have= ( Q A µ X)◦ (ΦX) ◦(BQ A µ X)◦ (BΦX)Φ= ( Q A µ X)◦(QA A µ X)◦ (ΦAX) ◦ (BΦX)Xmodule= ( Q A µ X)◦ (QmA X) ◦ (ΦAX) ◦ (BΦX)(Q A µ X)◦ (ΦX) ◦ (mB QX) = B µe QX◦ (m B QX) .B µe QX◦ (u B QX) = ( Q A µ X)◦ (ΦX) ◦ (uB QX)propertyofΦ=(Q A µ X)◦ (QuA X) Xmodule= QX.Now, let f : ( X, A µ X)→(Y, A µ Y)a morphism of left A-modules, that is a morphismf : X → Y in A such thatA µ Y ◦ (Af) = f ◦ A µ X .We have to prove that ˜Q( ) ( )(f) : ˜QX = QX, B µ QX → ˜QY = QX, B µ QY is amorphism of left B-modules. We set ˜Q (f) = Q (f) and we compute(B µe ˜Qf) ( )?QY◦ B = ˜Qfi.e. by definition of the functor ˜Qin fact◦ B µe QXB µ QY ◦ (BQf) ? = (Qf) ◦ B µ QXB µ QY ◦ (BQf) = ( Q A µ Y)◦ (ΦY ) ◦ (BQf) Φ = ( Q A µ Y)◦ (QAf) ◦ (ΦX)fmorphA-mod= (Qf) ◦ ( Q A µ X)◦ (ΦX) = (Qf) ◦ B µ QX .Let now check that ˜Q is a lifting of Q. Let ( X, A µ X)∈ A A and computeBU ˜Q (( X, A µ X))= B U ( QX, B µ QX)= QX = QA U (( X, A µ X))and thus on the objectsBU ˜Q = Q A U.Let f : ( ) ( )X, A µ X → Y, A µ Y ∈ A A be a morphism, we haveBU ˜Q (f) : QX → QY = Q A U (f) : QX → QY.Therefore ˜Q is a lifting of the functor Q.25

26We have to prove that it is a bijection. Let us start with ˜Q : A A → B B a liftingof the functor Q : A → B. Then we construct Φ : BQ → QA given by( )Φ = BUλ B ˜QA F ◦ ( B U B F Qu A )and using this functorial morphism we define a functor Q : A A → B B as follows: forevery ( X, A µ X)∈ A AQ (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX)).Since both ˜Q and Q are lifting of Q, we have that B U ˜Q = Q A U = B UQ. We haveto prove that B U ( λ B Q ) ( )= B U λ B ˜Q . Let Z ∈ A A. We computeBU ( λ B QZ ) ()= (Q A Uλ A Z) ◦ BUλ B ˜QA F A UZ ◦ ( B U B F Qu AA UZ)(eQliftingQ= BU ˜Qλ) ()A Z ◦ BUλ B ˜QA F A UZ ◦ ( B U B F Qu AA UZ)) (λ=B(BUλ B ˜QZ ◦ BU B F B U ˜Qλ)A Z ◦ ( B U B F Qu AA UZ))=(BUλ B ˜QZ ◦ ( B U B F [Q A Uλ A Z ◦ Qu AA UZ]) (u A,λ A )adj= BUλ B ˜QZ.Conversely, let us start with a functorial morphism Φ : BQ → QA satisfying Φ ◦(m B Q) = (Qm A ) ◦ (ΦA) ◦ (BΦ) and Φ ◦ (u B Q) = Qu A . Then we construct a functor˜Q : A A → B B by setting, for every ( X, A µ X)∈ A A,˜Q (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX))which lifts Q : A → B. Now, we define a functorial morphism Ψ : BQ → QA givenby( )Ψ = BUλ B ˜QA F ◦ ( B U B F Qu A ) .Then we have( )Ψ = BUλ B ˜QA F◦ ( B U B F Qu A ) def e Q= (Q A Uλ AA F ) ◦ (Φ A F ) ◦ ( B U B F Qu A )= (Qm A ) ◦ (ΦA) ◦ (BQu A ) Φ = (Qm A ) ◦ (QAu A ) ◦ Φ Amonad= Φ.Corollary 3.25. Let X , A be categories, let A = (A, m A , u A ) be a monad on a categoryA and let F : X → A be a functor. Then there exists a bijective correspondencebetween the following collections of data:H Left A-module actions A µ F : AF → FG Functors A F : X → A A such that A U A F = F ,given byã : H → G where A Uã (A µ F)= F and A Uλ A ã (A µ F)= A µ F i.e.ã (A µ F)(X) =(F X, A µ F X ) and ã (A µ F)(f) = F (f)˜b : G → H where ˜b (A F ) = A Uλ AA F : AF → F.□

Proof. Apply Proposition 3.24 to the case A = X , B = A, A = Id X and B = A.Then ˜Q = A F is the lifting of F and Φ = A µ F satisfies A µ F ◦(m A F ) = A µ F ◦ ( )A A µ Fand A µ F ◦ (u A F ) = F that is ( )F, A µ F is a left A-module functor.□Corollary 3.26. Let (L, R) be an adjunction with L : B → A and R : A → Band let A = (A, m A , u A ) be a monad on B. Then there is a bijective correspondencebetween the following collections of dataK Functors K : A → A B such that A U ◦ K = R,L functorial morphism α : AR → R such that (R, α) is a left module functorfor the monad Agiven byΦ : K → L where Φ (K) = A Uλ A K : AR → RΩ : L → K where Ω (α) (X) = (RX, αX) and A UΩ (α) (f) = R (f) .Proof. Apply Corollary 3.25 to the case ”F ” = R : A → B where (L, R) is anadjunction with L : B → A and R : A → B and A = (A, m A , u A ) a monad onB. □In the following Proposition we give a more precise version of Lemma 3 in [J].Proposition 3.27. Let A = (A, m A , u A ) be a monad on a category A and letB = (B, m B , u B ) be a monad on a category B. Let Q : A → B be a functor andlet ˜Q : A A → B B be a lifting of Q (i.e. BU ˜Q = Q A U) and Φ : BQ(→ QA)asin Proposition 3.24. Then Φ is an isomorphism if and only if φ = λ B ˜QA F ◦( B F Qu A ) : B F Q → ˜Q A F is an isomorphism.Proof. By construction in Proposition 3.24 we have that Φ = B Uφ. Assume thatΦ is an isomorphism. Since, by Proposition 3.18, B U reflects isomorphisms, φ :BF Q → ˜Q A F is an isomorphism. Conversely, assume that φ : B F Q → ˜Q A F is anisomorphism. Then Φ = B Uφ is also an isomorphism.□Corollary 3.28. Let (L, R) be an adjunction where L : B → A and R : A → Band let B = (B, m B , u B ) be a monad on B. Let K : A → B B be a functor such thatBU ◦ K = R and let (R, α) be a left B-module functor as in Corollary 3.26. Then αis an isomorphism if and only if λ B K : B F R → K is an isomorphism.Proof. Apply Proposition 3.27 with Q = R, A = Id A . Then ˜Q = K is the lifting ofR and Φ = α : BR → R, given by α = B Uφ = B Uλ B K.□<strong>Some</strong> <strong>results</strong> in the following part of this section can be found in the literature(see e.g. [BM] and [BMV]). To introduce our main tools of investigation, for thereader’s sake, we give here a full description.Lemma 3.29. Let A = (A, m A , u A ) be a monad over a category A with coequalizers.Let Q : B → A be a left A-module functor with functorial morphisms A µ Q : AQ → Q.Then there exists a unique functor A Q : B → A A such thatAU ◦ A Q = Q and A Uλ AA Q = A µ Q .27

28Moreover if ϕ : Q → T is a functorial morphism between left A-module functors andϕ satisfiesA µ T ◦ (Aϕ) = ϕ ◦ (A µ Q)then there is a unique functorial morphism A ϕ : A Q → A T such thatAU A ϕ = ϕ.Proof. Corollary 3.25 applied to the case where F = Q gives us the first statement.Let B ∈ B. Then we have( Aµ T B ) ◦ (AϕB) = (ϕB) ◦ (A µ Q B )which means that ϕB yields a morphism A ϕB in A A.Proposition 3.30. Let A = (A, m A , u A ) be a monad over a category A and letB = (B, m B , u B ) be a monad over a category B. Assume that both A and B havecoequalizers and that A preserves coequalizers. Let Q : B → A be a functor and letA µ Q : AQ → Q and µ B Q : QB → Q be functorial morphisms. Assume that A µ Q isassociative and unital and that A µ Q ◦ ( Aµ B Q)= µBQ ◦ (A µ Q B ) . Set(6) (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B)Then Q B : B B → A is a left A-module functor where A µ QB : AQ B → Q B is uniquelydetermined by(7) p Q ◦ (A µ QB U ) = A µ QB ◦ (Ap Q ) .Moreover there exists a unique functor A (Q B ) : B B → A A such that(8) AU A (Q B ) = Q B and A Uλ AA (Q B ) = A µ QB .Proof. By Lemma <strong>2.</strong>7 we can consider (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B). SinceA µ Q ◦ ( Aµ B Q)= µBQ ◦ (A µ Q B )we deduce that(9) ( A µ QB U) ◦ ( Aµ B QBU ) = ( µ B QBU ) ◦ (A µ Q B B U ) .Also, in view of the naturality of A µ Q , we have(10) ( A µ QB U) ◦ (AQ B Uλ B ) = (Q B Uλ B ) ◦ (A µ Q B B U ) .We computeand hence we obtainp Q ◦ ( A µ QB U) ◦ (AQ B Uλ B ) (10)= p Q ◦ (Q B Uλ B ) ◦ (A µ Q B B U )p Q coeq= p Q ◦ (µ B QBU) ◦ (A µ Q B B U ) (9)= p Q ◦ ( A µ QB U) ◦ ( Aµ B QBU )p Q ◦ ( A µ QB U) ◦ (AQ B Uλ B ) = p Q ◦ ( A µ Q BU) ◦ ( Aµ B QBU ) .Since A preserves coequalizers, we get(AQ B , Ap Q ) = Coequ Fun(AµBQ BU, AQ B Uλ B).Hence there exists a unique functorial morphism A µ QBp Q ◦ (A µ Q BU ) = A µ QB ◦ (Ap Q ) .: AQ B → Q B such that□

Since Q is a left A-module functor, by Lemma 3.17, also Q B U is a left A-modulefunctor. Now p Q is an epimorphism and hence, since A preserves coequalizers, alsoAAp Q is an epimorphism. Therefore we can apply Lemma 3.23 to ”ϕ” = A µ QB ,”h” = p Q and ”µ” = A µ Q BU and hence we obtain that ( )Q B , A µ QB is a left A-module functor that is A µ QB is associative and unital. By Lemma 3.29 appliedto ( )Q B , A µ QB there exists a functor A (Q B ) : B B → A A such that A U A (Q B ) =Q B and A Uλ AA (Q B ) = A µ QB . Moreover A (Q B ) is unique with respect to theseproperties.□Proposition 3.3<strong>1.</strong> Let A = (A, m A , u A ) be a monad over a category A and letB = (B, m B , u B ) be a monad over a category B. Assume that both A and B havecoequalizers and A preserves them. Let Q : B → A be an A-B-bimodule functorwith functorial morphisms A µ Q : AQ → Q and µ B Q : QB → Q. Then the functorAQ : B → A A is a right B-module functor via µ B AQ : AQB → A Q where µ B AQ isuniquely determined by(11) AUµ B AQ = µ B Q.Let (( A Q) B, p A Q) = Coequ Fun(µBAQBU, A Q B Uλ B). Then we have( A Q) B= A (Q B ) : B B → A A.Proof. Since Q is endowed with a left A-module structure, by Lemma 3.29 thereexists a unique functor A Q : B → A A such that A U A Q = Q and A Uλ AA Q =A µ Q . Note that, since Q is an A-B-bimodule functor, in particular the compatibilityconditionA µ Q ◦ ( )Aµ B Q = µBQ ◦ (A µ Q B ) .holds. Thus, by Lemma 3.29, there exists a functorial morphism µ B AQ : AQB → A Qsuch thatAUµ B = µAQ B Q.By the associativity and unitality properties of µ B Q and since AU is faithful, weget that also µ B AQ is associative and unital, so that ( AQ, µ AQ) B is a right B-modulefunctor. Thus we can consider the coequalizer(12) AQB B Uµ B A Q BUAQ B Uλ B AQ B Up A Q ( A Q) Bso that we get a functor ( A Q) B: B B → A A. Since A preserves coequalizers, byLemma 3.21, also A U preserves coequalizers. Then, by applying the functor A U to12 we still get a coequalizer29that can be written as AU A QB B U A Uµ B A Q BUAU A Q B UAU A Q B Uλ BAUp A QAU ( A Q) BQB B Uµ B QBUQ B Uλ B Q B UAUp A QAU ( A Q) B

30Since, by Proposition 3.30, (Q B , p Q ) = Coequ Fun(µBQ BU,Q B Uλ B), we get thatAU ( A Q) B= Q B and A Up A Q = p Q .MoreoverAUλ A ( A Q) B: A A U ( A Q) B= AQ B → A U ( A Q) B= Q B .By Proposition 3.30, we know that ( )Q B , A µ QB is a left A-module functor andAUλ AA (Q B ) = A µ QB . Hence we geti.e.AUλ A ( A Q) B= A µ QB(8)= A Uλ AA (Q B )( A Q) B= A (Q B ) .Notation 3.3<strong>2.</strong> Let A = (A, m A , u A ) be a monad over a category A and letB = (B, m B , u B ) be a monad over a category B. Assume that both A and B havecoequalizers and A preserves them. Let Q : B → A be an A-B-bimodule functor. Inview of Proposition 3.31, we setAQ B = ( A Q) B= A (Q B ) .Lemma 3.33. Let B = (B, m B , u B ) be a monad over a category B and assume that Bhave coequalizers. Let ( Q : B → A, µ B Q)be a right B-module functor. With notationsof Proposition 3.30 we have that(13) Q B λ BB F = µ B Q.Furthermore, if we assume that the functors Q, B preserve coequalizers we also have(14) Q B λ BB P = p Q BP .Proof. Let us consider the following diagramQB B U B F B U B Fµ B QBU B F B U B F Q B Uλ B B F B U B FQ B U B F B U B F p QBF B U B F Q BB F B U B F□QB B Uλ B B FQ B Uλ B B FQ B λ B B FQB B U B Fµ B QBU B FQ B Uλ B B F Q B U B Fp QB F Q BB FNote that QB B Uλ BB F = QBm B and Q B Uλ BB F = Qm B so that the left squareserially commutes because of the associativity of m B and of µ B Q . Both the rows arecoequalizers in view of the dual version of Lemma <strong>2.</strong>10 so that, by the universal propertyof coequalizers, there exists a unique functorial morphism ζ : Q BB F B U B F →Q BB F such that ζ ◦ (p QB F B U B F ) = (p QB F ) ◦ (Q B Uλ BB F ). Since p Q : Q B U → Q B isa functorial morphism, we know that Q B λ BB F makes the right square be commutative,but since by (15) we have p QB F = µ B Q we also have that µB Q makesthe right square commute. Therefore, we deduce that ζ = Q B λ BB F = µ B Q . Assumingthat Q and B preserve coequalizers, by Lemma 3.21, we get that B Ualso preserves coequalizers so that, in view of Corollary <strong>2.</strong>12 we also have that

(Q BB F, p QB F ) = Coequ Fun(µBQ B, Qm B)=(Q, µBQ).□(Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B)preserves them. Hence, using that(Q B , Q B λ B ) = Coequ Fun (Q B λ BB F B U, Q BB F B Uλ B ), in view of Lemmas <strong>2.</strong>10 and3.29, we have(Q BB P , Q B λ BB P ) = Coequ Fun (Q B λ BB F B U B P,Q BB F B Uλ BB P )= Coequ Fun(µBQ BU B P,Q B µ P)so that we get Q B λ BB P = p QB P .= (Q BB P , p QB P )Proposition 3.34. Let A = (A, m A , u A ) be a monad over a category A and letB = (B, m B , u B ) be a monad over a category B. Assume that both A and Bhave coequalizers and let Q : B → A be an A-B-bimodule functor. Then, withnotations(of Proposition)3.30, we can consider the functor Q B where (Q B , p Q ) =Coequ Fun µBQ BU, Q B Uλ B . Then(15) Q BB F = Q and p QB F = µ B Q.Proof. By construction we have that (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B). By applyingit to the functor B F we get that(Q BB F, p QB F ) = Coequ Fun(µBQ BU B F, Q B Uλ BB F )= Coequ Fun(µBQ B, Qm B).Since Q is a right B-module functor, by Proposition 3.16 we have that(Q, µBQ)= CoequFun(µBQ B, Qm B)so that we get31□Proposition 3.35. Let B = (B, m B , u B ) be a monad on a category B with coequalizerssuch that B preserves coequalizers. Let G : B B → A be a functor preservingcoequalizers. SetQ = G ◦ B F and let µ B Q = Gλ BB FThen ( Q, µ B Q)is a right B-module functor and(16) Q B = (G ◦ B F ) B= G.Proof. We computeandµ B Q ◦ ( µ B QB ) = (Gλ BB F ) ◦ (Gλ BB F B) λ B= (Gλ BB F ) ◦ (G B F B Uλ BB F )= (Gλ BB F ) ◦ (G ◦ B F m B ) = µ B Q ◦ (Qm B )µ B Q ◦ (Qu B ) = (Gλ BB F ) ◦ (G B F u B ) adj= G ◦ B F = Q.Thus ( Q, µ B Q)is a right B-module functor. Recall that (see Proposition 3.30)(Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B)

32and by Proposition 3.34 we have Q BB F = Q and p QB F = µ B Q . In particular we getQ BB F = Q = G B F.In order to prove that Q B = G it suffices to prove that(G, Gλ B ) = Coequ Fun(µBQ BU, Q B Uλ B). In fact, by Corollary 3.15, (B U, B Uλ B ) =Coequ Fun (B B Uλ B , m BB U) and, since by Lemma 3.20 B U reflects coequalizers, wehave(Id B B, λ B ) = Coequ Fun ( B F B Uλ B , λ BB F B U) .Since G preserves coequalizers, we get that(G, Gλ B ) = Coequ Fun (G B F B Uλ B , Gλ BB F B U)= Coequ Fun(QB Uλ B , µ B QBU ) = (Q B , p Q ) .Proposition 3.36. Let A = (A, m A , u A ) be a monad on a category A with coequalizerssuch that A preserves coequalizers. Let H : B → A A be a functor preservingcoequalizers. SetQ = A U ◦ H and let A µ Q = A Uλ A HThen ( Q, A µ Q)is a left A-module functor and(17) AQ = A ( A U ◦ H) = H.Proof. First we want to prove that A µ Q = A Uλ A H is associative. We haveA µ Q ◦ ( A A µ Q)= (A Uλ A H) ◦ (A A Uλ A H) λ A= ( A Uλ A H) ◦ ( A Uλ AA F A UH)□so that we get= ( A Uλ A H) ◦ (m AA UH) = A µ Q ◦ (m A Q)A µ Q ◦ ( A A µ Q)= A µ Q ◦ (m A Q) .Now we prove that A µ Q = A Uλ A H is unital. We computeso that we getA µ Q ◦ (u A Q) = ( A Uλ A H) ◦ (u AA UH) adj= A UH = QA µ Q ◦ (u A Q) = Q.Thus ( Q, A µ Q)is a left A-module functor. Recall that (see Lemma 3.29) there existsa unique functor A Q : B → A A such thatAU ◦ A Q = Q and A Uλ AA Q = A µ Q .Thus we haveAU ◦ A Q = Q = A U ◦ HandAUλ AA Q = A µ Q = A Uλ A Hso that, by Proposition 3.12, we obtain thatAQ = H.□

Theorem 3.37. Let B = (B, m B , u B ) be a monad on a category B with coequalizerssuch that B preserves coequalizers. Then there exists a bijective correspondencebetween the following collections of data:F B right B-module functors Q : B → A such that QB preserves coequalizers(A ← B B) functors G : B B → A preserving coequalizersgiven byν B : F B → (A ← B B) where ν B((Q, µBQ))= QBκ B : (A ← B B) → F B where κ B (G) = (G B F,Gλ BB F )where Q B is uniquely determined by (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B).Proof. Let Q : B → A be a right B-module functor. Then we can consider Q B :BB → A defined by (6) as(Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B).Since by assumption QB preserves coequalizers, by Lemma 3.19 also Q preservescoequalizers. Moreover, since B preserves coequalizers, by Lemma 3.21 also thefunctor B U preserves coequalizers. Thus both QB B U and Q B U preserve coequalizers.By Corollary <strong>2.</strong>12 we get that also Q B : B B → A preserves coequalizers.Conversely, let us consider a functor G : B B → A that preserves coequalizers. ByProposition 3.35 we can consider the right B-module functor defined as followsQ = G ◦ B F and let µ B Q = Gλ BB F .Since B F is left adjoint to B U, in particular B F preserves coequalizers and sinceby assumption G preserves coequalizers, we get that also Q = G ◦ B F preservescoequalizers and so does QB.Now, we want to prove that ν B and κ B determine a bijective correspondence betweenF B and (A ← B B). Let us start with a right B-module functor ( Q : B → A, µ B Q).Then we haveMoreover we have(κ B ◦ ν B ) (( Q, µ Q)) B = κB (Q B ) = (Q BB F,Q B λ BB F )= ( )Q BB F,µ B (15)Q B B F = ( Q,µ Q) B .(ν B ◦ κ B ) (G) = ν B ((G B F,Gλ BB F )) = (G B F ) B(16)= G.Theorem 3.38. Let A = (A, m A , u A ) be a monad on a category A with coequalizerssuch that A preserves coequalizers. Then there exists a bijective correspondencebetween the following collections of data:AF left A-module functors Q : B → A such that AQ preserves coequalizers( A A ← B) functors H : B → A A preserving coequalizersgiven byAν : AF → ( A A ← B) where A ν (( Q, A µ Q))= A QAκ : ( A A ← B) → A F where A κ (H) = ( A U ◦ H, A Uλ A H)33□

34where A Q : B → A A is the functor defined in Lemma 3.29.Proof. Let ( Q : B → A, A µ Q)be a left A-module functor. Then, by Lemma 3.29,there exists a unique functor A Q : B → A A such thatAU ◦ A Q = Q and A Uλ AA Q = A µ Q .Note that, since AQ preserves coequalizers, by Lemma 3.19, Q = A U ◦ A Q preservescoequalizers. Then, by Lemma 3.20, also A Q preserves coequalizers. Conversely, ifH : B → A A is a functor preserving coequalizers, we get that A U ◦ H : B → A.Moreover, by Lemma 3.21, A U preserves coequalizers and thus also A U ◦H preservescoequalizers. Now, let us prove that A ν and A κ determine a bijective correspondencebetween A F and ( A A ← B). We compute( A κ ◦ A ν) (( Q, A µ Q))= A κ ( A Q) = ( A U A Q, A Uλ AA Q) = ( Q, A µ Q).On the other hand we have( A ν ◦ A κ) (H) = A ν (( A U ◦ H, A Uλ A H)) = A ( A U ◦ H) (17)= H.Theorem 3.39. Let A = (A, m A , u A ) be a monad on a category A with coequalizerssuch that A preserves coequalizers. Let B = (B, m B , u B ) be a monad on a categoryB with coequalizers such that B preserves coequalizers. Then there exists a bijectivecorrespondence between the following collections of data:AF B A-B-bimodule functors Q : B → A such that AQ and QB preserve coequalizers( A A ← B B) functors G : B B → A A preserving coequalizersgiven byAν B : AF B → ( A A ← B B) where A ν B((Q, A µ Q , µ B Q))= A Q BAκ B : ( A A ← B B) → A F B where A κ B (G) = ( A U ◦ G ◦ B F , A Uλ A G B F , A UGλ BB F ) .Proof. Let us consider an A-B-bimodule functor ( Q : B → A, A µ Q , µ Q) B such that AQand QB preserve coequalizers. In particular, ( Q, µ Q) B is a right B-module functor,so that we (( can apply )) the map ν B : F B → (A ← B B) of Theorem 3.37 and we get afunctor ν B Q, µBQ = QB : B B → A which preserves coequalizers. By Proposition3.30, ( )Q B , A µ QB is a left A-module functor so that we can also apply the mapAν : A F → ( A A ← B) of Theorem 3.38 where the category B is B B. The map A νis defined by A ν (( ))Q B , A µ QB = A (Q B ) = A Q B : B B → A A and A Q B preservescoequalizers. Conversely, let us consider a functor G : B B → A A which preservescoequalizers. By Theorem 3.38, we get a left A-module functor given byAκ (G) = ( A U ◦ G, A Uλ A G)where A U ◦ G : B B → A and A A UG preserves coequalizers. By Lemma 3.19, alsoAU ◦ G : B B → A preserves coequalizers. Thus, we can apply Theorem 3.37 and weget a right B-module functorκ B ( A UG) = ( A UG B F, A UGλ BB F )□

where A UG B F : B → A is such that A UG B F B preserves coequalizers. Clearly, sinceAUG preserves coequalizers, B F is a left adjoint and A preserves coequalizers byassumption, we deduce that also A A UG B F preserves coequalizers. Now, we wantto prove that A ν B : A F B → ( A A ← B B) and A κ B : ( A A ← B B) → A F B determine abijection. We haveand( A κ B ◦ A ν B ) (( Q, A µ Q , µ B Q))= A κ B ( A Q B )= ( A U ◦ A Q B ◦ B F , A Uλ AA Q BB F , A U A Q B λ BB F )= (Q, A Uλ AA Q, Q B λ BB F ) = ( Q, A µ Q , µ B Q B B F)=(Q, A µ Q , µ B Q( A ν B ◦ A κ B ) (G) = A ν B (( A U ◦ G ◦ B F , A Uλ A G B F , A UGλ BB F ))= A ( A U ◦ G ◦ B F ) B = A (( A U ◦ G ◦ B F ) B )(16)= A ( A U ◦ G) (17)= G.Proposition 3.40. Let A = (A, m A , u A ) be a monad over a category A with coequalizersand assume that A preserves coequalizers. Let Q : A → A be an A-bimodulefunctor. Then there exists a unique lifted functor A Q A : A A → A A such thatAU A Q AA F = Q.Proof. By Proposition 3.31 there exists a unique functor A Q A : A A → A A such thatAU A Q A = Q A . Now, by Proposition 3.34 we also get that Q AA F = Q so that weobtainAU A Q AA F = Q.□Proposition 3.4<strong>1.</strong> Let A = (A, m A , u A ) be a monad over a category A with coequalizersand assume that A preserves coequalizers. Let B = (B, m B , u B ) be a monadover a category B with coequalizers and let Q : B → A be an A-B-bimodule functor.Then there exists a unique lifted functor A Q B : B B → A A such thatAU A Q BB F = Q.Proof. By Proposition 3.31 there exists a unique functor A Q B : B B → A A such thatAU A Q B = Q B . Now, by Proposition 3.34 we also get that Q BB F = Q so that weobtainAU A Q BB F = Q.□Proposition 3.4<strong>2.</strong> Let A = (A, m A , u A ) be a monad over a category A with coequalizersand assume that A preserves coequalizers. Let B = (B, m B , u B ) be amonad over a category B with coequalizers and let P, Q : B → A be A-B-bimodulefunctors. Let f : P → Q be a functorial morphism of left A-module functors andof right B-module functors. Then there exists a unique functorial morphism of leftA-module functorsf B : P B → Q B)35□

36satisfyingf B ◦ p P = p Q ◦ (f B U) .Then we can considerAf B : A P B → A Q Bsuch thatAU A f B = f B .Proof. Consider the following diagramP B B Uµ B P BUP B Uλ B P B U p PP Bf BfB B UQB B Uµ B QBUf B UQ B Uλ B Q B U pQ Q BSince f is a functorial morphism and it is a functorial morphism of right B-modulefunctors, the left square serially commutes. Note thatp Q ◦ (f B U) ◦ ( µ B P BU ) = p Q ◦ (f B U) ◦ (P B Uλ B )so that, by the universal property of the coequalizer, there exists a unique morphismf B : P B → Q B such that(18) f B ◦ p P = p Q ◦ (f B U) .We now want to prove that f B is a functorial morphism of left A-module functor.In fact we havef B ◦ A µ PB ◦ (Ap P ) (7)= f B ◦ p P ◦ (A µ P B U )(18)= p Q ◦ (f B U) ◦ (A µ P B U ) fleftAlin= p Q ◦ (A µ QB U ) ◦ (Af B U)(7)= A µ QB ◦ (Ap Q ) ◦ (Af B U) (18)= A µ QB ◦ (Af B ) ◦ (Ap P )and since A preserves coequalizers Ap Q is an epimorphism so that we getf B ◦ A µ PB = A µ QB ◦ (Af B ) .Then there exists a functorial morphism A f B : A P B → A Q B such thatAU A f B = f B .3.<strong>2.</strong> The category of balanced bimodule functors. We will construct here themonoidal category of balanced bimodule functors with respect to a monad.Definition 3.43. Let A = (A, m A , u A ) be a monad over a category A such thatAhas coequalizers and the underlying functor A preserves coequalizers. Let us definethe category ( A A ← A A) of balanced bimodule functors as followsOb Objects are functors A Q A : A A→ A A where Q : A → A is an A-A-bimodulefunctor such that Q A preserves coequalizers.M Morphisms are functorial morphisms A f A : A P A → A Q A where f : P → Q isa functorial morphism of A-A-bimodule functors.□

Proposition 3.44. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers and let A P A , A Q A ∈ Ob (( A A ← A A)).Then the functor A P AA Q A ∈ Ob (( A A ← A A)).Proof. We will prove that A P AA Q A = A (P AA Q) A . Let us consider the functorP AA Q : A → A. Since A Q is a right A-module functor by Proposition 3.31, thenalso P AA Q is a right A-module functor by Lemma 3.17. Thus, we can consider((P AA Q) A, p PAA Q) = Coequ Fun(µAPAA QAU, P AA Q A Uλ A)i.e.Lem3.17= Coequ Fun(PA µ A AQAU, P AA Q A Uλ A)P A preserves coequ= (P AA Q A , P A p A Q)(19) ((P AA Q) A, p PAA Q) = (P AA Q A , P A p A Q) .Now, observe that (P AA Q) Ais a left A-module functor by Proposition 3.30 andP AA Q A is a left A-module functor by Lemma 3.17. So we can consider both liftingfunctors : A (P AA Q) A and A (P A ) A Q A and we haveandHenceAU A ((P AA Q) A ) Pro3.31= (P AA Q) A(19)= P AA Q APro3.31= A U A (P A ) A Q AAUλ AA ((P AA Q) A) Pro3.30= A µ A (P AA Q) A(19)= µ PAA Q ALem3.17= A µ PA AQ APro3.30= A Uλ AA (P A ) A Q A .A ((P AA Q) A ) Pro3.31= A (P AA Q) A = A (P A ) A Q APro3.31= A P AA Q A .ThusAP AA Q A = A (P AA Q) Awhere P AA Q : A → A is an A-bimodule functor satisfying the required conditions.□Proposition 3.45. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers. Then A A A ∈ Ob (( A A ← A A)) and itis the unit element for the category ( A A ← A A).Proof. Since A is a monad, in particular an A-bimodule functor. Then we canconsider A A A ∈ Ob (( A A ← A A)) as the object coming from the endofunctor A :A → A. By definition we have(A A , p A ) = Coequ Fun (m AA U,A A Uλ A ) = ( A U, A Uλ A )and it is a left A-module functor by Proposition 3.13.considerAA A = A ( A U) = Id A A37By Lemma 3.29, we can

38as the unique functor which satisfiesAU A A A = A UId A A = A U = A AandAUλ AA A A = A Uλ A Id A A = A Uλ A = A µ A U = A µ AA .Clearly A A A = Id A A is the identity element for the category ( A A ← A A).Corollary 3.46. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers. Then we havefor every F ∈ Ob (( A A ← A A)).AA A ◦ F = F and F ◦ A A A = FProof. By Proposition 3.45 we have that A A A = Id A A is the identity element for thecategory ( A A ← A A). Therefore, in particular, we have thatfor every F ∈ Ob (( A A ← A A)).AA A ◦ F = F and F ◦ A A A = FProposition 3.47. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers, let A P A , A Q A , A T A ∈ Ob (( A A ← A A))and let A f A : A P A → A Q A , A g A : A Q A → A T A be morphisms in ( A A ← A A). ThenAg A ◦ A f A is still a morphism in the category ( A A ← A A) and(20) A (g ◦ f) A = A g A ◦ A f A .Proof. We will prove that A g A ◦ A f A = A (g ◦ f) A where g ◦ f is an A-bilinear functorialmorphism as composite of A-bilinear functorial morphisms. By assumption,using notations of Proposition 3.42 we have the following serially commutative diagramP A A UfA A UQA A UgA A UT A A Uµ A P AUP A Uλ A µ A QAUQ A Uλ Aµ A T AUThen A f A is the unique morphism such thatwhereP A U p Pf A UQ A U pQ g A UP Af AQ Ag A T A U p TT AT A Uλ AAU A f A = f A(21) f A ◦ p P = p Q ◦ (f A U)and A g A is the unique morphism such thatwhereAU A g A = g A(22) g A ◦ p Q = p T ◦ (g A U) .□□

Note that, since f and g are A-bilinear morphism, g ◦ f is still an A-bilinear morphism,so that we can also consider (g ◦ f) Asuch that(23) (g ◦ f) A◦ p P = p T ◦ [(g ◦ f) A U] = p T ◦ (g A U) ◦ (f A U) .First we prove that (g ◦ f) A= g A ◦ f A . In fact we have(g ◦ f) A◦ p P(23)= p T ◦ (g A U) ◦ (f A U)(22)= g A ◦ p Q ◦ (f A U) (21)= g A ◦ f A ◦ p Pand since p P is an epimorphism we obtain(g ◦ f) A= g A ◦ f A .The, we can both consider A (g ◦ f) A = A ((g ◦ f) A ) such thatAU A (g ◦ f) A = A U A ((g ◦ f) A ) = (g ◦ f) Aand the composite of the liftings A g A ◦ A f A such thatWe haveAU [ A g A ◦ A f A ] = ( A U A g A ) ◦ ( A U A f A ) = g A ◦ f A .AU A (g ◦ f) A ◦ p P = A U A ((g ◦ f) A ) ◦ p P = (g ◦ f) A ◦ p P(23)= p T ◦ (g A U) ◦ (f A U) = g A ◦ p Q ◦ (f A U) = g A ◦ f A ◦ p Pand since p P is an epimorphism we deduce thatSince A U reflects we conclude thatAU A (g ◦ f) A = g A ◦ f A = A U A g A ◦ A U A f A .A (g ◦ f) A = A g A ◦ A f Awhere A U A (g ◦ f) A = (g ◦ f) A and (g ◦ f) A◦ p P = p T ◦ [(g ◦ f) A U] .Proposition 3.48. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers, let A P A , A Q A , A T A ∈ Ob (( A A ← A A))and let A f A : A P A → A Q A , A g A : A Q A → A T A , A h A : A T A → A W A be morphisms in( A A ← A A). ThenAh A ◦ ( A g A ◦ A f A ) = ( A h A ◦ A g A ) ◦ A f A .Proof. By Proposition 3.47 we have that, for every morphisms A f A : A P A → A Q A ,Ag A : A Q A → A T A in ( A A ← A A) , also the morphism A g A ◦ A f A is in ( A A ← A A)and A (g ◦ f) A = A g A ◦ A f A . Hence we have that39□Ah A ◦ ( A g A ◦ A f A ) (20)= A h A ◦ ( A (g ◦ f) A ) (20)= ( A (h ◦ (g ◦ f)) A )Astrictly monoidal= ( A ((h ◦ g) ◦ f) A ) (20)= A (h ◦ g) A ◦ A f A(20)= ( A h A ◦ A g A ) ◦ A f A .□

40Theorem 3.49. Let A = (A, m A , u A ) be a monad over a category A such thatAhas coequalizers and the underlying functor A preserves coequalizers. The category( A A ← A A) of balanced bimodule functors is a strict monoidal category.Proof. By Proposition 3.44, we defined a composition of the objects of the category( A A ← A A). Moreover, by Proposition 3.45, A A A is the unit for the category( A A ← A A). Since the composition of functors is associative and by Corollary 3.46,it is easy to prove that ( A A ← A A) is a strict monoidal category.□3.3. The comparison functor for monads.Proposition 3.50. Let (L, R) be an adjunction where L : B → A and R : A → Bwith unit η and counit ɛ and let A = (A, m A , u A ) be a monad on the category B.There exists a bijective correspondence between the following collections of data:M monad morphisms ψ : A = (A, m A , u A ) → RL = (RL, RɛL, η)R functorial morphism r : LA → L such that (L, r) is a right module functorfor the monad AL functorial morphism l : AR → R such that (R, l) is a left module functor forthe monad Agiven byΘ : M → R where Θ (ψ) = (ɛL) ◦ (Lψ)Ξ : R → M where Ξ (r) = (Rr) ◦ (ηA)Γ : M → L where Γ (ψ) = (Rɛ) ◦ (ψR)Λ : L → M where Λ (l) = (lL) ◦ (Aη) .Theorem 3.5<strong>1.</strong> Let (L, R) be an adjunction where L : B → A and R : A → Band let A = (A, m A , u A ) be a monad on the category B. There exists a bijectivecorrespondence between the following collections of data:K Functors K : A → A B such that A U ◦ K = RM monad morphisms ψ : A = (A, m A , u A ) → RL = (RL, RɛL, η)given byΨ : K → M where Ψ (K) = ([ A Uλ A K] L) ◦ (Aη)Υ : M → K where Υ (ψ) (X) = (RX, (RɛX) ◦ (ψRX)) and Υ (ψ) (f) = Rf.Remark 3.5<strong>2.</strong> When A = RL = (RL, RɛL, η) and ψ = Id RL the functor K =Υ (ψ) : A → RL B such that RL U ◦ K = R is called the Eilenberg-Moore comparisonfunctor.Corollary 3.53. Let A = (A, m A , u A ) and B = (B, m B , u B ) be monads on acategory B. There exists a bijective correspondence between the following collectionsof data:K Functors K : A B → B B such that B U ◦ K = A U,M monad morphisms ψ : A → Bgiven byΨ : K → M where Ψ (K) = ([ A Uλ A K] A F ) ◦ (Au A )Υ : M → K where Υ (ψ) (X) = ( A UX, ( A Uλ A X) ◦ (ψ A UX)) and Υ (ψ) (f) = A U (f) .

Proposition 3.54. Let (L, R) be an adjunction where L : B → A and R : A → B,let A = (A, m A , u A ) be a monad on the category B and let ψ : A = (A, m A , u A ) →RL = (RL, RɛL, η) be a monad morphism. Let r = Θ (ψ) = (ɛL) ◦ (Lψ). Then thefunctor K ψ = Υ (ψ) : A → A B has a left adjoint D ψ : A B → A if and only if, forevery ( Y, A µ Y)∈ A B, there exists Coequ A(rY, L A µ Y). In this case, there exists afunctorial morphism d ψ : L A U → D ψ such thatand thus(D ψ , d ψ ) = Coequ Fun (r A U, L A Uλ A )[Dψ((Y, A µ Y)), dψ(Y, A µ Y)]= CoequA(rY, L A µ Y).Corollary 3.55. Let (L, R) be an adjunction where L : B → A and R : A →B. Let r = Θ (Id RL ) = ɛL. Then the functor K = Υ (Id RL ) : A → RL B has aleft adjoint D : RL B → A if and only, for every ( Y, RL µ Y)∈ RL B, there existsCoequ A(ɛLY, L RL µ Y). In this case, there exists a functorial morphism d : LRL U →D such that(D, d) = Coequ Fun (ɛL RL U, L RL Uλ RL )and thus[D((Y, RL µ Y)), d(Y, RL µ Y)]= CoequA(ɛLY, L RL µ Y).Remark 3.56. In the setting of Proposition 3.54, for every X ∈ A, we note thatthe counit of the adjunction (D ψ , K ψ ) is given by( )˜ɛX = ã −1X,K ψ X IdKψ X : Dψ K ψ (X) → X.We will consider the diagram( (( )) )(24) Hom A Dψ Y, A ã X,Y (( )µ Y , X Hom A B Y, A µ Y , Kψ X )41Hom A(d ψ((Y, A µ Y )),X)Hom A (LY, X)Hom A (rY,X)Hom A(L A µ Y ,X)Hom A (LAY, X)a X,Ya X,AYHom B (Y, RX)(Γ(ψ)X)◦(A−)Hom B( A µ Y ,RX) HomB (AY, RX)defining ã −,Y in the particular case of ( Y, A µ Y)= Kψ X. Note that, since K ψ X =(RX, (RɛX) ◦ (ψRX)) = (RX, lX) , we havei.e.(D ψ K ψ (X) , d ψ K ψ (X)) = (D ψ (RX, lX) , d ψ K ψ (X)) = Coequ B (rRX, LlX)= Coequ B ((ɛLRX) ◦ (LψRX) , (LRɛX) ◦ (LψRX))(25) (D ψ K ψ (X) , d ψ K ψ (X)) = Coequ B (rRX, LlX)where l = Γ (ψ) = (Rɛ) ◦ (ψR) . We compute(˜ɛX) ◦ (d ψ K ψ X) = Hom A (d ψ K ψ X, X) ((˜ɛX))( (= Hom A (d ψ K ψ X, X) ã −1X,K ψ X IdKψ X) )

42[= Hom A (d ψ K ψ X, X) ◦ ã −1 (IdKψ )X,K ψ X]X(24)= a −1X,K ψ X AU ( ) (Id Kψ X = a−1X,K ψ X IdA UK ψ X)= a −1X,K ψ X (Id RX) = ɛXso that(˜ɛX) ◦ (d ψ K ψ X) = ɛX.(Since ˜ɛX = ã −1X,K ψ X IdKψ X)and ã−1X,K ψ Xis an isomorphism, we deduce that ˜ɛX :D ψ K ψ (X) → X is defined as the unique morphism such that(26) (˜ɛX) ◦ (d ψ K ψ X) = ɛX.On the other hand, for every ( )Y, A µ Y ∈ A B, the unit of the adjunction (D ψ , K ψ ) ,˜η : A B → K ψ D ψ , is given by˜η ( ) ( ) ( ) (( ))Y, A µ Y = ãDψ (Y, A µ Y ),Y IdDψ (Y, A µ Y ) : Y, A µ Y → Kψ D ψ Y, A µ Y .Then by commutativity of the diagram (24), we deduce thatAU ˜η ( ) (Y, A µ Y = A Uã Dψ (Y, A µ Y ),Y IdDψ (Y, A µ Y ))( (( )) ( )) ( )= a Dψ (Y, A µ Y ),Y ◦ Hom A dψ Y, A µ Y , Dψ Y, A µ Y IdDψ (Y, A µ Y )( (( ))) ( ( ))= a Dψ (Y, A µ Y ),Y dψ Y, A µ Y = Rdψ Y, A µ Y ◦ (ηY ) .Thus we obtain that(27) AU ˜η ( Y, A µ Y)=(Rdψ(Y, A µ Y))◦ (ηY ) .Observe that, for every Y ∈ B we have that A F (Y ) = (AY, m A Y ) . Moreoverso that we get(D ψA F (Y ) , d ψA F (Y )) = (D ψ (AY, m A Y ) , d ψ (AY, m A Y ))= Coequ A (rAY, Lm A Y ) (2)= (LY, rY )(28) (D ψA F , d ψA F ) = (L, r) .In particular(29) d ψ (AY, m A Y ) = rY.Theorem 3.57. Let (L, R) be an adjunction where L : B → A and R : A → B, letA = (A, m A , u A ) be a monad on the category B and let ψ : A = (A, m A , u A ) → RL =(RL, RɛL, η) be a monad morphism. Let r = Θ (ψ) = (ɛL) ◦ (Lψ). Assume that, forevery ( Y, A µ Y)∈ A B, there exists Coequ A(rY, L A µ Y). Then we can consider thefunctor K ψ = Υ (ψ) : A → A B. Its left adjoint D ψ : A B → A is full and faithful ifand only if1) R preserves the coequalizer(D ψ , d ψ ) = Coequ Fun (r A U, L A Uλ A )2) ψ : A → RL is a monad isomorphism.

Corollary 3.58. Let (L, R) be an adjunction where L : B → A and R : A → B.Let r = Θ (Id RL ) = ɛL. Assume that, for every ( )( )Y, RL µ Y ∈ RL B, there existsCoequ A ɛLY, L RL µ Y . Then we can consider the functor K = Υ (IdRL ) : A → RL B.Its left adjoint D : RL B → A is full and faithful if and only if R preserves thecoequalizer(D, d) = Coequ Fun (ɛL RL U, L RL Uλ RL ) .Theorem 3.59. Let (L, R) be an adjunction where L : B → A and R : A → B,let A = (A, m A , u A ) be a monad on the category B and let ψ : A = (A, m A , u A ) →RL = (RL, RɛL, η) be a monad morphism. Let r = Θ (ψ) = (ɛL) ◦ (Lψ) andl = Γ (ψ) = (Rɛ) ◦ (ψR). Assume that, for every ( )( )Y, A µ Y ∈ A B, there existsCoequ A rY, L A µ Y . Then we can consider the functor Kψ = Υ (ψ) : A → A B andits left adjoint D ψ : A B → A. The functor K ψ is an equivalence of categories if andonly if1) R preserves the coequalizer(D ψ , d ψ ) = Coequ Fun (r A U, L A Uλ A )2) R reflects isomorphisms and3) ψ : A → RL is a monad isomorphism.Definition 3.60. Let A = (A, m A , u A ) be a monad on the category B and let(R, A µ R)be a left A-module functor. We say that(R, A µ R)is a left A-coGaloisfunctor if R has a left adjoint L and if the canonical morphismcocan := (A µ R L ) ◦ (Aη) : A → RLis a monad isomorphism, where η denotes the unit of the adjunction (L, R).Corollary 3.6<strong>1.</strong> Let ( R, A µ R)be a left A-coGalois functor where R : A → Bpreserves coequalizers, R reflects isomorphisms and A = (A, m A , u A ) is a monad onB. Assume that, for every ( Y, A µ Y)∈ A B, there exists Coequ A(rY, L A µ Y)wherer = (ɛL) ◦ (Lcocan) where L is the left adjoint of R and ɛ is the counit of theadjunction (L, R). Then we can consider the functor K cocan : A → A B and its leftadjoint D cocan : A B → A. Then the functor K cocan is an equivalence of categories.Theorem 3.62 ( Beck’s Theorem for monads). Let (L, R) be an adjunction whereL : B → A and R : A → B. Let r = Θ (Id RL ) = ɛL and assume that, for every(Y, RL µ Y)∈ RL B, there exists Coequ A(ɛLY, L RL µ Y). Then we can consider thefunctor K = Υ (Id RL ) : A → RL B and its left adjoint D : RL B → A. The functor Kis an equivalence of categories if and only if1) R preserves the coequalizer2) R reflects isomorphisms.(D, d) = Coequ Fun (ɛL RL U, L RL Uλ RL ) .Definition 3.63. Let A = (A, m A , u A ) be a monad on the category B and letR : A → B be a functor. The functor R is called ψ-monadic if it has a left adjointL : B → A for which there exists ψ : A → RL a monad morphism such that thefunctor K ψ = Υ (ψ) : A → A B is an equivalence of categories.43

44Definition 3.64. Let R : A → B be a functor. The functor R is called monadic ifit has a left adjoint L : B → A for which the functor K = Υ (Id RL ) : A → RL B is anequivalence of categories.The following is a slightly improved version of Theorem 3.14 p. 101 [BW].Theorem 3.65 (Generalized Beck’s Precise Tripleability Theorem). Let R : A → Bbe a functor and let A = (A, m A , u A ) be a monad on the category B. Then R isψ-monadic if and only if1) R has a left adjoint L : B → A,2) ψ : A → RL is a monads isomorphism where RL = (RL, RɛL, η) with η andɛ unit and counit of (L, R) ,3) for every ( Y, A µ Y)∈ A B, there exist Coequ A(rY, L A µ Y), where r = Θ (ψ) =(ɛL) ◦ (Lψ) , and R preserves the coequalizerCoequ Fun (r A U, L A Uλ A ) ,4) R reflects isomorphisms.In this case in A there exist coequalizers of R-contractible coequalizer pairs and Rpreserves them.Corollary 3.66 (Beck’s Precise Tripleability Theorem). Let R : A → B be afunctor. Then R is monadic if and only if1) R has a left adjoint L : B → A,2) for every ( Y, RL µ Y)∈ RL B, there exist Coequ A(ɛLY, L RL µ Y)and R preservesthe coequalizerCoequ Fun (ɛL RL U, L RL Uλ RL ) ,3) R reflects isomorphisms.In this case in A there exist coequalizers of R-contractible coequalizer pairs and Rpreserves them.Theorem 3.67 (Generalized Beck’s Theorem for Monads). Let (L, R) be an adjunctionwhere L : B → A and R : A → B, let A = (A, m A , u A ) be a monad onthe category B and let ψ : A = (A, m A , u A ) → RL = (RL, RɛL, η) be a monadsmorphism such that ψY is an epimorphism for every Y ∈ B. Let K ψ = Υ (ψ) =(R, (Rɛ) ◦ (ψR)) and A UK ψ (f) = A UΥ (ψ) (f) = R (f) for every morphism f in A.Then K ψ : A → A B is full and faithful if and only if for every X ∈ A we have that(X, ɛX) = Coequ A (LRɛX, ɛLRX) .Corollary 3.68 (Beck’s Theorem for Monads). Let (L, R) be an adjunction whereL : B → A and R : A → B. Then K = Υ (Id RL ) : A → RL B is full and faithful ifand only if for every X ∈ A we have that (X, ɛX) = Coequ A (LRɛX, ɛLRX) .4. ComonadsDefinition 4.<strong>1.</strong> A comonad on a category A is a triple C = ( C, ∆ C , ε C) , whereC : A → A is a functor, ∆ C : C → CC and ε C : C → A are functorial morphismssatisfying the coassociativity and the counitality conditions(∆ C C ) ◦ ∆ C = ( C∆ C) (◦ ∆ C and) CεC◦ ∆ C = C = ( ε C C ) ◦ ∆ C .

Definition 4.<strong>2.</strong> A morphism between two comonads C = ( C, ∆ C , ε C) and D =(D, ∆ D , ε D) on a category A is a functorial morphism ϕ : C → D such that∆ C ◦ ϕ = (ϕϕ) ◦ ∆ D and ε C ◦ ϕ = ε D .Example 4.3. Let ( C, ∆ C , ε C) an A-coring where A is a ring. Then• C is an A-A-bimodule• ∆ C : C → C ⊗ A C is a morphism of A-A-bimodules• ε C : C → A is a morphism of A-A-bimodules satisfying the following(∆ C ⊗ A C ) ◦∆ C = ( C ⊗ A ∆ C) ◦∆ C , ( C ⊗ A ε C) (◦∆ C = r −1Cand ε C ⊗ A C ) ◦∆ C = l −1Cwhere r C : C⊗ A A → C and l C : A⊗ A C → C are the right and left constraints.LetC = − ⊗ A C : Mod-A → Mod-A∆ C = − ⊗ A ∆ C : − ⊗ A C → − ⊗ A C ⊗ A Cε C = r − ◦ ( − ⊗ A ε C) : − ⊗ A C → − ⊗ A A → −Then, dually to the case of the R-ring, C = ( C, ∆ C , ε C) is a comonad on thecategory Mod-A.Proposition 4.4 ([H]). Let (L, R) be an adjunction with unit η and counit ɛ whereL : B → A and R : A → B. Then LR = (LR, LηR, ɛ) is a comonad on the categoryA.Proof. Dual to the proof of Proposition 3.4.Definition 4.5. A left comodule functor for a comonad C = ( C, ∆ C , ε C) on acategory A is a pair ( Q, C ρ Q)where Q : B → A is a functor and C ρ Q : Q → CQ isa functorial morphism such that(C C ρ Q)◦ C ρ Q = ( ∆ C Q ) ◦ C ρ Q and Q = ( ε C Q ) ◦ C ρ Q .Definition 4.6. A right comodule functor for a comonad C = ( C, ∆ C , ε C) on acategory A is a pair ( P, ρ C P)where P : A → B is a functor and ρCP : P → P C is afunctorial morphism such that(ρCP C ) ◦ ρ C P = ( P ∆ C) ◦ ρ C P and P = ( P ε C) ◦ ρ C P .45□Definition 4.7. For two comonads C = ( (C, ∆ C , ε C) on a category A and D =D, ∆ D , ε D) on a category B, a C-D-bicomodule functor is a triple ( )Q, C ρ Q , ρ D Q ,where Q : B → A is a functor and ( ) ( )Q, C ρ Q is a left C-comodule, Q, ρDQ is a rightD-comodule such that in addition(Cρ D Q) ◦ C ρ Q = (C ρ Q D ) ◦ ρ D Q.Definition 4.8. A morphism between two left C-comodule functors ( Q, C ρ Q)and(Q ′ , C ρ Q)is a morphism f : Q → Q ′ in A such thatC ρ Q ◦ f = (Cf) ◦ C ρ Q .

46Definition 4.9. A comodule for a comonad C = ( C, ∆ C , ε C) on a category A is apair ( X, C ρ X)where X ∈ A and C ρ X : X → CX is a morphism in A such that(C C ρ X)◦ C ρ X = ( ∆ C X ) ◦ C ρ X and X = ( ε C X ) ◦ C ρ X .A morphism between two C-comodules ( X, C ρ X)and(X ′ , C ρ X ′)is a morphism f :X → X ′ in A such thatC ρ X ◦ f = (Cf) ◦ C ρ X .We denote by C A the category of C-comodule and their morphisms.Definition 4.10. Corresponding to a comonad C = ( C, ∆ C , ε C) on A, there is anadjunction (C U, C F ) where C U is the forgetful functor and C F is the free functorC U : C A → A C F : A → C A(X, C ρ X)→ X X →(CX, ∆ C X )f → f f → CfNote that C U C F = C. The counit of the adjunction is given by the counit ε C of thecomonad Cε C : C = C U C F → A.The unit γ C : C A → C F C U of this adjunction is defined by settingC U ( γ C ( X, C ρ X))= C ρ X for every ( X, C ρ X)∈ C A.Therefore we have(εC C U ) ◦ (C Uγ C) = C U and( CF ε C) ◦ ( γ C C F ) = C F .Proposition 4.1<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad on a category A and letZ, W ∈ C A. Then Z = W if and only if C U (Z) = C U (W ) and C U ( γ C Z ) =C U ( γ C W ) . In particular, if F, G : X → C A are functors, we haveF = G if and only if C UF = C UG and C U ( γ C F ) = C U ( γ C G ) .Proposition 4.1<strong>2.</strong> Let C = ( (C, ∆ C , ε C) be a comonad on a category A.CU, (C Uγ C)) is a left C-comodule functor.ThenProof. We have to prove these two equalities(C C Uγ C) ◦ (C Uγ C) = ( ∆ C C U ) ◦ (C Uγ C)(εC C U ) ◦ (C Uγ C) = C ULet us consider ( X, C ρ X)∈ C A, we have to show that(C C Uγ C) ( X, C ρ X)◦( CUγ C) ( X, C ρ X)=(∆C C U ) ( X, C ρ X)◦( CUγ C) ( X, C ρ X)and thati.e.(εC C U ) ( X, C ρ X)◦( CUγ C) ( X, C ρ X)= C U ( X, C ρ X)(C C ρ X)◦ C ρ X = ( ∆ C X ) ◦ C ρ X and (ε C X) ◦ C ρ X = Xwhich both hold in view of the definition of C-comodule.□

Proposition 4.13. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let(X, C ρ X)be a comodule for C. Then we have(X, C ρ X)= EquA(C C ρ X , ∆ C X ) .In particular if ( Q, C ρ Q)is a left C-comodule functor, then(Q, C ρ Q)= EquFun(C C ρ Q , ∆ C Q ) .47Corollary 4.14. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let( CU, C F ) be the associated adjunction. Then (C U, (C Uγ C)) is a left C-comodulefunctor and( CU, (C Uγ C)) = Equ Fun(C C Uγ C , ∆ C C U ) .Proof. By Proposition 4.12 (C U, (C Uγ C)) is a left C-comodule functor. By Proposition4.13 we get that (C U, (C Uγ C)) (= Equ Fun C C Uγ C , ∆ C C U ) .□Proposition 4.15. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let(P, ρCP)where P : A → B a right C-comodule functor. Then we have(P, ρCP)= EquFun(ρCP C, P ∆ C) .Proof. By definition we have that(ρCP C ) ◦ ρ C P = ( P ∆ C) ◦ ρ C P .Now, let ζ : Z → P C be a functorial morphism such that ( ρ C P C) ◦ ζ = ( P ∆ C) ◦ ζand consider ζ := ( P ε C) ◦ ζ : Z → P. Then we haveρ C P ◦ ζ = ρ C P ◦ ( P ε C) ◦ ζ ρC P= ( P Cε C) ◦ ( ρ C P C ) ◦ ζ == ( P Cε C) ◦ ( P ∆ C) ◦ ζ Ccomonad= ζ.Moreover, let ζ ′ : Z → P C be another functorial morphism such that ( ρ C P C) ◦ζ ′ = ζ.Thenζ ′ = ( P ε C) ◦ ρ C P ◦ ζ ′ = ( P ε C) ◦ ζ = ζso that ζ is the unique functorial morphism such that ( ρ C P C) ◦ ζ = ζ.Lemma 4.16. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let ( Q, C ρ Q)bea left and ( P, ρ C P)be a right C-comodule functors where Q : Q → A and P : A → P.Let F : X → Q and G : P → B be functors. Then(1) ( QF, C ρ Q F ) is a left C-comodule functor and(2) ( GP, Gρ C P)is a right C-comodule functor.Proposition 4.17. Let C = ( C, ∆ C , ε C) be a comonad on A and let (C U, C F ) bethe adjunction associated. Then C U reflects isomorphisms.Proof. Let f : ( X, C ρ X)→(Y, C ρ Y)be a morphism in C A such that C Uf has atwo-sided inverse f −1 in A. SinceC ρ Y ◦ f = (Cf) ◦ C ρ X□

48we get that(Cf−1 ) ◦ C ρ Y = C ρ X ◦ f −1 .Lemma 4.18. Let C = ( C, ∆ C , ε C) be a comonad on a category A, let ( P, ρ C P)be aright C-comodule functor where P : A → B and let ( Q, C ρ Q)be a left C-comodulefunctor. Then any equalizer preserved by P C is also preserved by P and any equalizerpreserved by CQ is also preserved by Q.Proof. Consider the following equalizer□Xx Yfg Zin the category A and assume that P C preserves it. Applying to it functors P Cand P we get the following diagrams in BP ε C XP XP CXρ C P XP xP CxP ε C YP YP CYρ C P YP fP gP CfP CgP ε C ZP ZP CZρ C P ZBy assumption, the second row is an equalizer. Assume that there exists a morphismh : H → P Y such that(P f) ◦ h = (P g) ◦ h.Then, by composing with ρ C P Z we get(ρCP Z ) ◦ (P f) ◦ h = ( ρ C P Z ) ◦ (P g) ◦ hand since ρ C Pis a functorial morphism we obtain(P Cf) ◦ ( ρ C P Y ) ◦ h = (P Cg) ◦ ( ρ C P Y ) ◦ h.Since (P CX, P Cx) = Equ B (P Cf, P Cg), there exists a unique morphism k : H →P CX such that(30) (P Cx) ◦ k = ( ρ C P Y ) ◦ h.By composing with P ε C Y we get(P ε C Y ) ◦ (P Cx) ◦ k = ( P ε C Y ) ◦ ( ρ C P Y ) ◦ hand thus(P x) ◦ ( P ε C X ) ◦ k = h.Let l := ( P ε C X ) ◦ k : H → P X. Then we have(P x) ◦ l = (P x) ◦ ( P ε C X ) ◦ k εC = ( P ε C Y ) ◦ (P Cx) ◦ k(30)= ( P ε C Y ) ◦ ( ρ C P Y ) ◦ h = h.Let l ′ : H → P X be another morphism such that(P x) ◦ l ′ = h.

Then we have(P Cx) ◦ ( ρ C P X ) ◦ l ′ = ( ρ C P Y ) ◦ (P x) ◦ l ′ = ( ρ C P Y ) ◦ h= ( ρ C P Y ) ◦ (P x) ◦ l = (P Cx) ◦ ( ρ C P X ) ◦ l.Since P C preserves equalizers, we have that P Cx is a monomorphism. Since ρ C P X isalso a monomorphism, we deduce that l = l ′ . Therefore we obtain that (P X, P x) =Equ B (P f, P g) . The second statement can be proved similarly.□Lemma 4.19. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let f, g :(X, C ρ X)→(Y, C ρ Y)be morphisms in C A. Assume that there exists (E, e) =Equ A( CUf, C Ug ) and assume that CC preserves equalizers. Then there exists (Ξ, ξ) =EquC A (f, g) and C U (Ξ, ξ) = (E, e) .Proof. Since CC preserves equalizers and ( C, ∆ C) is a right C-comodule functor,also C preserves equalizers by Lemma 4.18, in particular, C preserves (E, e) . Since(C C Uf ) ◦ C ρ X ◦ e f∈C A= C ρ Y ◦ (C Uf ) ◦ eeequ= C ρ Y ◦ (C Ug ) ◦ e g∈C A= ( C C Ug ) ◦ C ρ X ◦ eby the universal property of the equalizer (CE, Ce) there exists a unique morphismC ρ E : E → CE such that(Ce) ◦ C ρ E = C ρ X ◦ e.Moreover, by composing with ε C X the first term of this equality we get(ε C X ) ◦ (Ce) ◦ C ρ Eε C = e ◦ ( ε C E ) ◦ C ρ Ewhereas the second term becomes(ε C X ) ◦ C ρ X ◦ e = eso that we obtain the following equalitye ◦ ( ε C E ) ◦ C ρ E = e.Since e is a monomorphism we deduce that(ε C E ) ◦ C ρ E = E.Now, consider the following serially commutative diagram49EC ρ ECE∆ C EC C ρ ECCEeCeCCeXC ρ XCX∆ C XC C ρ XCCXC UfC UgC C UfC C UgCC C UfCC C UgYC ρ Y CY∆ C YC C ρ Y CCY

50Since we already observed that the columns are equalizers and also the second andthe third row are equalizers by Proposition 4.13, in view of Lemma <strong>2.</strong>13 also thefirst row is an equalizer, so that (E, e) has a left C-comodule structure, i.e. thereexists (Ξ, ξ) ∈ C A such that (Ξ, ξ) = EquC A (f, g) and C U (Ξ, ξ) = (E, e) . □Lemma 4.20. Let C = ( C, ∆ C , ε C) be a comonad on a category A with equalizers andlet (C U, C F ) be the adjunction associated. The following statements are equivalent:(i) C : A → A preserves equalizers(ii) CC : A → A preserves equalizers(iii) C A has equalizers and they are preserved by C U : C A → A(iv) C U : C A → A preserves equalizers.Proof. (i) ⇒ (ii) and (iii) ⇒ (iv) are clear.(ii) ⇒ (iii) follows by Lemma 4.19.(iv) ⇒ (i) Note that C F is a right adjoint, so that in particular it preserves equalizers.Then C U C F = C also preserves equalizers.□Lemma 4.2<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A and assumethat C preserves coequalizers. Then C F preserves coequalizers where (C U, C F ) isthe adjunction associated to the comonad.Proof. Dual to proof of Lemma 3.2<strong>2.</strong> LetXfgbe a coequalizer in A. Let us consider the fork obtained by applying the functor C Fto the coequalizeri.e.C F XC F fC F g Y C F Yk KC F k C F K(CX, ∆ C X ) Cf (CY, ∆ C Y ) Ck ( CK, ∆ C K )CgNow, let ( Z, C ρ Z)∈ C A and z : ( CY, ∆ C Y ) → ( Z, C ρ Z)be a morphism in C A suchthat z ◦ (Cf) = z ◦ (Cg). Since C preserves coequalizers, we know that (CK, Ck) =Coequ A (Cf, Cg) . By the universal property of the coequalizer (CK, Ck) in A, thereexists a unique morphism z ′ : CK → Z in A such that z ′ ◦ (Ck) = z. We now wantto prove that z ′ is a morphism in C A, i.e. that (Cz ′ ) ◦ ( ∆ C K ) = C ρ Z ◦ z ′ . Since zis a morphism in C A we have that(Cz) ◦ ( ∆ C Y ) = C ρ Z ◦ zand since also Ck is a morphism in C A we have thatThen we have(CCk) ◦ ( ∆ C Y ) = ( ∆ C K ) ◦ (Ck) .(Cz ′ ) ◦ ( ∆ C K ) ◦ (Ck) = (Cz ′ ) ◦ (CCk) ◦ ( ∆ C Y )propz= (Cz) ◦ ( ∆ C Y ) = C ρ Z ◦ z propz= C ρ Z ◦ z ′ ◦ (Ck)

and since C preserves coequalizers, Ck is an epimorphism, so that we get(Cz ′ ) ◦ ( ∆ C K ) = C ρ Z ◦ z ′ .Lemma 4.2<strong>2.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A, let L, N :B → A be functors and let ρ : L → CL be a coassociative and counital functorialmorphism, that is (L, ρ) is a left C-comodule functor. Let u : N → L and letφ : N → CN be functorial morphisms such that(31) ρ ◦ u = (Cu) ◦ φ.If CCu and u are monomorphisms, then φ is coassociative and counital, that is(N, φ) is a left C-comodule functor.Proof. Let us prove that φ is coassociative(CCu) ◦ (Cφ) ◦ φ (31)= (Cρ) ◦ (Cu) ◦ φ (31)= (Cρ) ◦ ρ ◦ uρcoass= ( ∆ C L ) ◦ ρ ◦ u (31)= ( ∆ C L ) ◦ (Cu) ◦ φ ∆C= (CCu) ◦ ( ∆ C N ) ◦ φ.Since CCu is a monomorphism we get thatLet us prove that φ is counital(Cφ) ◦ φ = ( ∆ C N ) ◦ φ.u ◦ ( ε C N ) ◦ φ εC = ( ε C L ) ◦ (Cu) ◦ φ (31)= ( ε C L ) ◦ ρ ◦ u ρcounit= u.Since u is a monomorphism we conclude.4.<strong>1.</strong> Lifting of comodule functors. This subsection collects the dual <strong>results</strong> forliftings of module functors so that one can skip reading all the proofs we keep herein order to give details of the <strong>results</strong> we use in the following.Proposition 4.23 ([W] 3.5). Let C = ( C, ∆ C , ε C) be a comonad on a category A,let D = ( D, ∆ D , ε D) be a comonad on a category B and let T : A → B be a functor.Then there is a bijection between the following collections of dataF functors ˜T : C A → D B that are liftings of T (i.e. D U ˜T = T C U)M functorial morphisms Ξ : T C → DT such that(∆ D T ) ◦ Ξ = (DΞ) ◦ (ΞC) ◦ ( T ∆ C) (and ε D T ) ◦ Ξ = T ε Cgiven by( )a : F → M where a ˜T = (D U D F T ε C) ◦( )D Uγ D ˜T C Fb : M → F where D Ub (Ξ) = T C U and D Uγ D b (Ξ) = Ξ ◦ ( T C Uγ C) i.e.b (Ξ) (( X, C ρ X))=(T X, (ΞX) ◦(T C ρ X))and b (Ξ) (f) = T (f) .Proof. Let ˜T : C A → D B be a lifting of the functor T : A → B (i.e. D U ˜T = T C U).Define a functorial morphism ξ : ˜T C F → D F T as the compositeξ := (D F T ε C) ( )◦ γ D ˜T C F51□□

52where ε C : C = C U C F → A is also the counit of the adjunction (C U, C F ) andγ D : D B → D F D U is the unit of the adjunction (D U, D F ) . Let now definethat isΞ def= D Uξ : D U ˜T C F = T C U C F = T C → D U D F T = DTΞ = D Uξ = (D U D F T ε C) ◦( )D Uγ D ˜T C F .Dually to Proposition 3.24 you can prove that Ξ is a functorial morphism satisfying(∆ D T ) ◦ Ξ = (DΞ) ◦ (ΞC) ◦ ( T ∆ C) and ( ε D T ) ◦ Ξ = T ε C .Conversely, let Ξ be a functorial morphism satisfying ( ∆ D T ) ◦ Ξ = (DΞ) ◦ (ΞC) ◦(T ∆C ) and ( ε D T ) ◦ Ξ = T ε C . We define ˜T : C A → D B by setting, for every(X, C ρ X)∈ C A,˜T (( X, C ρ X))=(T X, (ΞX) ◦(T C ρ X))and for every f : ( X, C ρ X)→(Y, C ρ Y)∈ C A,˜T (f) = T (f) .Dually to Proposition 3.24 you can prove that ˜T is a functor between C A → D B whichlifts T and that a : F → M and b : M → F define a bijective correspondence. □Corollary 4.24. Let X , A be categories and let C = ( C, ∆ C , ε C) be a comonad ona category A and let F : X → A be a functor. Then there is a bijection between thefollowing collections of data:F Functors C F : X → C A such that C U C F = F ,G Left C-comodule coactions C ρ F : F → CFgiven byα : F → G where α (C F ) = C Uγ C C F : F → CFβ : G → F where C Uβ (C ρ F)= F and C Uγ C β (C ρ F)= C ρ F i.e.β : G → F where β (C ρ F)(X) =(F X, C ρ F X ) and β (C ρ F)(f) = F (f) .Proof. Apply Proposition 4.23 to the case A = X , B = A, C = Id X , D = C. Then˜T = C F is the lifting of F and Ξ = C ρ F : F → CF satisfies ( ∆ C F ) ◦ C ρ F =(C C ρ F)◦ C ρ F and ( ε C F ) ◦ C ρ F = F that is ( F, C ρ F)is a left C-comodule functor. □Corollary 4.25. Let (L, R) be an adjunction where L : B → A, R : A → B andlet C = ( C, ∆ C , ε C) be a comonad on a category A. Then there exists a bijectivecorrespondence between the following collections of data:K Functors K : B → C A such that C U ◦ K = L,L Functorial morphism β : L → CL such that (L, β) is a left comodule functorfor the comonad Cgiven byΦ : K → L where Φ (K) = C U ( γ C K ) : L → CLΩ : L → K where Ω (β) (Y ) = (LY, βY ) and C UΩ (β) (f) = L (f) .

Proof. Apply Corollary 4.24 to the case ”F ” = L : B → A where (L, R) is anadjunction and C = ( C, ∆ C , ε C) a comonad on A.□Proposition 4.26. Let C = ( C, ∆ C , ε C) be a comonad on a category A and letD = ( D, ∆ D , ε D) be a comonad on a category B. Let T : A → B be a functor,let ˜T : C A → D B be a lifting of T (i.e.D U ˜T = T C U) and let Ξ : T C → DTas in Proposition 4.23. Then Ξ is an isomorphism if and only if ξ = (D ( )F T ε C) ◦γ D ˜T C F : ˜T C F → D F T is an isomorphism.Proof. By construction in Proposition 4.23 we have that Ξ = D Uξ. Assume thatΞ is an isomorphism. Since, by Proposition 4.17, D U reflects isomorphisms, ξ :˜T C F → D F T is an isomorphism. Conversely, assume that ξ : ˜T C F → D F T is anisomorphism. Then D Uξ is also an isomorphism.□Corollary 4.27. Let (L, R) be an adjunction where L : B → A and R : A → Band let C = ( C, ∆ C , ε C) be a comonad on B. Let K : B → C A be a functor suchthat C U ◦ K = L and let (L, β) be a left C-comodule functor as in Corollary 4.25.Then β is an isomorphism if and only if γ C K : K → C F L is an isomorphism.Proof. Apply Proposition 4.26 with T = L so that the categories A and B areinterchanged, C = Id B and D = C. Then ˜T = K is the lifting of L and Ξ = β : L →CL, given by β = C Uξ = C Uγ C K.□Lemma 4.28. Let C = ( C, ∆ C , ε C) be a comonad over a category A with equalizers.Let Q : B → A be a left C-comodule functor with functorial morphisms C ρ Q : Q →CQ. Then there exists a unique functor C Q : B → C A such thatC U C Q = Q and C Uγ C C Q = C ρ Q .Moreover if ψ : Q → T is a functorial morphism between left C-module functors andψ satisfiesC ρ Q ◦ (Cψ) = ψ ◦ (C ρ T)then there is a unique functorial morphism C ψ : C Q → C T such thatC U C ψ = ψ.Proof. Corollary 4.24 applied to the case where F = Q and C ρ F = C ρ Q gives us thefirst statement. Let B ∈ B. Then we have( Cρ Q B ) ◦ (CψB) = (ψB) ◦ (C ρ T B )which means that ψB yields a morphism C ψB in C A.Proposition 4.29. Let C = ( C, ∆ C , ε C) be a comonad over a category A and letD = ( D, ∆ D , ε D) be a comonad over a category B. Assume that both A and B haveequalizers and that C preserves equalizers. Let Q : B → A be a functor and letC ρ Q : Q → CQ and ρ D Q : Q → QD be functorial morphisms. Assume that C ρ Q iscoassociative and counital and that ( CρQ) D ◦ C ρ Q = (C ρ Q D ) ◦ ρ D Q . Set((32)Q D , ι Q) (= Equ Fun ρD DQ U, Q D Uγ D) .53□

54Then Q D : D B → A is a left C-comodule functor where C ρ Q D : Q D → CQ D isuniquely determined by((33)Cρ D Q U ) ◦ ι Q = ( Cι Q) ◦ C ρ Q D.Moreover there exists a unique functor ( C Q D) : D B → C A such that(34)C U C ( Q D) = Q D and C Uγ C C ( Q D) = C ρ Q D.Proof. By Lemma <strong>2.</strong>8 we can consider ( Q D , ι Q) = Equ Fun(ρDQ D U, Q D Uγ D) . Since(CρDQ)◦ C ρ Q = (C ρ Q D ) ◦ ρ D Qwe deduce that(35)(CρDQ D U ) ◦ (C ρ Q D U ) = (C ρ Q D D U ) ◦ ( ρ D Q D U ) .Also, in view of the naturality of C ρ Q , we have(36)(CQ D Uγ D) ◦ (C ρ Q D U ) = (C ρ Q D D U ) ◦(Q D Uγ D) .We compute(CQ D Uγ D) ◦ (C ρ D Q U ) ◦ ι Q (36)= (C ρ Q D D U ) (◦ Q D Uγ D) ◦ ι Qι Q equ= (C ρ Q D D U ) ◦ ( ρ D Q D U ) ◦ ι Q (35)= ( Cρ D Q D U ) ◦ (C ρ Q D U ) ◦ ι Q .Since C preserves equalizers, we have(CQ D , Cι Q) = Equ Fun(CρDQ D U, CQ D Uγ D)hence there exists a unique functorial morphism C ρ Q D : Q D → CQ D such that( ) CιQ◦ C ρ Q D = (C ρ D Q U ) ◦ ι Q .Since Q is a left C-comodule functor, by Lemma 4.16, also Q D U is a left C-comodulefunctor. Now ι Q is a monomorphism and hence, since C preserves equalizers, alsoCCι Q is a monomorphism. Therefore we can apply Lemma 4.22 to ”φ” = C ρ Q D,”u” = ι Q and ”ρ” = C ρ D Q U and hence we obtain that ( Q D , C ρ Q D)is a left C-comodule functor that is C ρ Q D is coassociative and counital. By Lemma 4.28 appliedto ( Q D , C ρ Q D) ( there exists a functorCQ D) : D B → C A such that C U ( C Q D) =Q D and C ρ Q D = C Uγ ( C C Q D) . Moreover ( C Q D) is unique with respect to theseproperties.□Proposition 4.30. Let C = ( C, ∆ C , ε C) be a comonad over a category A and letD = ( D, ∆ D , ε D) be a comonad over a category B. Assume that both A and B haveequalizers and C preserves them. Let Q : B → A be a C-D-bicomodule functorwith functorial morphisms C ρ Q : Q → CQ and ρ D Q : Q → QD. Then the functorC Q : B → C A is a right D-comodule functor via ρ DC Q : C Q → C QD where ρ DC Q isuniquely determined by(37)C Uρ DC Q = ρD Q.

Let( (CQ ) ()D, ι Q) C = Equ Fun ρ DC C QU, C Q C Uγ D . Then we have( CQ ) D=C ( Q D) : D B → C A.55Proof. Since Q is endowed with a left C-comodule structure, by Lemma 4.28 thereexists a unique functor C Q : B → C A such that C U C Q = Q and C Uγ C C Q = C ρ Q .Note that, since Q is a C-D-bicomodule functor, in particular the compatibilitycondition(Cρ D Q) ◦ C ρ Q = (C ρ Q D ) ◦ ρ D Qholds, that is ρ D Q : Q = C U C Q → QD = C U C QD is a morphism in C A. Thus, thereexists a functorial morphism ρ DC Q : C Q → C QD such thatC Uρ DC Q = ρD Q.By the coassociativity and counitality properties of ρ( )D Q we get that also ρDC Q iscoassociative and counital, so thatC Q, ρ DC Qis a right D-comodule functor. Thuswe can consider the equalizer(38)( CQ ) D ι C Q C Q D Uρ D C Q D UC Q D Uγ D C QD D Uso that we get a functor (C Q ) D: D B → C A. Since C preserves equalizers, by Lemma4.20 also C U preserves equalizers. Then, by applying the functor C U to (38) we stillget an equalizerC U (C Q ) DC Uι C Q C U C Q D UC Uρ D C Q D UC U C Q D Uγ D C U C QD D Uthat isC U (C Q ) DC Uι C Q Q D Uρ D Q D UQ D Uγ D QD D UBy Proposition 4.29 ( Q D , ι Q) (= Equ Fun ρD DQ U, Q D Uγ D) , then we haveC U (C Q ) D= Q D and C Uι CQ = ι Q .MoreoverC Uγ C ( CQ ) D: C U (C Q ) D= Q D → C C U (C Q ) D= CQDso that, using Proposition 4.29 where we prove that ( Q D , C ρ Q D)is a left C-comodulefunctor and that C Uγ C C ( Q D) = C ρ Q D, we getC Uγ C ( CQ ) D= C ρ Q D = C Uγ C C ( Q D)i.e.( CQ ) D=C ( Q D) .□

56Notation 4.3<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A and letD = ( D, ∆ D , ε D) be a comonad over a category B. Assume that both A and B haveequalizers and A preserves them. Let Q : B → A be a C-D-bicomodule functor. Inview of Proposition 4.30, we setC Q D = (C Q ) D=C ( Q D) .Proposition 4.3<strong>2.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A and letD = ( D, ∆ D , ε D) be a comonad over a category B. Assume that both A and Bhave equalizers and let Q : B → A be an C-D-bicomodule functor. Then, withnotations of Proposition 4.29, we can consider the functor Q D where ( (Q D , ι Q) =Equ Fun ρD DQ U, Q D Uγ D) . Then(39) Q DD F = Q and ι QD F = ρ D Q.Proof. By construction we have that ( Q D , ι Q) = Equ Fun(ρDQ D U, Q D Uγ D) . By applyingit to the functor D F we get that(Q DD F , ι QD F ) = Equ Fun(ρDQ D U D F , Q D Uγ DD F )= Equ Fun(ρDQ D, Q∆ D) .Since Q is a right D-comodule functor, by Proposition 4.15 we have that(Q, ρDQ)= EquFun(ρDQ D, Q∆ D)so that we get(Q DD F , ι QD F ) (= Equ Fun ρDQ D, Q∆ D) = ( )Q, ρ D Q .□Proposition 4.33. Let D = ( D, ∆ D , ε D) be a comonad over a category B withequalizers such that D preserves equalizers. Let G : D B → A be a functor preservingequalizers. SetQ = G ◦ D F and let ρ D Q = Gγ DD FThen ( Q, ρ D Q)is a right D-comodule functor and(40) Q D = ( G ◦ D F ) D= G.Proof. We compute(ρDQ D ) ◦ ρ D Q = ( Gγ DD F D ) ◦ ( Gγ DD F ) γ D = ( G D F D Uγ DD F ) ◦ ( Gγ DD F )and= ( G D F ∆ D) ◦ ( Gγ DD F ) = ( Q∆ D) ◦ ρ D Q(QεD ) ◦ ρ D Q = ( G D F ε D) ◦ ( Gγ DD F ) adj= G D F = Q.Thus ( Q, ρ D Q)is a right D-comodule functor. Recall that (see Proposition 4.29)(Q D , ι Q) = Equ Fun(ρDQ D U, Q D Uγ D)and by Proposition 4.32 we have Q DD F = Q and ι QD F = ρ D Q . In particular we getQ DD F = Q = G D F .

58F D right D-comodule functors Q : B → A such that QD preserves equalizers.(A ← D B ) functors G : D B → A preserving equalizersgiven byν D : F D → ( A ← D B ) where ν (( )) D Q, ρ D Q = QD(κ D : A ← D B ) → F D where κ D (G) = ( G D F , Gγ DD F )where Q D is uniquely determined by ( Q D , ι Q) = Equ Fun(ρDQ D U, Q D Uγ D) .Proof. Let Q : B → A be a right D-comodule functor. Then we can considerQ D : D B → A defined by (32) as(Q D , ι Q) (= Equ Fun ρD DQ U, Q D Uγ D) .Since by assumption QD preserves equalizers, by Lemma 4.18 also Q preservesequalizers. Moreover, since D preserves equalizers, by Lemma 4.20 also the functorD U preserves equalizers. Thus both QD D U and Q D U preserve equalizers. ByCorollary <strong>2.</strong>14 we get that also Q D : D B → A preserves equalizers.Conversely, let us consider a functor G : D B → A that preserves equalizers. ByProposition 4.33 we can consider the right D-comodule functor defined as followsQ = G ◦ D F and let ρ D Q = Gγ DD F .Since D F is right adjoint to D U in particular D F preserves equalizers and since byassumption G preserves equalizers, we get that also Q = G ◦ D F preserves equalizersand so does QD.Now, we want to prove that ν D and κ D determine a bijective correspondencebetween F D and ( A ← D B ) . Let us start with a right D-comodule functor(Q : B → A, ρDQ). Then we have(κ D ◦ ν D) (( Q, ρ D Q))= κD ( Q D) = ( Q DD F , Q D γ DD F )= ( Q DD F , ρ D Q DD F) (39)= ( Q, ρ D Q).Moreover we have(ν D ◦ κ D) (G) = ν D (( G D F , Gγ DD F )) = ( G D F ) D (40)= G.Theorem 4.36. Let C = ( C, ∆ C , ε C) be a comonad on a category A with equalizerssuch that C preserves equalizers. Then there exists a bijective correspondencebetween the following collections of data:C F left C-comodule functors Q : B → A such that CQ preserves equalizers( CA ← B ) functors H : B → C A preserving equalizers□given byC ν :C κ :C F → (C A ← B ) where C ν (( ))Q, C ρ Q = C Q( CA ← B ) → C F where C κ (H) = (C U ◦ H, C Uγ C H )where C Q : B → C A is the functor defined in Lemma 4.28.

Proof. Let ( Q : B → A, C ρ Q)be a left C-comodule functor. Then, by Lemma 4.28,there exists a unique functor C Q : B → C A such thatC U ◦ C Q = Q and C Uγ C C Q = C ρ Q .Note that, since CQ preserves equalizers, by Lemma 4.18, Q = C U ◦ C Q preservesequalizers. Then, by Lemma 4.19, also C Q preserves equalizers. Conversely, ifH : B → C A is a functor preserving equalizers, we get that C U ◦ H : B → A.Moreover, by Lemma 4.20, C U preserves equalizers and thus also C U ◦ H preservesequalizers. Now, let us prove that C ν and C κ determine a bijective correspondencebetween C F and (C A ← B ) . We compute( Cκ ◦ C ν ) (( Q, C ρ Q))= C κ (C Q ) = (C U C Q, C Uγ C C Q ) = ( Q, C ρ Q).On the other hand we have( Cν ◦ C κ ) (H) = C ν ((C U ◦ H, C Uγ C H )) = C (C U ◦ H ) (41)= H.Theorem 4.37. Let C = ( C, ∆ C , ε C) be a comonad on a category A with equalizerssuch that C preserves equalizers. Let D = ( D, ∆ D , ε D) be a comonad on a categoryB with equalizers such that D preserves equalizers. Then there exists a bijectivecorrespondence between the following collections of data:( C F D C-D-bimodule functors Q : B → A such that CQ and QD preserve equalizersCA ← D B ) functors G : D B → C A preserving equalizersgiven byC ν D :C F D → (C A ← D B ) where C ν D (( Q, C ρ Q , ρ D Q))= C Q D59□C κ D :( CA ← D B ) → C F D where C κ D (G) = (C U ◦ G ◦ D F, C Uγ C G D F, C UGγ DD F ) .Proof. Let us consider a C-D-bicomodule functor ( Q : B → A, C ρ Q , ρQ) D such thatCQ and QD preserve equalizers. In particular, ( Q, ρQ) D is a right D-comodule functor,so that we can apply the map ν D : F D → ( A ← D B ) of Theorem 4.35 and weget a functor ν (( D Q, ρQ)) D = Q D : D B → A which preserves equalizers. By Proposition4.29, ( Q D , C ρ Q D)is a left C-comodule functor so that we can also apply themap C ν : C F → (C A ← B ) of Theorem 4.36 where the category B is D B. The mapC ν is defined by C ν (( Q D , C ρ Q D)) ( =CQ D) = C Q D : D B → C A and C Q D preservesequalizers. Conversely, let us consider a functor G : D B → C A which preservesequalizers. By Theorem 4.36, we get a left C-comodule functor given byC κ (G) = (C U ◦ G, C Uγ C G )where C U ◦ G : D B → A and C C UG preserves equalizers. By Lemma 4.18, alsoC U ◦ G : D B → A preserves equalizers. Thus, we can apply Theorem 4.35 and weget a right D-comodule functorκ D (C UG ) = (C UG D F , C UGγ DD F )where C UG D F : B → A is such that C UG D F D preserves equalizers. Clearly, sinceC UG preserves equalizers, D F is a right adjoint and C preserves equalizers by assumption,we deduce that also C C UG D F preserves equalizers. Now, we want to

60prove that C ν D : C F D → (C A ← D B ) and C κ D : (C A ← D B ) → C F D determine abijection. We have(Cκ D ◦ C ν D) (( ))Q, C ρ Q , ρ D Q = C κ D (C Q D)= (C U ◦ C Q D ◦ D F , C Uγ C C Q DD F , C U C Q D γ DD F ) = ( Q, C Uγ C C Q, Q D γ DD F )= ( ) ( )Q, C ρ Q , ρ D Q DD F = Q, C ρ Q , ρ D Qand( Cν D ◦ C κ D) (G) = C ν D ((C U ◦ G ◦ D F , C Uγ C G D F , C UGγ DD F ))= C (C U ◦ G ◦ D F ) D=C ((C U ◦ G ◦ D F ) D )(40)= C (C U ◦ G ) (41)= G.Proposition 4.38. Let C = ( C, ∆ C , ε C) be a comonad over a category A with equalizersand assume that C preserves equalizers. Let D = ( D, ∆ D , ε D) be a comonadover a category B with equalizers and let Q : B → A be a C-D-bicomodule functor.Then there exists a unique lifted functor C Q D : D B → C A such thatC U C Q DD F = Q.Proof. By Proposition 4.30 there exists a unique functor C Q D : D B → C A such thatC U C Q D = Q D . Now, by Proposition 4.32 we also get that Q DD F = Q so that weobtainC U C Q DD F = Q.Corollary 4.39. Let C = ( C, ∆ C , ε C) be a comonad over a category A with equalizersand assume that C preserves equalizers and let Q : A → A be a C-bicomodulefunctor. Then there exists a unique lifted functor C Q C : C A → C A such thatC U C Q C C F = Q.Proof. We can apply Proposition 4.38 to the case D = C and B = A.Proposition 4.40. Let C = ( C, ∆ C , ε C) be a comonad over a category A with equalizersand assume that C preserves equalizers. Let D = ( D, ∆ D , ε D) be a comonadover a category B with equalizers and let P, Q : B → A be C-D-bicomodule functors.Let f : P → Q be a functorial morphism of left C-comodule functors and ofright D-comodule functors. Then there exists a unique functorial morphism of leftC-comodule functorsf D : P D → Q Dsatisfyingι Q ◦ f D = ( f D U ) ◦ ι P .Then we can considerC f D : C P D → C Q Dsuch thatD U C f D = f D .□□□

61Proof. Consider the following diagramP D ιP P D Uf Df D UQ D ιQ Q D Uρ D P D UP D Uγ D ρ D Q D UQ D Uγ DP D D UfD D U QD D USince f is a functorial morphism and it is a functorial morphism of right D-comodulefunctors, the right square serially commutes. Note that(ρDQ D U ) ◦ ( f D U ) ◦ ι P = ( Q D Uγ D) ◦ ( f D U ) ◦ ι Pso that, by the universal property of the equalizer, there exists a unique morphismf D : P D → Q D such that((42)f D U ) ◦ ι P = ι Q ◦ f D .We now want to prove that f D is a functorial morphism of left C-comodule functor.In fact we have(CιQ ) ◦ C ρ Q D ◦ f D (33)= (C ρ Q D U ) ◦ ι Q ◦ f D(42)= (C ρ D Q U ) ◦ ( f D U ) ◦ ι P(Cf D U ) ◦ (C ρ D P U ) ◦ ι PfleftCcolin=(33)= ( Cf D U ) ◦ ( Cι P ) ◦ C ρ P D(42)= ( Cι Q) ◦ ( Cf D) ◦ C ρ P Dand since C preserves equalizers Cι Q is a monomorphism so that we getC ρ Q D ◦ f D = ( Cf D) ◦ C ρ P D.Then there exists a functorial morphism C f D : C P D → C Q D such thatC U C f D = f D .Corollary 4.4<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad over a category A with equalizersand assume that C preserves equalizers and let P, Q : B → A be C-bicomodulefunctors. Let f : P → Q be a functorial morphism of C-bicomodule functors. Thenthere exists a unique functorial morphism of left C-comodule functorssatisfyingThen we can considersuch thatf C : P C → Q Cι Q ◦ f C = ( f C U ) ◦ ι P .C f C : C P C → C Q CC U C f C = f C .□Proof. We can apply Proposition 4.40 to the case D = C and B = A.□

624.<strong>2.</strong> The comparison functor for comonads.Proposition 4.42 ([GT, Proposition <strong>2.</strong>1]). Let (L, R) be an adjunction where L :B → A and R : A → B and let C = ( C, ∆ C , ε C) be a comonad on a category A.There exists a bijective correspondence between the following collections of data:M comonad morphisms ϕ : LR = (LR, LηR, ɛ) → C = ( C, ∆ C , ε C)R functorial morphism α : R → RC such that (R, α) is a right comodule functorfor the comonad CL functorial morphism β : L → CL such that (L, β) is a left comodule functorfor the comonad Cgiven byProof. For a given ϕ ∈ M, we computeΘ : M → R where Θ (ϕ) = (Rϕ) ◦ (ηR)Ξ : R → M where Ξ (α) = (ɛC) ◦ (Lα)Γ : M → L where Γ (ϕ) = (ϕL) ◦ (Lη)Λ : L → M where Λ (β) = (Cɛ) ◦ (βR) .(Θ (ϕ) C) ◦ Θ (ϕ) = (RϕC) ◦ (ηRC) ◦ (Rϕ) ◦ (ηR)η= (RϕC) ◦ (RLRϕ) ◦ (ηRLR) ◦ (ηR)η,ϕϕmorphcom (= (Rϕϕ) ◦ (RLηR) ◦ (ηR) =) R∆C◦ (Rϕ) ◦ (ηR) = ( R∆ C) ◦ Θ (ϕ)and( ) RεC◦ Θ (ϕ) = ( Rε C) ◦ (Rϕ) ◦ (ηR) ϕmorphcom= (Rɛ) ◦ (ηR) = R.Therefore we deduce that Θ (ϕ) ∈ R. For a given α ∈ R, we compute(Ξ (α) Ξ (α)) ◦ (LηR) Ξ(α)= (Ξ (α) C) ◦ (LRΞ (α)) ◦ (LηR)= (ɛCC) ◦ (LαC) ◦ (LRɛC) ◦ (LRLα) ◦ (LηR)η= (ɛCC) ◦ (LαC) ◦ (Lα) (R,α)= (ɛCC) ◦ ( LR∆ C) ◦ (Lα)ɛ= ∆ C ◦ (ɛC) ◦ (Lα) = ∆ C ◦ Ξ (α)andε C ◦ Ξ (α) = ε C ◦ (ɛC) ◦ (Lα) = ɛ ɛ ◦ ( LRε C) ◦ (Lα) (R,α)= ɛ.Therefore we deduce that Ξ (α) ∈ M. For a given ϕ ∈ M, we computeϕ[CΓ (ϕ)] ◦ Γ (ϕ) = (CϕL) ◦ (CLη) ◦ (ϕL) ◦ (Lη)= (ϕCL) ◦ (LRϕL) ◦ (LRLη) ◦ (Lη) = η (ϕCL) ◦ (LRϕL) ◦ (LηRL) ◦ (Lη)= (ϕϕL) ◦ (LηRL) ◦ (Lη) ϕmorphcom (= ∆ C L ) ◦ (ϕL) ◦ (Lη) = ( ∆ C L ) ◦ Γ (ϕ)and(ε C L ) ◦ Γ (ϕ) = ( ε C L ) ◦ (ϕL) ◦ (Lη) ϕmorphcom= (ɛL) ◦ (Lη) = L.Therefore we deduce that Γ (ϕ) ∈ L. For a given β ∈ L, we compute(Λ (β) Λ (β)) ◦ (LηR) Λ(β)= (CΛ (β)) ◦ (Λ (β) LR) ◦ (LηR)

= (CCɛ) ◦ (CβR) ◦ (CɛLR) ◦ (βRLR) ◦ (LηR)β= (CCɛ) ◦ (CβR) ◦ (CɛLR) ◦ (CLηR) ◦ (βR) = (CCɛ) ◦ (CβR) ◦ (βR)(L,β)= (CCɛ) ◦ ( ∆ C LR ) ◦ (βR) =∆ C= ∆ C ◦ (Cɛ) ◦ (βR) = ∆ C ◦ Λ (β)andε C ◦ Λ (β) = ε C ◦ (Cɛ) ◦ (βR) εC = ɛ ◦ ( ε C LR ) ◦ (βR) (L,β)= ɛ.Therefore we deduce that Λ (β) ∈ M. Let now ϕ ∈ M and let us calculateΞΘ (ϕ) = (ɛC) ◦ (LRϕ) ◦ (LηR) ɛ = ϕ ◦ (ɛLR) ◦ (LηR) = ϕ.Let now α ∈ R and let us calculateΘΞ (α) = (RΞ (α)) ◦ (ηR) = (RɛC) ◦ (RLα) ◦ (ηR) (RɛC) ◦ (ηRC) ◦ α = α.Let now ϕ ∈ M and let us calculateΛΓ (ϕ) = (Cɛ) ◦ (Γ (ϕ) R) = (Cɛ) ◦ (ϕLR) ◦ (LηR) ϕ = ϕ ◦ (LRɛ) ◦ (LηR) = ϕ.Let now β ∈ L and let us calculateΓΛ (β) = (Λ (β) L) ◦ (Lη) = (CɛL) ◦ (βRL) ◦ (Lη) β = (CɛL) ◦ (CLη) ◦ β = β.Theorem 4.43 ([D, Theorem II.<strong>1.</strong>1] and [GT, Theorem <strong>1.</strong>2]). Let (L, R) be an adjunctionwhere L : B → A and R : A → B and let C = ( C, ∆ C , ε C) be a comonadon a category A. There exists a bijective correspondence between the following collectionsof data:K Functors K : B → C A such that C U ◦ K = L,M comonad morphisms ϕ : LR = (LR, LηR, ɛ) → C = ( C, ∆ C , ε C)given byΨ : K → M where Ψ (K) = (Cɛ) ◦ ([C U ( γ C K )] R )Υ : M → K where Υ (ϕ) (Y ) = (LY, (ϕLY ) ◦ (LηY )) and Υ (ϕ) (f) = L (f) .Proof. By Corollary 4.25, there exists a bijective correspondence between K and thecollection L of functorial morphisms β : L → CL such that (L, β) is a left comodulefunctor for the comonad C given byΦ : K → L where Φ (K) = C U ( γ C K ) : L → CLΩ : L → K where Ω (β) (Y ) = (LY, βY ) and Ω (β) (f) = L (f) .By Proposition 4.42, there exists a bijective correspondence between L and thecollection M of comonad morphisms ϕ : LR = (LR, LηR, ɛ) → C = ( C, ∆ C , ε C)given byWe computeΛ : L → M where Λ (β) = (Cɛ) ◦ (βR)Γ : M → L where Γ (ϕ) = (ϕL) ◦ (Lη) .(Λ ◦ Φ) (K) = (Cɛ) ◦ ([C U ( γ C K )] R ) = Ψ (K)63□

64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY, (ϕLY ) ◦ (LηY )) = Υ (ϕ) (Y ) and [(Ω ◦ Γ) (ϕ)] (f) = Lf.Remark 4.44. When C = LR = (LR, LηR, ɛ) and ϕ = Id LR the functor K =Υ (ϕ) : B → LR A such that LR U ◦ K = L is called the Eilenberg-Moore comparisonfunctor.Corollary 4.45. Let C = ( C, ∆ C , ε C) and D = ( D, ∆ D , ε D) be comonads on acategory A. There exists a bijective correspondence between the following collectionsof data:K Functors K : C A → D A such that D U ◦ K = C U,M comonad morphisms ϕ : C → Dgiven byΨ : K → M where Ψ (K) = (Cɛ) ◦ ([D U ( γ D K )] C F )Υ : M → K where Υ (ϕ) (Y ) = (C UY, ( ϕ C UY ) ◦ (C Uγ C Y )) and Υ (ϕ) (f) = C U (f) .Proof. Apply Theorem 4.43 to the case L = C U : C A → A and R = C F : A → C Aand note that (LR, LηR, ɛ) = (C U C F , C Uγ C C F , ε C) = ( C, ∆ C , ε C) .□Proposition 4.46. Let (L, R) be an adjunction where L : B → A and R : A →B , let C = ( C, ∆ C , ε C) be a comonad on the category A and let ϕ : LR =(LR, LηR, ɛ) → C = ( C, ∆ C , ε C) be a comonad morphism. Let α = Θ (ϕ) =(Rϕ) ◦ (ηR). Then the isomorphism a X,Y : Hom A (LY, X) → Hom B (Y, RX) of theadjunction (L, R) induces an isomorphismâ X,Y: HomC A(Kϕ Y, ( X, C ρ X))→ EquSets(HomB (Y, αX) , Hom B(Y, R C ρ X)).Proof. Leta X,Y : Hom A (LY, X) → Hom B (Y, RX)be the isomorphism of the adjunction (L, R) for every Y ∈ B and for every X ∈ A.Recall that a X,Y (ξ) = (Rξ) ◦ (ηY ) and a −1X,Y(ζ) = (ɛX) ◦ (Lζ) . Let us check thatwe can apply Lemma <strong>2.</strong>15 to the case Z = Hom A (L−, X) , Z ′ = Hom B (−, RX) ,W = Hom A (L−, CX) , W ′ = Hom( B (−, RCX), ) a = C ρ X ◦ −, b = C − ◦Γ (ϕ) Y,a ′ = Hom( B (−, αX), b ′ = Hom B −, R C ρ X and ϕ = aX,− , ψ = a CX,− , E =Equ CFun ρ X ◦ −, C (−) ◦ Γ (ϕ) Y ) (and( ))E ′ = Equ Fun HomB (−, αX) , Hom B −, R C ρ X(Equ CFun ρ X ◦ −, C (−) ◦ Γ (ϕ) − ) â ( ( ))X,− Equ Fun HomB (−, αX) , Hom B −, R C ρ X□iZ = Hom A (L−, X)a= C ρ X ◦−b=C−◦Γ(ϕ)−W = Hom A (L−, CX)a X,−a CX,−Z ′ = Hom B (−, RX)a ′ =Hom B (−,αX)i ′b ′ =Hom B(−,R C ρ X) W ′ = Hom B (−, RCX)

For every Y ∈ B, X ∈ A and for every ξ ∈ Hom A (LY, X) , let us computeHom B (Y, αX) ◦ a X,Y (ξ) = αX ◦ a X,Y (ξ) = (RϕX) ◦ (ηRX) ◦ a X,Y (ξ)defa= (RϕX) ◦ (ηRX) ◦ (Rξ) ◦ (ηY )η= (RϕX) ◦ (RLRξ) ◦ (RLηY ) ◦ (ηY ) defa= a CX,Y [(ϕX) ◦ (LRξ) ◦ (LηY )] ϕ == a CX,Y [(Cξ) ◦ (ϕLY ) ◦ (LηY )]Since Γ (ϕ) = (ϕL) ◦ (Lη) we have obtained thatLet us calculateHom B (Y, αX) ◦ a X,Y = a CX,Y ◦ [(C−) ◦ (Γ (ϕ) Y )] .( )Hom B Y, R C ρ X ◦ aX,Y (ξ) = ( )R C ρ X ◦ aX,Y (ξ) defa= ( )R C ρ X ◦ (Rξ) ◦ (ηY )defa (= a CCX,Y ρ X ◦ ξ )Therefore we get thatHom B(Y, R C ρ X)◦ aX,Y = a CX,Y ◦ (C ρ X ◦ − )Since K ϕ (Y ) = Υ (ϕ) (Y ) = (LY, (ϕLY ) ◦ (LηY )), for every χ ∈ Hom A (LY , X) wehave65and[C (−) ◦ Γ (ϕ) Y ] (χ) = Γ (ϕ) Y = (Cχ) ◦ (ϕLY ) ◦ (LηY ) = (Cχ) ◦ C ρ LY[ Cρ X ◦ − ] (χ) = C ρ X ◦ χso thatThus we get[C (−) ◦ Γ (ϕ) Y ] (χ) = [C ρ X ◦ − ] (χ) if and only ifχ ∈ HomC A(((LY ) , (ϕLY ) ◦ (LηY )) ,(X, C ρ X)).Equ HomA (LY ,X)( Cρ X ◦ −, C (−) ◦ Γ (ϕ) Y )= { f ∈ Hom A (LY , X) | C ρ X ◦ f = (Cf) ◦ (Γ (ϕ) Y ) }= { f ∈ Hom A (LY , X) | C ρ X ◦ f = (Cf) ◦ (ϕLY ) ◦ (LηY ) }= { (f ∈ Hom CA U (K ϕ Y ) , C U ( )) }X, C ρ X | C ρ X ◦ f = (Cf) ◦ C ρC U(K ϕY )(= HomC A Kϕ Y, ( ))X, C ρ Xso that Equ Fun( Cρ X ◦ −, C (−) ◦ Γ (ϕ) − ) = HomC A(Kϕ −, ( X, C ρ X)). □Part of the following Proposition is already in [GT], Proposition <strong>2.</strong>3.Proposition 4.47. Let (L, R) be an adjunction where L : B → A and R : A → B,let C = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR = (LR, LηR, ɛ) →C = ( C, ∆ C , ε C) be a comonad morphism. Let α = Θ (ϕ) = (Rϕ) ◦ (ηR). Thenthe functor K ϕ = Υ (ϕ) : B → C A has a right adjoint D ϕ : C A → B if and only if,

66for every ( X, C ρ X)∈ C A, there exists Equ B(αX, R C ρ X). In this case there exists afunctorial morphism d ϕ : D ϕ → R C U such thatand thus(D ϕ , d ϕ ) = Equ Fun(α C U, R C Uγ C) .[Dϕ((X, C ρ X)), dϕ(X, C ρ X)]= EquB(αX, R( Cρ X)).Proof. Assume first that, for every ( ) ( )X, C ρ X ∈ C A, there exists Equ B αX, R C ρ X .By Proposition 4.46, the isomorphism a X,Y : Hom A (LY, X) → Hom B (Y, RX) ofthe adjunction (L, R) induces an isomorphismâ X,Y : HomC A(Kϕ Y, ( X, C ρ X))→ EquSets(HomB (Y, αX) , Hom B(Y, R C ρ X)).Let ( D ϕ((X, C ρ X)), dϕ((X, C ρ X)))denote the equalizerD ϕ(X, C ρ X) dϕ RXR C ρ XαX RCXwhere d ϕ(X, C ρ X): Dϕ((X, C ρ X))→ RX is the canonical embedding. Then, byLemma <strong>2.</strong>17 we have(HomB(Y, Dϕ((X, C ρ X))), HomB(Y, dϕ((X, C ρ X))))= Equ Sets(HomB (Y, αX) , Hom B(Y, R C ρ X)).Thus, for every ( )(X, C ρ X ∈ C A and for every Y ∈ B, a X,Y induces an isomorphismâ X,Y : HomC A Kϕ Y, ( )) ( ( ))X, C ρ X → HomB Y, Dϕ X, C ρ X such that the followingdiagram is commutative((43) HomC A Kϕ Y, ( ))X, C â X,Y ( ( ))ρ X Hom B Y, Dϕ X, C ρ XHom A (LY, X)a X,YHom B (Y,d ϕ)Hom B (Y, RX)C ρ X ◦−(C−)◦(βX)Hom A (LY, CX)a CX,YHom B(Y,R C ρ X) HomB (Y, RCX)Hom B (Y,αX)i.e. (K ϕ , D ϕ ) is an adjunction.Conversely, assume now that the functor K ϕ = Υ (ϕ) : B → C A has a right adjointD ϕ : C A → B. Let ̂ɛ : K ϕ D ϕ → IdC A be the counit of the adjunction (K ϕ , D ϕ ) andletd ϕ = aC U,D ϕ( CÛɛ ) = ( R C Ûɛ ) ◦ (ηD ϕ ) : D ϕ → R C U.We will prove that(D ϕ , d ϕ ) = Equ Fun(α C U, R C Uγ C) .First of all let us compute(α C U ) ◦ d ϕ = ( α C U ) ◦ ( R C Ûɛ ) ◦ (ηD ϕ )= ( Rϕ C U ) ◦ ( ηR C U ) ◦ ( R C Ûɛ ) ◦ (ηD ϕ )η= ( Rϕ C U ) ◦ ( RLR C Ûɛ ) ◦ (RLηD ϕ ) ◦ (ηD ϕ )

67and alsoϕ= ( RC C Ûɛ ) ◦ (RϕLD ϕ ) ◦ (RLηD ϕ ) ◦ (ηD ϕ )(R C Uγ C) ◦ d ϕ = ( R C Uγ C) ◦ ( R C Ûɛ ) ◦ (ηD ϕ )bɛmorph C A=(RC C Ûɛ ) ◦ ( R C Uγ C K ϕ D ϕ)◦ (ηDϕ )so thatdefK ϕ=(RC C Ûɛ ) ◦ (RϕLD ϕ ) ◦ (RLηD ϕ ) ◦ (ηD ϕ )(α C U ) ◦ d ϕ = ( R C Uγ C) ◦ d ϕ .Now, we will prove that the following diagram is commutative( HomC A Kϕ Y, ( ))X, C ba (X, C ρX ),Yρ XC UHom A (LY, X)Hom B(Y, Dϕ(X, C ρ X))a X,Y HomB (Y, RX) .Hom B(Y,d ϕ(X, C ρ X))In fact, for every ζ ∈ HomC A(Kϕ Y, ( X, C ρ X)), we have[HomB(Y, dϕ(X, C ρ X))◦ â(X, C ρ X ),Y](ζ)defba= Hom B(Y, dϕ(X, C ρ X))[(Dϕ ζ) ◦ (̂ηY )]= ( d ϕ(X, C ρ X))◦ (Dϕ ζ) ◦ (̂ηY )defd ϕ=(R C Ûɛ ( X, C ρ X))◦(ηDϕ(X, C ρ X))◦ (Dϕ ζ) ◦ (̂ηY )η= ( R C Ûɛ ( X, C ρ X))◦ (RLDϕ ζ) ◦ (RL̂ηY ) ◦ (ηY )defK ϕ=(R C Ûɛ ( X, C ρ X))◦(R C UK ϕ D ϕ ζ ) ◦ ( R C UK ϕ̂ηY ) ◦ (ηY )bɛ= ( R C Uζ ) ◦ ( R C ÛɛK ϕ Y ) ◦ ( R C UK ) ϕ̂ηY ◦ (ηY ) (Kϕ,Dϕ)= ( R C Uζ ) ◦ (ηY )and on the other hand(aX,Y ◦ C U ) (ζ) = a X,Y( CUζ ) defa= ( R C Uζ ) ◦ (ηY )so that, for every ( X, C ρ X)∈ C A we have( ( ))Hom B −, dϕ X, C ρ X ◦ â(X, C ρ X ),− = a X,− ◦ C U.( ( ))Since a X,− and â (X, C ρ X ),− are isomorphisms, we deduce that Hom B −, dϕ X, C ρ Xis ( mono. ) Applying the commutativity of this diagram in the particular case ofX, C ρ X = Kϕ Y, we get that(d ϕ K ϕ Y ) ◦ (̂ηY ) = Hom B (Y, d ϕ K ϕ Y ) (̂ηY )= Hom B (Y, d ϕ K ϕ Y ) ( ( ))â KϕY,Y IdKϕY= [ ] ( )Hom B (Y, d ϕ K ϕ Y ) ◦ â KϕY,Y IdKϕY= [ aC UK ϕY,Y ◦ C U ] ( )Id KϕY = (aLY,Y ) (C )UId KϕY= (a LY,Y ) ( IdC UK ϕY)= aLY,Y (Id LY ) = ηY

68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂ηY ) = ηY.Now, we have to prove the universal property of the equalizer. Let Z ∈ B and letζ : Z → RX be a morphism such that (αX) ◦ ζ = ( R C ρ X)◦ ζ, i.e.(RϕX) ◦ (ηRX) ◦ ζ = ( )R C ρ X ◦ ζ.( ( )) (This means ζ ∈ Equ Sets HomB (Y, αX) , Hom B Y, R C ρ X ≃ HomC A Kϕ Y, ( ))X, C ρ Xby Proposition 4.46. Then,a −1X,Z (ζ) = (ɛX) ◦ (Lζ) ∈ ( ( ))HomC A (LZ, (ϕLZ) ◦ (LηZ)) , X, C ρ X= HomC A(Kϕ Z, ( X, C ρ X)).We want to prove that there exists ζ ′ : Z → D ϕ(X, C ρ X)such that dϕ(X, C ρ X)◦ζ ′ =ζ. By hypothesis the map( HomC A Kϕ Y, ( ))X, C ba (X, C ρX ( ( ))ρ ),Y X Hom B Y, Dϕ X, C ρ Xis bijective. Hence, given (ɛX) ◦ (Lζ) ∈ HomC A(Kϕ Z, ( X, C ρ X)),â (X, C ρ X ),Z ((ɛX) ◦ (Lζ)) = (D ϕ ɛX) ◦ (D ϕ Lζ) ◦ (̂ηZ) ∈ Hom B(Z, Dϕ(X, C ρ X)). Wewant to prove that(dϕ (X, C ρ X))◦ (Dϕ ɛX) ◦ (D ϕ Lζ) ◦ (̂ηZ) = ζ.We compute(dϕ(X, C ρ X))◦ (Dϕ ɛ) ◦ (D ϕ Lζ) ◦ (̂ηZ) dϕ = (RɛX) ◦ (RLζ) ◦ (d ϕ K ϕ Z) ◦ (̂ηZ)(44)= (RɛX) ◦ (RLζ) ◦ (ηZ) η = (RɛX) ◦ (ηRX) ◦ ζ (L,R)= ζ.Let us denote by ζ ′ = (D ϕ ɛX)◦(D ϕ Lζ)◦(̂ηZ) the morphism such that ( d ϕ(X, C ρ X))◦ζ ′ = ζ. We have to prove that ζ ′ is unique with respect to this property. Letζ ′′ : Z → D ϕ(X, C ρ X)be another morphism in B such that(dϕ(X, C ρ X))◦ ζ ′′ = ζ.Then we haveHom B(Z, dϕ(X, C ρ X))(ζ ′′ ) = ( d ϕ(X, C ρ X))◦ ζ ′′ = ζ= ( d ϕ(X, C ρ X))◦ ζ ′ = Hom B(Z, dϕ(X, C ρ X))(ζ ′ )and since Hom B(Z, dϕ(X, C ρ X))is mono we deduce thatζ ′′ = ζ ′ .Corollary 4.48. Let (L, R) be an adjunction where L : B → A and R : A → B.Let α = Θ (Id LR ) = ηR. Then the functor K = Υ (Id LR ) : B → LR A has a rightadjoint D : LR A → B if and only if, for every ( X, LR ρ X)∈ LR A, there existsEqu B(ηRX, R LR ρ X). In this case there exists a functorial morphism d : D → R LR Usuch that(D, d) = Equ Fun(ηR LR U, R LR Uγ LR) .□

69and thus[D((X, LR ρ X)), d(X, LR ρ X)]= EquB(ηRX, R( LRρ X)).Proof. We can apply Proposition 4.47 with ”ϕ” = Id LR .Remark 4.49 ([GT]). In the setting of Proposition 4.47, for every Y ∈ B, we notethat the unit of the adjunction (K ϕ , D ϕ ) is given by( )̂ηY = â KϕY,Y IdKϕY : Y → Dϕ K ϕ (Y ) .We will consider the diagram (43) in the particular case of ( X, C ρ X)= Kϕ Y. Notethat since K ϕ Y = (LY, (ϕLY ) ◦ (LηY )) = (LY, βY ) we havei.e.(D ϕ K ϕ (Y ) , d ϕ K ϕ (Y )) = (D ϕ ((LY, βY )) , d ϕ K ϕ (Y )) = Equ B (αLY, RβY )(45) (D ϕ K ϕ (Y ) , d ϕ K ϕ (Y )) = Equ B (αLY, RβY )where β = Γ (ϕ) = (ϕL) ◦ (Lη). We compute(d ϕ K ϕ Y ) ◦ (̂ηY ) = Hom B (Y, d ϕ K ϕ Y ) (̂ηY ) = Hom B (Y, d ϕ K ϕ Y ) ( ( ))â KϕY,Y IdKϕY= [ ] ( ) (43)Hom B (Y, d ϕ K ϕ Y ) ◦ â KϕY,Y IdKϕY = a C KϕY,Y U ( )Id KϕY( )= a KϕY,Y IdC UK ϕY = aKϕY,Y (Id RY ) = ηYso that(46) (d ϕ K ϕ Y ) ◦ (̂ηY ) = ηY.On the other hand, for every ( X, C ρ X)∈ C A, the counit of the adjunction (K ϕ , D ϕ )is given bŷɛ ( X, C ρ X)= â−1X,D ϕ(X, C ρ X )(IdDϕ(X, C ρ X )): Kϕ D ϕ((X, C ρ X))→(X, C ρ X).Then we have thatâ X,Dϕ(X, C ρ X )(̂ɛ(X, C ρ X))= IdDϕ(X, C ρ X ).By commutativity of the diagram (43), we deduce that( ) ( ) ( )))d ϕ X, C ρ X = dϕ X, C ρ X ◦(âX,Dϕ(X, C ρ X )(̂ɛ X, C ρ XThus we obtain that(47)= a X,Dϕ(X, C ρ X )( CÛɛ ( X, C ρ X)).C Ûɛ ( X, C ρ X)= a−1X,D ϕ(X, C ρ X )(dϕ(X, C ρ X))= (ɛX) ◦(Ldϕ(X, C ρ X)).Observe that, for every X ∈ A, we have that C F (X) = ( CX, ∆ C X ) ∈ C A. Moreover(Dϕ C F (X) , d ϕ C F (X) ) = ( D ϕ(CX, ∆ C X ) , d ϕ(CX, ∆ C X )) =□so that we get(48)= Equ B(αCX, R∆ C X ) (4.15)= (RX, αX)(Dϕ C F , d ϕ C F ) = (R, α) .

70In particular(49) d ϕ(CX, ∆ C X ) = αX.Corollary 4.50. In the setting of Proposition 4.47, assume that, for every ( X, C ρ X)∈C A, there exists Equ B(αX, R C ρ X). Then for every X ∈ A we haveC Ûɛ ( CX, ∆ C X ) = ϕXand henceC Ûɛ C F = ϕwhere ̂ɛ is the counit of the adjunction (K ϕ , D ϕ ).Proof. Let us calculateC Ûɛ ( CX, ∆ C X ) (47)= (ɛCX) ◦ ( Ld ϕ(CX, ∆ C X ))(48)= (ɛCX) ◦ (LαX) = Ξ (α) (X) = ϕX.Corollary 4.5<strong>1.</strong> In the setting of Proposition 4.47, assume that, for every ( X, C ρ X)∈C A, there exists Equ B(αX, R C ρ X). Then, the functor Dϕ is full and faithful if andonly if ̂ɛ is a functorial isomorphism.Proof. By Proposition 4.47, (K ϕ , D ϕ ) is an adjunction with counit ̂ɛ : K ϕ D ϕ → C A.Then we can apply Proposition <strong>2.</strong>3<strong>2.</strong>□Lemma 4.52 ([GT, Lemma <strong>2.</strong>5]). In the setting of Proposition 4.47, assume that, forevery ( X, C ρ X)∈ C A, there exists Equ B(αX, R C ρ X). Then, for every(X, C ρ X)∈C A the following diagram□LD ϕ(X, C ρ X) C Ubɛ(X, C ρ X)C U ( X, C ρ X)Ld ϕ(X, C ρ X) LR C U ( )X, C ϕ C U(X, C ρ X)ρ XC U C γ(X, C ρ X)C C U ( )X, C ρ XLα C U(X, C ρ X) LR C U C γ(X, C ρ X) ∆ C C U(X, C ρ X) C C U C γ(X, C ρ X)LRC C U ( )X, C ϕC C U(X, C ρ X)ρ X CC C U ( )X, C ρ Xserially commutes. Therefore we get( CUγ C) ◦ (C Ûɛ ) = ( ϕ C U ) ◦ (Ld ϕ ) andProof. Let us compute(∆C C U ) ◦ ( ϕ C U ) = ( ϕC C U ) ◦ ( Lα C U ) .C ρ X ◦ C Ûɛ ( X, C ρ X) (47)= C ρ X ◦ (ɛX) ◦ ( Ld ϕ(X, C ρ X))ϕmorphcomonads=C ρ X ◦ ( ε C X ) ◦ (ϕX) ◦ ( Ld ϕ(X, C ρ X))ε C = ( ε C CX ) ◦ ( C C ρ X)◦ (ϕX) ◦(Ldϕ(X, C ρ X))ϕ= ( ε C CX ) ◦ (ϕCX) ◦ ( ) ( ( ))LR C ρ X ◦ Ldϕ X, C ρ X

defd ϕ=(ε C CX ) ◦ (ϕCX) ◦ (LαX) ◦ ( Ld ϕ(X, C ρ X))defα= ( ε C CX ) ◦ (ϕCX) ◦ (LRϕX) ◦ (LηRX) ◦ ( Ld ϕ(X, C ρ X))ϕ= ( ε C CX ) ◦ (ϕϕX) ◦ (LηRX) ◦ ( ( ))Ld ϕ X, C ρ X(ε C CX ) ◦ ( ∆ C X ) ◦ (ϕX) ◦ ( ( ))Ld ϕ X, C ρ Xϕmorphcomonads=so that we deduce thatCcomonad= (ϕX) ◦ ( Ld ϕ(X, C ρ X))C ρ X ◦ (C Ûɛ ( X, C ρ X))= (ϕX) ◦(Ldϕ(X, C ρ X))71and thus( CUγ C) ◦ (C Ûɛ ) = ( ϕ C U ) ◦ Ld ϕ .Let us calculate(∆C C U ) ◦ ( ϕ C U ) ϕcomonadmorph (= ϕϕ C U ) ◦ ( LηR C U )= ( ϕC C U ) ◦ ( LRϕ C U ) ◦ ( LηR C U ) defα= ( ϕC C U ) ◦ ( Lα C U ) .Theorem 4.53 ([GT] Theorem <strong>2.</strong>6). Let (L, R) be an adjunction where L : B → Aand R : A → B, let C = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR =(LR, LηR, ɛ) → C = ( C, ∆ C , ε C) be a comonad morphism. Let α = Θ (ϕ) = (Rϕ) ◦(ηR) and assume that, for every ( ) ( )X, C ρ X ∈ C A, there exists Equ B αX, R C ρ X .Then we can consider the functor K ϕ = Υ (ϕ) : B → C A and its right adjointD ϕ : C A → B. D ϕ is full and faithful if and only if1) L preserves the equalizer(D ϕ , d ϕ ) = Equ Fun(α C U, R C Uγ C) .2) ϕ : LR → C is a comonad isomorphism.Proof. Recall that, by Corollary 4.50,(50)C Ûɛ C F = ϕ.By Corollary 4.51, D ϕ is full and faithful if and only if ̂ɛ is a functorial isomorphism.Let us assume that ̂ɛ is a functorial isomorphism, hence ϕ is an isomorphism too.Recall that, by Lemma4.52, we have((51)CUγ C) ◦ (C Ûɛ ) = ( ϕ C U ) ◦ (Ld ϕ )so that(52)Let us consider the diagramC Uγ C = ( ϕ C U ) ◦ (Ld ϕ ) ◦ (C Ûɛ −1) .d ϕLD ϕLR C U LRC Uγ C LRC C ULα C U□

72We have to prove that (LD ϕ , Ld ϕ ) = Equ Fun(Lα C U, LR C Uγ C) . Since L is a functor,we clearly have ( Lα C U ) ◦(Ld ϕ ) = ( LR C Uγ C) ◦(Ld ϕ ). Let Z : Z → C A be a functorand let ξ : Z → LR C U be a functorial morphism such that(Lα C U ) ◦ ξ = ( LR C Uγ C) ◦ ξ.Recall that ( ε C C U ) ◦ (C Uγ C) = C U and (C F ε C) ◦ ( γ C C F ) = C F . We compute(ϕ C U ) ◦ ξ = Id C C U ◦ ϕ C U ◦ ξ Ccomonad (= ε C C C U ) ◦ ( ∆ CC U ) ◦ ( ϕ C U ) ◦ ξ =ϕcomonadmorp (= ε C C C U ) ◦ ( ϕϕ C U ) ◦ ( LηR C U ) ◦ ξ == ( ε C C C U ) ◦ ( ϕC C U ) ◦ ( LRϕ C U ) ◦ ( LηR C U ) ◦ ξ == ( ε C C C U ) ◦ ( ϕC C U ) ◦ ( Lα C U ) ◦ ξ = ( ε C C C U ) ◦ ( ϕC C U ) ◦ ( LR C Uγ C) ◦ ξ =ϕ= ( ε C C C U ) ◦ ( C C Uγ C) ◦ ( ϕ C U ) ◦ ξ εC = (C Uγ C) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ =Since ϕ is iso, we get(52)= ( ϕ C U ) ◦ (Ld ϕ ) ◦ (C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ.ξ = (Ld ϕ ) ◦ [(C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ ] .Let now w : Z → LD ϕ be a functorial morphism such thatWe computeξ = (Ld ϕ ) ◦ w.( CUγ C) ◦ (C Ûɛ ) ◦ [(C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ ] =(51)= ( ϕ C U ) ◦ (Ld ϕ ) ◦ [(C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ ] == ( ϕ C U ) ◦ ξ = ( ϕ C U ) ◦ (Ld ϕ ) ◦ w (51)= (C Uγ C) ◦ (C Ûɛ ) ◦ wand since C Uγ C is a monomorphism (since it is an equalizer) and ̂ɛ is an isomorphismwe obtain that( CÛɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ = w.Conversely, assume that 1) and 2) hold. Then ϕ is a functorial isomorphism. Considerthe diagramLD ϕ(X, C ρ X) C Ubɛ(X, C ρ X)C U ( X, C ρ X)Ld ϕ(X, C ρ X) LR C U ( )X, C ϕ C U(X, C ρ X)ρ XC U C γ(X, C ρ X)C C U ( )X, C ρ XLα C U(X, C ρ X) LR C U C γ(X, C ρ X) ∆ C C U(X, C ρ X) C C U C γ(X, C ρ X)LRC C U ( )X, C ϕC C U(X, C ρ X)ρ X CC C U ( )X, C ρ Xof Lemma 4.52 where the last row is always an equalizer (see Proposition 4.13) andthe first row is also an equalizer by the assumption 1). Then we can apply Lemma

<strong>2.</strong>15 and hence we get that C Ûɛ is a functorial isomorphism. Since,by Proposition4.17, C U reflects isomorphism we deduce that ̂ɛ is a functorial isomorphism. □Corollary 4.54. Let (L, R) be an adjunction where L : B → A and R : A → B.Let α = Θ (Id LR ) = ηR and assume that, for every ( X, LR ρ X)∈ LR A, there existsEqu B(ηRX, R LR ρ X). Then we can consider the functor K = Υ (IdLR ) : B → LR Aand its right adjoint D : LR A → B. D is full and faithful if and only if L preservesthe equalizer(D, d) = Equ Fun(ηR LR U, R LR Uγ LR) .Proof. We can apply Theorem 4.53 with ”ϕ” = Id LR .Theorem 4.55 ([GT, Theorem <strong>2.</strong>7]). Let (L, R) be an adjunction where L : B → Aand R : A → B, let C = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR =(LR, LηR, ɛ) → C = ( C, ∆ C , ε C) be a comonad morphism. Let α = Θ (ϕ) = (Rϕ) ◦(ηR) and assume that, for every ( ) ( )X, C ρ X ∈ C A, there exists Equ B αX, R C ρ X .Then we can consider the functor K ϕ = Υ (ϕ) : B → C A and its right adjointD ϕ : C A → B. The functor K ϕ is an equivalence of categories if and only if1) L preserves the equalizer((D ϕ , d ϕ ) = Equ Fun α C U, R C Uγ C)2) L reflects isomorphisms and3) ϕ : LR → C is a comonad isomorphism.Proof. If K ϕ is an equivalence then, by Lemma <strong>2.</strong>33, D ϕ is an equivalence of categoriesso that, by Theorem 4.53, 1) and 3) hold. By Proposition 4.17, the functorC U reflects isomorphisms. Since L = C UK ϕ we get that 2) holds.Conversely assume that 1), 2) and 3) hold. By Theorem 4.53 , D ϕ is full andfaithful and hence by Corollary 4.51 ̂ɛ is a functorial isomorphism. Let us provethat ̂η is an isomorphism as well. Since L reflects isomorphisms, it is enough toprove that L̂η is an isomorphism. As observed in Remark 4.49, by (44), ̂ηY is theunique morphism such thatHence we getso that(d ϕ K ϕ Y ) ◦ (̂ηY ) = ηY.(Ld ϕ K ϕ Y ) ◦ (L̂ηY ) = LηY(ɛLY ) ◦ (Ld ϕ K ϕ Y ) ◦ (L̂ηY ) = (ɛLY ) ◦ (LηY ) = LY.We now want to prove that (ɛLY ) ◦ (Ld ϕ K ϕ Y ) is also a right inverse for L̂ηY . Wecompute(Ld ϕ K ϕ Y ) ◦ (L̂ηY ) ◦ (ɛLY ) ◦ (Ld ϕ K ϕ Y ) (44)= (LηY ) ◦ (ɛLY ) ◦ (Ld ϕ K ϕ Y )Since L preserves the equalizer(L,R)adj= (Ld ϕ K ϕ Y ) .(D ϕ , d ϕ ) = Equ Fun(α C U, R C Uγ C)73□

74we have that Ld ϕ K ϕ Y is mono and hence we obtainso that L̂η is a functorial isomorphism.(L̂ηY ) ◦ (ɛLY ) ◦ (Ld ϕ K ϕ Y ) = LD ϕ K ϕ YDefinition 4.56. Let ( L, C ρ L)be a left comodule functor for a comonad C =(C, ∆ C , ε C) such that L has a right adjoint R. Then we can consider a canonicalcomonad morphismcan := (Cɛ) ◦ (C ρ L R ) : LR → Cwhere ɛ denotes the counit of the adjunction (L, R) . A left C-Galois functor is aleft C-comodule functor ( L, C ρ L)with a right adjoint R such that can is a comonadisomorphism.Corollary 4.57. Let ( L, C ρ L)be a left C-Galois comodule functor such that Lpreserves equalizers, L reflects isomorphisms and let C = ( C, ∆ C , ε C) be a comonadon A. Assume that, for every ( X, C ρ X)∈ C A, there exists Equ B(αX, R C ρ X)whereα = (Rcan) ◦ (ηR) where R is the right adjoint of L and η is the unit of theadjunction. Then we can consider the functor K can : B → C A. Then the functorK can is an equivalence of categories.Proof. We can apply Theorem 4.55 to the case ϕ = can.Theorem 4.58 (Beck’s Theorem). Let (L, R) be an adjunction where L : B → A andR : A → B. Let α = Θ (Id LR ) = ηR and assume that, for every ( X, LR ρ X)∈ LR A,there exists Equ B(ηRX, R LR ρ X). Then we can consider the functor K = Υ (IdLR ) :B → LR A and its right adjoint D : LR A → B. The functor K is an equivalence ofcategories if and only if1) L preserves the equalizer2) L reflects isomorphisms.(D, d) = Equ Fun(ηR LR U, R LR Uγ LR) .Proof. Apply Theorem 4.55 taking ϕ = Id LR and thus α = Θ (Id LR ) = ηR.Definition 4.59. Let C = ( C, ∆ C , ε C) be a comonad on the category A and letL : B → A. The functor L is called ϕ-comonadic if it has a right adjoint R : A → Bfor which there exists ϕ : LR → C a comonad morphism such that the functorK ϕ = Υ (ϕ) : B → C A is an equivalence of categories with D ϕ : C A → B which isright adjoint.Definition 4.60. Let L : B → A be a functor. The functor L is called comonadicif it has a right adjoint R : A → B such that the functor K = Υ (Id LR ) : B → LR Ais an equivalence of categories with right adjoint D : LR A → B.Lemma 4.6<strong>1.</strong> Let L : B → A be a ϕ-comonadic functor and let(53) X ′ d 0 Xd 1□□□

e a L-contractible equalizer pair in B. Then (53) has an equalizer d : X ′′ → X ′ inB andLX ′′ Ld LX ′ Ld 0 LXLd 1is an equalizer in A.Proof. Since L is a ϕ-comonadic functor we know that K ϕ = Υ (ϕ) : B → C A is anequivalence of categories. Then, instead of consideringin the category B, we can considerX ′ d 0 Xd 1K ϕ X ′ K ϕd 0 K ϕ XK ϕd 1in C A which is a C U-contractible equalizer pair. Let us denote by ( Z ′ , C ρ Z ′):= Kϕ X ′and ( )Z, C ρ Z := Kϕ X so that we can rewrite the C U-contractible equalizer pair asfollows(Z ′ , C K ϕd 0ρ Z ′) ( )Z, C ρ ZK ϕd 1We want to prove that this pair has an equalizer in C A. Since the pair (K ϕ d 0 , K ϕ d 1 )is a C U-contractible equalizer in C A, we have thatZ ′′ d Z ′sis a contractible equalizer and thus, by Proposition <strong>2.</strong>19, an equalizer in A. Let usconsider the following diagramC ρ Z ′′C C ρ Z ′′Ld 0tLd 1Z ′′ d Z ′C ρ Z ′CZ ′′ Cd CZ ′ ∆ C Z ′∆ C Z ′′C C ρ Z ′ ZLd 0Ld 1CLd 0CLd 1CCZ ′′ CCd CCZ ′ CCLd 0 CCLd 1 Z CZC C ρ ZC ρ Z CCZBy Proposition <strong>2.</strong>20, all the rows are contractible equalizers. Since Ld 0 = C UK ϕ d 0and Ld 1 = C UK ϕ d 1 where K ϕ d 0 and K ϕ d 1 are morphisms in C A, we have that theupper right square serially commutes. Moreover, since we also have that ∆ C is afunctorial morphism, the lower right square serially commutes. We also have thatC ρ Z ′◦d is a fork for (CLd 0 , CLd 1 ) and, since (CZ ′′ , CZ ′ , CZ, Cd, CLd 0 , CLd 1 , Cs, Ct)is a contractible equalizer, in particular (CZ ′′ , Cd) = Equ A (CLd 0 , CLd 1 ); by theuniversal property of the equalizer, there exists a unique morphism C ρ Z ′′ : Z ′′ →CZ ′′ such that(54)C ρ Z ′ ◦ d = (Cd) ◦ C ρ Z ′′.∆ C Z75

and since d is mono we get that(ε C Z ′′) ◦ C ρ Z ′′ = Z ′′76Let us prove that ( Z ′′ , C ρ Z ′′)∈ C A and thus formula (54) will say that d is amorphism in C A. Since ∆ C is a functorial morphism and by definition of C ρ Z ′′, thelower left square serially commutes. We have(CCd) ◦ ( C C ρ Z ′′)◦ C ρ Z ′′(54)= ( C C ρ Z ′)◦ (Cd) ◦ C ρ Z ′′(54)= ( )C C ρ Z ′ ◦ C ρ Z ′ ◦ d C ρ Z ′coass= ( ∆ C Z ′) ◦ C ρ Z ′ ◦ d(54)= ( ∆ C Z ′) ◦ (Cd) ◦ C ∆ρ CZ ′′ = (CCd) ◦ ( ∆ C Z ′′) ◦ C ρ Z ′′and since CCd is a monomorphism we get( )C C ρ Z ′′ ◦ C ρ Z ′′ = ( ∆ C Z ′′) ◦ C ρ Z ′′that is that C ρ Z ′′is coassociative. Moreover we haved ◦ ( ε C Z ′′) ◦ C ρ Z ′′ε C = ( ε C Z ′) ◦ (Cd) ◦ C ρ Z ′′(54)= ( ε C Z ′) ◦ C ρ Z ′ ◦ d C ρ Z ′counit= dso that C ρ Z ′′ is also counital. Therefore ( Z ′′ , C ρ Z ′′)∈ C A and d is a morphism inC A. Now we want to prove that it is an equalizer in C A. Let ( E, C ρ E)∈ C A andf : ( E, C ρ E)→(Z ′ , C ρ Z ′)be a morphism in C A such that (K ϕ d 0 ) ◦ f = (K ϕ d 1 ) ◦ f.Then, by regarding f as a morphism in A we also have that(Ld 0 ) ◦ f = (Ld 1 ) ◦ f.Since (Z ′′ , d) = Equ A (Ld 0 , Ld 1 ) , there exists a unique morphism h : E → Z ′′ suchthatd ◦ h = f.Now we want to prove that h is a morphism in C A. In fact, let us consider thefollowing diagramEC ρ EhC ρ Z ′′Z ′′ d Z ′C ρ Z ′CE Ch CZ ′′ Cd CZ ′ .Since d ∈ C A, the right square commutes. Since f ∈ C A we haveso that we have(Cd) ◦ (Ch) ◦ C ρ E = (Cf) ◦ C ρ E = C ρ Z ′ ◦ f = C ρ Z ′ ◦ d ◦ h(Cd) ◦ C ρ Z ′′ ◦ h (54)= C ρ Z ′ ◦ d ◦ h = (Cd) ◦ (Ch) ◦ C ρ Eand since Cd is a monomorphism, we deduce thatC ρ Z ′′ ◦ h = (Ch) ◦ C ρ E

i.e. h ∈ C A. Therefore (Z ′′ , d) = EquC A (K ϕ d 0 , K ϕ d 1 ). Now, since K ϕ : B → C A isan equivalence of categories, there exist X ′′ , e ∈ B such thatK ϕ X ′′ = ( Z ′′ , C ρ Z ′′)and Kϕ e = dand thus (X ′′ , e) = Equ B (d 0 , d 1 ). Moreover, sinceZ ′′ d Z ′sis a contractible coequalizer and (Z ′′ , d) = (C UK ϕ X ′′ , C UK ϕ e ) (, we deduce thatCUK ϕ X ′′ , C UK ϕ e ) (is a contractible coequalizer of (Ld 0 , Ld 1 ). Then (LX ′′ , Le) =CUK ϕ X ′′ , C UK ϕ e ) is a contractible coequalizer of (Ld 0 , Ld 1 ) so that (LX ′′ , Le) =Equ A (Ld 0 , Ld 1 ).□The following is a slightly improved version of Theorem 3.14 p. 101 [BW] for thedual case.Theorem 4.62 (Generalized Beck’s Precise Cotripleability Theorem). Let L : B →A be a functor, let C = ( C, ∆ C , ε C) be a comonad on a category A. Then L isϕ-comonadic if and only if1) L has a right adjoint R : A → B,2) ϕ : LR → C is a comonads isomorphism where LR = (LR, LηR, ɛ) with ηand ɛ unit and counit of (L, R) ,3) for every ( X, C ρ X)∈ C A, there exist Equ B(αX, R C ρ X), where α = (Rϕ) ◦(ηR), and L preserves the equalizerLd 0tLd 1 ZEqu Fun(α C U, R C Uγ C) ,4) L reflects isomorphisms.In this case in B there exist equalizers of reflexive L-contractible equalizer pairsand L preserves them.Proof. Assume first that L is ϕ-comonadic. Then by definition L has a right adjointR : A → B and a comonad morphism ϕ : LR → C such that the functor K ϕ =Υ (ϕ) : B → C A is an equivalence of categories. Let K ϕ ′ be an inverse of K ϕ . Thenin particular K ϕ ′ : C A → B is a right adjoint of K ϕ so that by Proposition 4.47 forevery ( ) ( )X, C ρ X ∈ C A, there exists Equ B αX, R C ρ X where α = Θ (ϕ) = (Rϕ) ◦(ηR) and thus ( (K ϕ, ′ k ϕ) ′ = EquFun α C U, R C Uγ C) . Then we can apply Theorem 4.55to get that L preserves the equalizer ( (K ϕ, ′ k ϕ) ′ = EquFun α C U, R C Uγ C) , L reflectsisomorphisms and ϕ : LR → C is a comonads isomorphism.Conversely, by assumption 1) L has a right adjoint R :( A → B so) that (L, R)(is an adjunction)and by assumption 2) there exist Equ B αX, R C ρ X , for everyX, C ρ X ∈ C A so that we can apply one direction of Proposition 4.47. Thus thefunctor K ϕ = Υ (ϕ) : B → C A has a right adjoint D ϕ : C A → B. Now, by applyingTheorem 4.55 in the converse direction, we deduce that K ϕ = Υ (ϕ) : B → C A is anequivalence of categories, i.e. L is ϕ-comonadic.In the case L is ϕ-comonadic, by Lemma 4.61, in B there exist equalizers ofreflexive L-contractible equalizer pairs and L preserves them.□77

78Corollary 4.63 (Beck’s Precise Cotripleability Theorem). Let L : B → A be afunctor. Then L is comonadic if and only if1) L has a right adjoint R : A → B,2) for every ( X, LR ρ X)∈ LR A, there exist Equ B(ηRX, R LR ρ X)and L preservesthe equalizerEqu Fun((ηR) ◦( LRU ) , R LR Uγ LR)3) L reflects isomorphisms.In this case in B there exist equalizers of reflexive L-contractible equalizer pairsand L preserves them.Proof. Apply Theorem 4.62 to the case ϕ = Id LR .Lemma 4.64. Let (L, R) be an adjunction, where L : B → A and R : A → B, withunit η and counit ɛ. Then for every Y ∈ B,(LY, LRLY, LRLRLY, LηY, LηRLY, LRLηY, ɛLY, ɛLRLY ) is a contractible equalizerand in particular, for every Y ∈ BProof. Consider the following diagram(LY, LηY ) = Equ A (LηRLY, LRLηY ) .□and let us computeLY LηY LRLYɛLYLηRLYɛLRLYLRLηY LRLRLY(ɛLRLY ) ◦ (LηRLY ) = Id LRLY(ɛLY ) ◦ (LηY ) = Id LY(ɛLRLY ) ◦ (LRLηY ) = (LηY ) ◦ (ɛLY ) = Id LRLY(LηRLY ) ◦ (LηY ) = (LRLηY ) ◦ (LηY ) .Thus (LY, LRLY, LRLRLY, LηY, LηRLY, LRLηY, ɛLY, ɛLRLY ) is a contractibleequalizer for every Y ∈ B and by Proposition <strong>2.</strong>19 we get that (LY, LηY ) =Equ A (LηRLY, LRLηY ) .□Lemma 4.65. Let (L, R) be an adjunction where L : B → A and R : A → B, letC = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR = (LR, LηR, ɛ) →C = ( C, ∆ C , ε C) be a comonad morphism. Let K ϕ = Υ (ϕ) = (L, (ϕL) ◦ (Lη)) andC UK ϕ (f) = L (f) for every morphism f in B. For every Y ∈ B we have(55) (K ϕ Y, K ϕ ηY ) = EquC A (K ϕ ηRLY, K ϕ RLηY ) .Proof. By Lemma 4.64 we have that (LY, LηY ) = Equ A (LηRLY, LRLηY ). Leth : Z → K ϕ RLY = (LRLY, (ϕLRLY ) ◦ (LηRLY )) be a morphism in C A such thatThen(K ϕ RLηY ) ◦ h = (K ϕ ηRLY ) ◦ h.(56) (LRLηY ) ◦ (C Uh ) = (LηRLY ) ◦ (C Uh )

and hence there exists a ζ : C UZ → LY = C UK ϕ Y such that((57)CUh ) = (LηY ) ◦ ζ = (C UK ϕ ηY ) ◦ ζ.Let us prove that ζ gives rise to a morphism in C A. Since h is a morphism in C Awe have that(58) (ϕLRLY ) ◦ (LηRLY ) ◦ (C Uh ) = ( C C Uh ) ◦ (C Uγ C Z ) .Let us computeso that(CLηY ) ◦ (Cζ) ◦ (C Uγ C Z ) (57)= ( C C Uh ) ◦ (C Uγ C Z )(58)= (ϕLRLY ) ◦ (LηRLY ) ◦ (C Uh )(56)= (ϕLRLY ) ◦ (LRLηY ) ◦ (C Uh )(57)= (ϕLRLY ) ◦ (LRLηY ) ◦ (LηY ) ◦ ζϕ= (CLηY ) ◦ (ϕLY ) ◦ (LηY ) ◦ ζ(CLηY ) ◦ (Cζ) ◦ (C Uγ C Z ) = (CLηY ) ◦ (ϕLY ) ◦ (LηY ) ◦ ζ.Since (CɛLY ) ◦ (CLηY ) = CLRLY , CLηY is mono and hence we get(Cζ) ◦ (C Uγ C Z ) = (ϕLY ) ◦ (LηY ) ◦ ζi.e. ζ : C UZ → LY = C UK ϕ Y is a morphism of C-comodules.Proposition 4.66. Let (L, R) be an adjunction where L : B → A and R : A → B,let C = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR = (LR, LηR, ɛ) →C = ( C, ∆ C , ε C) be a comonad morphism. Let K ϕ = Υ (ϕ) = (L, (ϕL) ◦ (Lη)) andC UK ϕ (f) = L (f) for every morphism f in B. If ϕX is a monomorphism forevery X ∈ A, the assignment K Y,RLY ′ : Hom B (Y, RLY ′ ) → HomC A (K ϕ Y, K ϕ RLY ′ )defined by settingK Y,RLY ′ (f) = K ϕ (f)is an isomorphism whose inverse is defined byK −1Y,RLY ′ (h) = (RɛLY ′ ) ◦ ( R C Uh ) ◦ (ηY ) .Proof. Let f ∈ Hom B (Y, RLY ′ ). We compute( )˜K −1Y,RLY˜KY,RLY ′ ′ (f) = (RɛLY ′ ) ◦ ( R C UK ϕ f ) ◦ (ηY ) = (RɛLY ′ ) ◦ (RLf) ◦ (ηY )η= (RɛLY ′ ) ◦ (ηRLY ′ ) ◦ f = f.Let h ∈ HomC A (K ϕ Y, K ϕ RLY ′ ). This means that(ϕLRLY ′ ) ◦ (LηRLY ′ ) ◦ (C Uh ) = ( C C Uh ) ◦ (ϕLY ) ◦ (LηY )ϕ= (ϕLRLY ′ ) ◦ ( LR C Uh ) ◦ (LηY ) .Since ϕX is a monomorphism for every X ∈ A, we deduce that(59) (LηRLY ′ ) ◦ (C Uh ) = ( LR C Uh ) ◦ (LηY ) .79□

80We compute(LRɛLY ′ ) ◦ ( LR C Uh ) ◦ (LηY ) (59)= (LRɛLY ′ ) ◦ (LηRLY ′ ) ◦ (C Uh )= C Uhand since L = C UK ϕ and C U reflects, we getThen we deduce thatK Y,RLY ′(K ϕ RɛLY ′ ) ◦ ( K ϕ R C Uh ) ◦ (K ϕ ηY ) = h.(K−1Y,RLY(h) ) (= K ′ Y,RLY ′ (RɛLY ′ ) ◦ ( R C Uh ) ◦ (ηY ) )= (K ϕ RɛLY ′ ) ◦ ( K ϕ R C Uh ) ◦ (K ϕ ηY ) = h.Proposition 4.67. Let (L, R) be an adjunction where L : B → A and R : A → B,let C = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR = (LR, LηR, ɛ) →C = ( C, ∆ C , ε C) be a comonad morphism. Let K ϕ = Υ (ϕ) = (L, (ϕL) ◦ (Lη)) andC UK ϕ (f) = L (f) for every morphism f in B. If K ϕ is full and faithful then, forevery Y ∈ B, we haveProof. By Lemma 4.65 we have(Y, ηY ) = Equ B (RLηY, ηRLY ) .(K ϕ Y, K ϕ ηY ) = EquC A (K ϕ RLηY, K ϕ ηRLY ) .Then we can apply Lemma <strong>2.</strong>16 and deduce that (Y, ηY ) = Equ B (RLηY, ηRLY ) .□Theorem 4.68 (Generalized Beck’s Theorem for comonads). Let (L, R) be an adjunctionwhere L : B → A and R : A → B, let C = ( C, ∆ C , ε C) be a comonadon a category A and let ϕ : LR = (LR, LηR, ɛ) → C = ( C, ∆ C , ε C) be a comonadsmorphism such that ϕX is a monomorphism for every X ∈ A. Let K ϕ =Υ (ϕ) = (L, (ϕL) ◦ (Lη)) and C UK ϕ (f) = L (f) for every morphism f in B. ThenK ϕ : B → C A is full and faithful if and only if for every Y ∈ B we have that(Y, ηY ) = Equ B (ηRLY, RLηY ) .Proof. If K ϕ is full and faithful then we can apply Proposition 4.67 to get that forevery Y ∈ B we have that (Y, ηY ) = Equ B (RLηY, ηRLY ) .Conversely assume that for every Y ∈ B we have that (Y, ηY ) = Equ B (ηRLY, RLηY ) .We want to prove that K Y,Y ′ is bijective for every Y, Y ′ ∈ B. Let us consider the□

81following diagram0Hom B (Y, Y ′ )Hom B (Y,ηY ′ )Hom B (Y, RLY ′ )K Y,Y ′K Y,RLY ′0HomC A (K ϕ Y, K ϕ Y ′ )Hom CA (K ϕY,K ϕηY ′ )HomC A (K ϕ Y, K ϕ RLY ′ )Hom B (Y,RLηY ′ )Hom B (Y,ηRLY ′ )Hom B (Y, RLRLY ′ )Hom CA (K ϕY,K ϕRLηY ′ )K Y,RLRLY ′Hom CA (K ϕY,K ϕηRLY ′ ) HomC A (K ϕ Y, K ϕ RLRLY ′ )Since (Y ′ , ηY ′ ) = Equ B (ηRLY ′ , RLηY ′ ) the left column of the diagram is exact byLemma <strong>2.</strong>17. By Lemma 4.65 we have (K ϕ Y, K ϕ ηY ) = EquC A (K ϕ ηRLY, K ϕ RLηY )so that also the right column is also exact by Lemma <strong>2.</strong>17. Let f ∈ Hom B (Y, Y ′ )and g ∈ Hom B (Y, RLY ′ ). SinceK ϕ (ηY ′ ◦ f) = (K ϕ ηY ′ ) ◦ (K ϕ f) ,K ϕ (ηRLY ′ ◦ g) = (K ϕ ηRLY ′ ) ◦ (K ϕ g) and K ϕ (RLηY ′ ◦ g) = (K ϕ RLηY ′ ) ◦ (K ϕ g)the diagram is serially commutative. By Proposition 4.66, K Y,RLY ′ and K Y,RLRLY ′are isomorphisms and so is K Y,Y ′ by Lemma <strong>2.</strong>15. □Corollary 4.69 (Beck’s Theorem for comonads). Let (L, R) be an adjunctionwhere L : B → A and R : A → B. Then K = Υ (Id LR ) : B → LR A is full andfaithful if and only if for every Y ∈ B we have that (Y, ηY ) = Equ B (ηRLY, RLηY ) .5.<strong>1.</strong> Distributive laws.5. Liftings and distributive lawsDefinition 5.<strong>1.</strong> Let A = (A, m, u) be a monad and C = (C, ∆, ε) be a comonadon the same category A. A functorial morphism Φ : AC → CA is called a mixeddistributive law (or in some papers an entwining) if• Φ ◦ (mC) = (Cm) ◦ (ΦA) ◦ (AΦ) and Φ ◦ (uC) = Cu• (∆A) ◦ Φ = (CΦ) ◦ (ΦC) ◦ (A∆) and (εA) ◦ Φ = Aε.Definition 5.<strong>2.</strong> Let A = (A, m, u) be a monad and C = (C, ∆, ε) be a comonadon the same category A. A functorial morphism Ψ : CA → AC is called an oppositemixed distributive law if• Ψ ◦ (Cm) = (mC) ◦ (AΨ) ◦ (ΨA) and Ψ ◦ (Cu) = uC• (A∆) ◦ Ψ = (ΨC) ◦ (CΨ) ◦ (∆A) and (Aε) ◦ Ψ = εA.Lemma 5.3. Let A = (A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be a comonadon the category A. Let Q : B → A be a functor such that ( Q, A µ Q)is left A-module functor. Assume that Φ : AC → CA is a mixed distributive law. Then(CQ, A µ CQ)=(CQ,(C A µ Q)◦ (ΦQ))is a left A-module functor.

82Proof. First of all we prove that A µ CQ = ( C A µ Q)◦ (ΦQ) is associative. In fact wehaveA µ CQ ◦ ( )A A defµ A µ CQ( ) ( )CQ = C A µ Q ◦ (ΦQ) ◦ AC A µ Q ◦ (AΦQ)Φ= ( C A µ Q)◦(CA A µ Q)◦ (ΦAQ) ◦ (AΦQ)A µ Q ass= ( C A µ Q)◦ (CmA Q) ◦ (ΦAQ) ◦ (AΦQ)Φmdl= ( C A µ Q)◦ (ΦQ) ◦ (mA CQ) defA µ CQ= A µ CQ ◦ (m A CQ) .Now we prove the unitality condition. We haveA µ CQ ◦ (u A CQ) defA µ CQ( )= C A µ Q ◦ (ΦQ) ◦ (uA CQ)Φmdl= ( C A µ Q)◦ (CuA Q) A µ CQ uni= CQ.Proposition 5.4. Let A = (A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be acomonad on the category A. Assume that Φ : AC → CA is a mixed distributive lawbetween them. Let F, G be left A-module functors and α : F → G be a functorialmorphism between them satisfyingA µ G ◦ (Aα) = α ◦ (A µ F),i.e. there exists a functorial morphism A α : A F → A G such that A U A α = α. Thenalso Cα is a functorial morphism between left A-module functors satisfyingA µ CG ◦ (ACα) = (Cα) ◦ A µ CFi.e. there exists a functorial morphism A (Cα) : A (CF ) → A (CG) such thatAU A (Cα) = Cα. Moreover we haveA (Cα) = ˜C A αwhere ˜C is the lifted comonad on the category A A, i.e. A U ˜C = C A U.Proof. By Lemma 3.29 there exists A α : A F → A G such that A U A α = α. Moreover,by Lemma 5.3, we know that ( CF, A µ CF)=(CF,(C A µ F)◦ (ΦF ))and(CG, A µ CG)=(CG,(C A µ G)◦ (ΦG))are left A-module functors. Then we have□A µ CG ◦ (ACα) defA µ=CG( )C A µ G ◦ (ΦG) ◦ (ACα)Φ= ( C A µ G)◦ (CAα) ◦ (ΦG)αmorpAmod= (Cα) ◦ ( C A µ F)◦ (ΦG)def A µ CF= (Cα) ◦ A µ CFi.e. Cα is a functorial morphism between left A-module functors. Then there existsa functorial morphism A (Cα) : A (CF ) → A (CG) such that A U A (Cα) = Cα. Sincewe also haveAU ˜C A α = C A U A α = Cαwe deduce thatAU A (Cα) = A U ˜C A α.

Since A U is faithful, this implies that A (Cα) = ˜C A α.Lemma 5.5. Let A = (A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be a comonadon the category A. Let Q : B → A be a functor such that ( Q, C ρ Q)is a left C-comodule functor. Assume that Φ : AC → CA is a mixed distributive law. Then(AQ, C ρ AQ)=(AQ, (ΦQ) ◦(A C ρ Q))is a left C-comodule functor.Proof. First of all we prove that C ρ AQ = (ΦQ) ◦ ( A C ρ Q)is coassociative. In fact wehave(C C ρ AQ)◦ C ρ AQdef C ρ AQ= (CΦQ) ◦(CA C ρ Q)◦ (ΦQ) ◦(A C ρ Q)Φ= (CΦQ) ◦ (ΦCQ) ◦ ( ) ( )AC C ρ Q ◦ A C ρ QC ρ Q coass= (CΦQ) ◦ (ΦCQ) ◦ ( A∆ C Q ) ◦ ( A C ρ Q)Φmdl= ( ∆ C AQ ) ◦ (ΦQ) ◦ ( A C ρ Q) def C ρ AQ=(∆ C AQ ) ◦ C ρ AQ .Now we prove the counitality condition. We have(ε C AQ ) ◦ C ρ AQdef C ρ AQ=(ε C AQ ) ◦ (ΦQ) ◦ ( A C ρ Q)Φmdl= ( Aε C Q ) ◦ ( A C ρ Q) C ρ Q couni= AQ.Proposition 5.6. Let A = (A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be acomonad on the category A. Assume that Φ : AC → CA is a mixed distributive lawbetween them. Let F, G be left C-comodule functors and α : F → G be a functorialmorphism between them satisfyingC ρ G ◦ α = (Cα) ◦ (C ρ F),i.e. there exists C α : C F → C G such that C U C α = α. Then also Aα is a morphismbetween left C-comodule functors satisfying(CAα) ◦ C ρ AF = C ρ AG ◦ (Aα)i.e. there exists a functorial morphism C (Aα) : C (AF ) → C (AG) such thatC U C (Aα) = Aα. Moreover we haveC (Aα) = ÃC αwhere à is the lifted monad on the category C A, i.e. C Uà = AC U.Proof. Since F, G are left C-comodule functors, by Lemma 5.5 we know that(AF, C ρ AF)=(AF, (ΦF ) ◦(A C ρ F))and(AG, C ρ AG)=(AG, (ΦG) ◦(A C ρ G))areleft C-comodule functors. Then we have(CAα) ◦ C defρ C ρ AF( )AF = (CAα) ◦ (ΦF ) ◦ A C ρ FΦ= (ΦG) ◦ (ACα) ◦ ( )A C ρ FαmorpCcom= (ΦG) ◦ ( A C ρ G)◦ (Aα)def C ρ AG= C ρ AG ◦ (Aα)83□□

84i.e. Aα is a functorial morphism between left C-comodule functors. Then thereexists a functorial morphism C (Aα) : C (AF ) → C (AG) such that C U C (Aα) = Aα.Since we also haveC UÃC α = A C U C α = Aαwe deduce thatC U C (Aα) = C UÃC α.Since C U is faithful, this implies that C (Aα) = ÃC α.□5.<strong>2.</strong> Liftings of monads and comonads.Theorem 5.7 ([Be, Proposition p. 122] and [Mesa, Theorem <strong>2.</strong>1]). Let A =(A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be a comonad on a category A.There is a bijection between the following collections of data:C liftings ( of C to a comonad ˜C on the category A A, that is comonads˜C = ˜C, ∆C e, ε e )Con A A such thatAU ˜C = C A U, AU∆ eC = ∆ C AU and AUε eC = ε C AUD mixed distributive laws Φ : AC → CAM liftings of A to a)monad à on the category C A, that is monadsà =(Ã, mA e, u A e on C A such thatC Uà = AC U,C Um e A= m A C Ugiven by( )a : C → D where a ˜C =andC Uu e A= u A C U( )BUλ B ˜CA F ◦ ( B U B F Cu A )b : D → C where A Ub (Φ) = C A U and A Uλ A b (Φ) = (C A Uλ A ) ◦ Φ i.e.b (Φ) (( )) ( ( ) )X, A µ X = CX, C A µ X ◦ (ΦX) and b (Φ) (f) = C (f))d : M → D where d(à = (C U C F Aε C) ( )◦C Uγ C à C Fm : D → M where C Um (Φ) = A C U and C Uγ C m (Φ) = Φ ◦ ( A C Uγ C) i.e.m (Φ) (( X, C ρ X))=(AX, (ΦX) ◦(A C ρ X))and m (Φ) (f) = A (f) .Proof. In order to prove the bijection between C and D, we apply Proposition 3.24,to the case (A, m A , u A ) = (B, m B , u B ) monad on A and Q = C. In particular wewill prove that the bijection a : F → M, b : M → F of Proposition 3.24 induces abijection between C and D.Let ˜C) ( )∈ C. We have to prove that Φ = a(˜C = AUλ A ˜CA F ◦ ( A U A F Cu A ) ∈ D.We have(CΦ) ◦ (ΦC) ◦ ( A∆ C) =()()C A Uλ A ˜CA F ◦ (C A U A F Cu A ) ◦ AUλ A ˜CA F C ◦ ( A U A F Cu A C) ◦ ( A∆ C)(= AU ˜Cλ)A ˜CA F ◦(AU ˜C) ()A F Cu A ◦ AUλ A ˜CA F C ◦ ( A U A F Cu A CX) ◦ ( A∆ C)

[( ) ( ) ( )= A U ˜CλA ˜CA F ◦ ˜CA F Cu A ◦ λ A ˜CA F C ◦ ( A F Cu A C) ◦ ( AF ∆ C)][( ) (λ=AA U λ A ˜C ˜CA F ◦ AF A U ˜Cλ)A ˜CA F ◦(AF A U ˜C)A F Cu A ◦ ( A F Cu A C) ◦ ( AF ∆ C)][( ) ()= A U λ A ˜C ˜CA F ◦ AF C A Uλ A ˜CA F ◦ ( A F C A U A F Cu A ) ◦ ( A F Cu A C) ◦ ( AF ∆ C)][( ) ()u=AA U λ A ˜C ˜CA F ◦ AF C A Uλ A ˜CA F ◦ ( A F Cu A CA) ◦ ( A F CCu A ) ◦ ( AF ∆ C)][( ) () (= A U λ A ˜C ˜CA F ◦ AF C A Uλ A ˜CA F ◦ AF Cu AA U ˜C)A F ◦ ( A F CCu A ) ◦ ( AF ∆ C)][( )(λ A ,u A )adj,∆= CAU λ A ˜C ˜CA F ◦ ( AF ∆ C A ) ]◦ ( A F Cu A )[( ) (= A U λ A ˜C ˜CA F ◦ AF A U∆ e ) ]CA F ◦ ( A F Cu A )[(λ=AA U ∆ e ) ( ) ]CA F ◦ λ A ˜CA F ◦ ( A F Cu A )(= AU∆ e ) ( )CA F ◦ AUλ A ˜CA F ◦ ( A U A F Cu A ) = ( ∆ C A ) ◦ (Φ)85so thatMoreoverso that(CΦ) ◦ (ΦC) ◦ ( A∆ C) = ( ∆ C A ) ◦ (Φ) .(ε C A ) ◦ (Φ) = ( ε C A ) ( )◦ AUλ A ˜CA F ◦ ( A U A F Cu A )(= AUε e ) ( )CA F ◦ AUλ A ˜CA F ◦ ( A U A F Cu A )[(= A U ε e ) ( ) ]CA F ◦ λ A ˜CA F ◦ ( A F Cu A )[ (λ=AA U (λ AA F ) ◦ AF A Uε e ) ]CA F ◦ ( A F Cu A )= A U [ (λ AA F ) ◦ ( AF ε C AU A F ) ◦ ( A F Cu A ) ]ε C = A U [ (λ AA F ) ◦ ( A F u A ) ◦ ( AF ε C)](λ A ,u A )adj= AU A F ε C = Aε C(ε C A ) ◦ (Φ) = Aε C .Therefore Φ is a mixed distributive law.Conversely let Φ ∈ D. Then we know that b (Φ) = ˜C is a functor ˜C : A A → A Athat is a lifting of C (i.e. AU ˜C = C A U). We have to prove that such a ˜C givesrise to a comonad on the category A A. Let us prove that ∆ C and ε C are A-modulesmorphisms. Indeed, for every ( X, A µ X)∈ A A, by Lemma 5.3 we haveA µ CX = ( C A µ X)◦ (ΦX)and alsoA µ CCX = ( C A µ CX)◦ (ΦCX) =(CC A µ X)◦ (CΦX) ◦ (ΦCX) .

86Then we haveA µ CCX ◦ ( A∆ C X ) = ( CC A µ X)◦ (CΦX) ◦ (ΦCX) ◦(A∆ C X )andΦm.d.l.= ( CC A µ X)◦(∆ C AX ) ◦ (ΦX) ∆C= ( ∆ C X ) ◦ ( C A µ X)◦ (ΦX)(ε C X ) ◦ (A µ CX)=(ε C X ) ◦ ( C A µ X)◦ (ΦX)ε C = A µ X ◦ ( ε C AX ) ◦ (ΦX) Φm.d.l.= A µ X ◦ ( Aε C X ) .Thus ∆ C and ε C lift to functorial morphisms ∆ eC and ε eC uniquely defined byWe compute(AU ˜C∆ e ) (C◦ AU∆ e ) (C=AU∆ eC = ∆ C AU and AUε eC = ε C AU.C A U∆ e )C◦ ( ∆ C AU ) = ( C∆ C AU ) ◦ ( ∆ C AU )= [( C∆ C) ◦ ∆ C] AU Ccomonad [(= ∆ C C ) ◦ ∆ C] AU = ( ∆ C C A U ) ◦ ( ∆ C AU )(= ∆ C AU ˜C) (◦ AU∆ e ) (C= AU∆ e ) (C ˜C ◦ AU∆ e )Cand since A U is faithful , we deduce( ) (˜C∆C e◦ ∆ eC = ∆ e )C ˜C ◦ ∆ eC .We compute(AU ˜Cε e ) (C◦ AU∆ e ) (C= C A Uε e )C◦ ( ∆ C AU ) = ( Cε C AU ) ◦ ( ∆ C AU )= [( Cε C) ◦ ∆ C] AU Ccomonad= C A U = A U ˜Cand since A U is faithful, we obtain(˜Cεe C)◦ ∆ eC = ˜C.Similarly we compute(AUε e ) (C ˜C ◦ AU∆ e )Cand since A U is faithful, we obtain(ε e C ˜C)=(ε C AU ˜C)◦ ( ∆ C AU ) = ( ε C C A U ) ◦ ( ∆ C AU )= [( ε C C ) ◦ ∆ C] AU Ccomonad= C A U = A U ˜C◦ ∆ eC = ˜C.Therefore ˜C =(˜C, ∆e C, ε e C)is a comonad on A A.Similarly, in order to prove the bijection between D and M, we apply Proposition4.23, taking both ( C, ∆ C , ε C) , ( D, ∆ D , ε D) = ( C, ∆ C , ε C) comonad on A and T =A. In particular we will prove that the bijection a : F → M, b : M → F ofProposition 4.23 induces a bijection between M and D.

)Let(Ãà ∈ M. We have to prove that Φ = a = (C U C F Aε C) ( )◦C Uγ C à C F ∈ D.We have(Cm A ) ◦ (ΦA) ◦ (AΦ) =(Cm A ) ◦ (C U C F Aε C A ) ()◦C Uγ C à C F A ◦ ( A C U C F Aε C) ()◦ A C Uγ C à C FAlift= (C ) (U C F m A ◦ CU C F Aε C A ) () (C ) ()◦C Uγ C à C F A ◦ UÃC F Aε C ◦C UÃγC à C F[ (C ) (= C U F m A ◦ CF Aε C A ) ( ) (ÃC ))]◦ γ C à C F A ◦ F Aε C ◦(Ãγ C à C F[γ= C (C ) ( C U F m A ◦ CF Aε C A ) (C ) () ( )]◦ F C UÃC F Aε C ◦C F C UÃγC à C F ◦ γ C ÃÃC F[Alift (C ) (= C U F m A ◦ CF Aε C A ) ◦ (C F A C U C F Aε C) () ( )]◦ C F A C Uγ C à C F ◦ γ C ÃÃC F[ε= C (C ) ( C U F m A ◦ CF AAε C) ◦ (C F Aε C AC ) () ( )]◦ C F A C Uγ C à C F ◦ γ C ÃÃC F[Alift (C ) (= C U F m A ◦ CF AAε C) () () ( )]◦ C F Aε C C UÃC F ◦C F A C Uγ C à C F ◦ γ C ÃÃC Fso that we get(ε C ,γ C )adj,m A[ (C=C U F Aε C) ◦ (C F m A C ) ( )]◦ γ C ÃÃC F[Alift (C= C U F Aε C) ◦ (C F C Um A e C F ) ( )]◦ γ C ÃÃC F[γ= C (C C U F Aε C) ( )◦ γ C à C F ◦ ( m A e C F )]= (C U C F Aε C) ( )◦C Uγ C à C F ◦ (C Um A e C F ) Alift= Φ ◦ (m A C)(Cm A ) ◦ (ΦA) ◦ (AΦ) = Φ ◦ (m A C) .Moreover we haveΦ ◦ (u A C) = (C U C F Aε C) ( )◦C Uγ C à C F ◦ (u A C)Alift= (C U C F Aε C) ( )◦C Uγ C à C F ◦ (C Uu A e C F )[ (C= C U F Aε C) ( )◦ γ C à C F ◦ ( u A e C F )]γ C = C U [(C F Aε C) ◦ (C F C Uu e A C F ) ◦ ( γ C C F )]Alift= C U [(C F Aε C) ◦ (C F u A C ) ◦ ( γ C C F )]u A= C U [(C F u A)◦( CF ε C) ◦ ( γ C C F )]87so that we get(ε C ,γ C )adj=C U C F u A = Cu AΦ ◦ (u A C) = Cu A .Therefore Φ is a mixed distributive law.Conversely let Φ ∈ D. Then we know that à = b (Φ) (with notations of Proposition

884.23) is a functor à : C A → C A that is a lifting of A (i.e. C Uà = AC U). We have toprove that such a à gives rise to a monad on the category C A. Let us prove that m Aand u A are C-comodule morphisms. Indeed, for every ( X, C ρ X)∈ C A, by Lemma5.5 we haveC ρ AX = (ΦX) ◦ ( A C ρ X)and alsoC ρ AAX = (ΦAX) ◦ ( A C ρ AX)= (ΦAX) ◦ (AΦX) ◦(AA C ρ X).Then we haveand(Cm A X) ◦ C ρ AAX = (Cm A X) ◦ (ΦAX) ◦ (AΦX) ◦ ( AA C ρ X)Φm.d.l.= (ΦX) ◦ (m A CX) ◦ ( AA C ρ X)m A= (ΦX) ◦ ( A C ρ X)◦ (mA X) = C ρ AX ◦ (m A X)C ρ AX ◦ (u A X) = (ΦX) ◦ ( A C ρ X)◦ (uA X)u A= (ΦX) ◦ (u A CX) ◦ C ρ XΦm.d.l.= (Cu A X) ◦ C ρ XThus m A and u A lift to functorial morphisms m e Aand u e Auniquely defined byC Um e A= m A C UandC Uu e A= u A C U.We compute( C) (Um A e ◦ C AliftUm A eÃ) = ( m C A U ) ( )◦ m C AliftA Uà = ( m C A U ) ◦ ( m A A C U )Amonad= ( m C A U ) ◦ ( Am C A U ) Alift= (C ) ( )Um A e ◦ A C AliftUm A e = (C ) ( )Um A e ◦ C UÃm A eand since C U is faithful , we deduce( )m A e ◦ m A eÃ= m A e ◦We computeand since C U is faithful, we obtain(Ãm e A).( C) (Um A e ◦ C AliftUu A eÃ) = ( m C A U ) ( )◦ u C A UÃAlift= ( m A C U ) ◦ ( u A A C U ) Amonad= A C U Alift= C UÃm e A◦( )u A eÃ= Ã.Similarly we compute( CUm e A)◦ (C UÃu e A)Alift= ( m A C U ) ◦ ( A C Uu e A) Alift= ( m A C U ) ◦ ( Au A C U )Amonad= A C U Alift= C UÃand since C U is faithful, we obtain)m A e ◦(ÃuA e = Ã.

Therefore à = (Ã, m e A, u e A)is a monad on C A.89□6. (Co)Pretorsors and (co)herdsIn this section we collect the material we need in the following or we want tointroduce in this thesis about them, starting from pretorsor and copretorsor, throughherds and coherds, concluding with the tame and cotame case. From time to timewe decide whether or not include the details of the <strong>results</strong>, proving at least one ofthe two cases and having in mind that the other could also be obtained by dualizingit. In general we give the proof only for the less-known case even if it is not the firstpresented.6.<strong>1.</strong> Pretorsors.Proposition 6.1 ([BM, Lemma 4.8]). Let A and B be categories with equalizersand let P : A → B, Q : B → A and A : A → A be functors. Assume thatall the functors P, Q and A preserve equalizers. Let u A : A → A be a functorialmonomorphism and assume that (A, u A ) = Equ Fun (u A A, Au A ). Let τ : Q → QP Qbe a functorial morphism such that(QP τ) ◦ τ = (τP Q) ◦ τ.and let σ A : QP → A be a functorial morphism such that(σ A Q ) ◦ τ = u A Q.Let ω l = ( QP σ A) ◦ (τP ) and ω r = QP u A : QP → QP A. Set(60) (C, i) = Equ Fun(ω l , ω r) .There exists a functorial morphism C ρ Q : Q → CQ such that(61) (iQ) ◦ C ρ Q = τ.There exist functorial morphisms ∆ C : C → CC and ε C : C → A such thatC = ( C, ∆ C , ε C) is a comonad over A and C preserves equalizers. The functorialmorphisms ∆ C and ε C are uniquely determined by((62)Cρ Q P ) ◦ i = (Ci) ◦ ∆ C and σ A ◦ i = u A ◦ ε Cor equivalently(63) (τP ) ◦ i = (ii) ◦ ∆ C and σ A ◦ i = u A ◦ ε C .Moreover (Q, C ρ Q ) is a left C-comodule functor.Proposition 6.2 ([BM, Lemma 4.8]). Let A and B be categories with equalizersand let P : A → B, Q : B → A, and B : B → B be functors. Assume thatall the functors P, Q and B preserve equalizers. Let u B : B → B be a functorialmonomorphism and assume that (B, u B ) = Equ Fun (u B B, Bu B ). Let τ : Q → QP Qbe a functorial morphism such that(QP τ) ◦ τ = (τP Q) ◦ τLet σ B : P Q → B be a functorial morphism such that(QσB ) ◦ τ = Qu B .

90Let θ l = ( σ B P Q ) ◦ (P τ) and θ r = u B P Q : P Q → BP Q. Set(64) (D, j) = Equ Fun(θ l , θ r) .There exists a functorial morphism ρ D Q : Q → QD such that(65) (Qj) ◦ ρ D Q = τ.There exist functorial morphisms ∆ D : D → DD and ε D : D → B such thatD = ( D, ∆ D , ε D) is a comonad over B and D preserves equalizers. The functorialmorphisms ∆ D and ε D are uniquely determined by(66) (jD) ◦ ∆ D = ( )P ρ D Q ◦ j and σ B ◦ j = u B ◦ ε Dor equivalently(67) (P τ) ◦ j = (jj) ◦ ∆ D and σ B ◦ j = u B ◦ ε D .Moreover ( Q, ρ D Q)is a right D-comodule functor.Definition 6.3. Let A and B be categories. A preformal dual structure is a eightupleΞ = ( A, B, P, Q, σ A , σ B , u A , u B)where A : A → A, B : B → B, P : A → B andQ : B → A are functors, σ A : QP → A, σ B : P Q → B, u A : A → A, u B : B → B arefunctorial morphisms. A pretorsor τ for Ξ is a functorial morphism τ : Q → QP Qsatisfying the following conditions.1) Associativity, in the sense that(68) (QP τ) ◦ τ = (τP Q) ◦ τ2) Unitality, in the sense that(69) (σ A Q) ◦ τ = u A Qand(70) (Qσ B ) ◦ τ = Qu B .Definition 6.4. A preformal dual structure Ξ = ( P, Q, A, B, σ A , σ B , u A , u B , ) willbe called regular whenever (A, u A ) = Equ Fun (u A A, Au A ) and(B, u B ) = Equ Fun (u B B, Bu B ). In this case a pretorsor for Ξ will be called a regularpretorsor.Theorem 6.5 ([BM, Lemma 4.8]). Let A and B be categories with equalizers and letτ : Q → QP Q be a regular pretorsor for Ξ = ( A, B, P, Q, σ A , σ B , u A , u B). Assumethat the underlying functors P, Q, A and B preserve equalizers. Let ω l = ( QP σ A) ◦(τP ) and ω r = QP u A : QP → QP A. Set(71) (C, i) = Equ Fun(ω l , ω r) .Then there exists a functorial morphism C ρ Q : Q → CQ such that(72) (iQ) ◦ C ρ Q = τ.There exist functorial morphisms ∆ C : C → CC and ε C : C → A such thatC = ( C, ∆ C , ε C) is a comonad over A and C preserves equalizers. The functorialmorphisms ∆ C and ε C are uniquely determined by(73) (τP ) ◦ i = (ii) ◦ ∆ C and σ A ◦ i = u A ◦ ε C .

Moreover (Q, C ρ Q ) is a left C-comodule functor.Let θ l = ( σ B P Q ) ◦ (P τ) and θ r = u B P Q : P Q → BP Q. Set(74) (D, j) = Equ Fun(θ l , θ r) .There exists a functorial morphism ρ D Q : Q → QD such that(75) (Qj) ◦ ρ D Q = τ.There exist functorial morphisms ∆ D : D → DD and ε D : D → B such thatD = ( D, ∆ D , ε D) is a comonad over B and D preserves equalizers. The functorialmorphisms ∆ D and ε D are uniquely determined by(76) (P τ) ◦ j = (jj) ◦ ∆ D and σ B ◦ j = u B ◦ ε D .Moreover ( Q, ρ D Q)is a right D-comodule functor.Finally ( Q, C ρ Q , ρ D Q)is a C-D-bicomodule functor.Proof. See the dual Theorem 6.29.Theorem 6.6. Let Ξ = ( P, Q, A, B, σ A , σ B , u A , u B , ) be a regular preformal dualstructure on categories A and B such that the functors P, Q, A, B preserve equalizersand let τ : Q → QP Q be a pretorsor for Ξ. Assume that A and B aremonads, ( ) ( )P, B µ P is a left B-module functor and P, µAP is a right A-module functor.Moreover assume that the functorial morphism σ A is right A-linear, that isσ A ◦ ( )Qµ A P = mA ◦ ( σ A A ) and the functorial morphism σ B is left B-linear that isσ B ◦ (B µ P Q ) = m B ◦ ( Bσ B) and that they are compatible in the sense that(77)B µ P ◦ ( σ B P ) = µ A P ◦ ( P σ A) .Then there exists a comonad C = ( C, ∆ C , ε C) on the category A together with afunctorial morphism C ρ Q : Q → CQ such that ( Q, C ρ Q)is a left C-comodule functorand a comonad D = ( D, ∆ D , ε D) together with a functorial morphism ρ D Q : Q → QDsuch that ( Q, ρ D Q)is a right D-comodule functor. The underlying functors are definedas follows(C, i) = Equ Fun((QP σA ) ◦ (τP ) , QP u A)and(D, j) = Equ Fun((σ B P Q ) ◦ (P τ) , u B P Q ) .satisfying(iQ) ◦ C ρ Q = τ and (Qj) ◦ ρ D Q = τ.Furthermore1) The morphism can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CA is an isomorphism.2) The morphism can 1 := ( σ B D ) ◦ ( P ρ D Q): P Q → BD is an isomorphisms.3)(QP σA ) ◦ (τP ) = (iA) ◦ can 1 and ( σ B P Q ) ◦ (P τ) = (Bj) ◦ can 191□(78)(79)4)i = can −11 ◦ (Cu A )j = (can 1 ) −1 ◦ (u B D)

925)σ A = ( ε C A ) ◦ ( Cσ A) ◦ (C ρ Q P )6)σ B = ( Bε D) ◦ ( σ B D ) ◦ ( )P ρ D QFrom the last equalities, we deduce that, σ A is a regular epimorphism if and onlyif so is ε C A and σ B is a regular epimorphism if and only if so is Bε D .Proof. See the dual Theorem 6.30.6.<strong>2.</strong> Herds. Following [BV], we recall some definition about herds.Definition 6.7. A formal dual structure on two categories A and B is a sextupleM = (A, B, P, Q, σ A , σ B ) where A = (A, m A , u A ) and B = (B, m B , u B ) are monadson A and B respectively and ( )A, B, P, Q, σ A , σ B , u A , u B is a preformal dual structure.Moreover ( P : A → B, B µ P : BP → P, µ A P : P A → P ) (andQ : B → A, A µ Q : AQ → Q, µ B Q : QB → Q) are bimodule functors; σ A : QP →A, σ B : P Q → B are subject to the following conditions: σ A is A-bilinear andσ B is B-bilinear(80) σ A ◦ (A µ Q P ) = m A ◦ ( Aσ A) and σ A ◦ ( )Qµ A P = mA ◦ ( σ A A )(81) σ B ◦ (B µ P Q ) = m B ◦ ( Bσ B) and σ B ◦ ( )P µ B Q = mB ◦ ( σ B B )and the associative conditions hold(82)A µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B) and B µ P ◦ ( σ B P ) = µ A P ◦ ( P σ A) .Definition 6.8. Consider a formal dual structure M = (A, B, P, Q, σ A , σ B ) in thesense of the previous definition. A herd for M is a pretorsor τ : Q → QP Q i.e.(83) (QP τ) ◦ τ = (τP Q) ◦ τ,(84) (σ A Q) ◦ τ = u A Qand(85) (Qσ B ) ◦ τ = Qu B .Definition 6.9. A formal dual structure M = (A, B, P, Q, σ A , σ B ) will be calledregular whenever ( A, B, P, Q, σ A , σ B , u A , u B)is a regular preformal dual structure.In this case a herd for M will be called a regular herd.Lemma 6.10. Let M = (A, B, P, Q, σ A , σ B ) be a formal dual structure and let τ :Q → QP Q be a herd for M. Assume that the underlying functors A and B reflectequalizers. Then τ is a regular herd.Proof. Since A and B are monads, we have m A ◦(Au A ) = Id A and m B ◦(Bu B ) = Id B .Thus, Au A and Bu B are split monomorphisms and thus monomorphisms. Since Aand B reflect equalizers, we deduce that also u A and u B are monomorphisms andthus (A, u A ) = Equ Fun (u A A, Au A ) and (B, u B ) = Equ Fun (u B B, Bu B ), i.e. τ is aregular herd.□□

Proposition 6.1<strong>1.</strong> Let M = (A, B, P, Q, σ A , σ B ) be a formal dual structure suchthat the lifted functors A Q B : B B → A A and B P A : A A → B B determine an equivalenceof categories. Then ( A Q, P A ) and ( B P , Q B ) are adjunctions.Proof. Since ( A F , A U) and ( B F , B U) are adjunctions, ( A Q BB F , B U B P A ) = ( A Q, P A )and ( B P AA F , A U A Q B ) = ( B P , Q B ) are also adjunctions.□6.3. Herds and comonads.Theorem 6.12 ([Bo]). Let A and B be categories in both of which the equalizer ofany pair of parallel morphisms exists. Let M = (A, B, P, Q, σ A , σ B ) be a formal dualstructure on two categories A and B. Then we have(1) If C = ( C, ∆ C , ε C) is a comonad on the category A and ( Q, C ρ Q : Q → CQ )is a left C-comodule functor such that(i) the functorial morphism can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CA is anisomorphism(ii) the functorial morphism can 2 := ( (Cµ Q) B ◦ Cρ Q B ) : QB → CQ is anisomorphismthen τ := ( can −11 Q ) ◦ (Cu A Q) ◦ C ρ Q : Q → QP Q is a pretorsor and thusa herd.(2) If D = ( D, ∆ D , ε D) is a comonad on the category B and ( Q, ρ D Q : Q → QD)is a right D-comodule functor such that(i) the functorial morphism can 1 := ( σ B D ) ◦ ( P ρQ) D : P Q → BD is anisomorphism(ii) the functorial morphism can 2 := (A µ Q D ) ◦ ( AρQ) D : AQ → QD is anisomorphismthen τ := (Qcan −1 1 ) ◦ (Qu B D) ◦ ρ D Q : Q → QP Q is a pretorsor and thus aherd.Proof. See the dual Theorem 6.36.Theorem 6.13 ([Bo]). Let A and B be categories in both of which the equalizerof any pair of parallel morphisms exists. Let M = (A, B, P, Q, σ A , σ B ) be a regularformal dual structure such that the underlying functors A, B, P and Q preserveequalizers, then the existence of the following structures are equivalent:(a) A herd τ : Q → QP Q in M;(b) A comonad C = ( C, ∆ C , ε C) on the category A such that the functor Cpreserves equalizers and ( Q, C ρ Q : Q → CQ ) is a left C-comodule functorsubject to the following conditions(i) the functorial morphism can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CA is anisomorphism(ii) the functorial morphism can 2 := ( Cµ B Q)◦( Cρ Q B ) : QB → CQ is anisomorphism;(c) A comonad D = ( D, ∆ D , ε D) on the category B such that the functor Dpreserves equalizers and ( Q, ρ D Q : Q → QD) is a right D-comodule functorsubject to the following conditions(i) the functorial morphism can 1 := ( σ B D ) ◦ ( P ρ D Q): P Q → BD is anisomorphism93□

94(ii) the functorial morphism can 2 := (A µ Q D ) ◦ ( Aρ D Q): AQ → QD is anisomorphism.Proof. See the dual Theorem 6.37.6.4. Herds and distributive laws.Proposition 6.14 ([Bo]). Let A and B be categories with equalizers and let τ : Q →QP Q be a regular herd for M = (A, B, P, Q, σ A , σ B ) where the underlying functorsP : A → B, Q : B → A, A : A → A and B : B → B preserve equalizers. LetC = ( C, ∆ C , ε C) and D = ( D, ∆ D , ε D) be the associated comonads constructed inProposition 6.1 and in Proposition 6.<strong>2.</strong> Then1) There exists a mixed distributive law between the comonad C and the monadA, Φ : AC → CA such that(iA) ◦ Φ = φ = ( QP σ A) ◦ (τP ) ◦ (A µ Q P ) ◦ (Ai) .2) There exists an opposite mixed distributive law between the comonad D andthe monad B, Ψ : DB → BD such that(Bj) ◦ Ψ = ψ = ( σ B P Q ) ◦ (P τ) ◦ ( P µ B Q)◦ (jB) .Proof. See the dual Proposition 6.38.6.5. Herds and Galois functors.Lemma 6.15. Let M = ( A, B, P, Q, σ A , σ B) be a formal dual structure where Q :B → A, P : A → B and A = (A, m A , u A ) is a monad on the category A andB = (B, m B , u B ) is a monad on B. Assume that both A and B have coequalizers andthat A, QB preserve them. Then σ A : QP → A induces a morphism σ A A : QP A → A Uin A A and hence there exists a morphism A σ A A : AQP A → Id A A such that(86) AU A σ A A = σ A A.Moreover σ A AAF = σ A : QP AA F = QP → A U A F = A.Proof. Let us consider the following diagram with notations of Proposition 3.30QP A A UAA A Uσ A A A UQµ A P AUQP A Uλ Am AA UA A Uλ A QP A U A A Uσ AA UQp P QPAσ A ′AUλ ASince by assumption QB preserves coequalizers, by Lemma 3.19 also Q preservescoequalizers. Since ( A Uλ A ) ◦ ( σ A AU ) coequalizes the pair ( Qµ A P AU, QP A Uλ A)and(QP A , Qp P ) = Coequ Fun(QµAP AU, QP A Uλ A), by the universal property of the coequalizer,there exists a unique morphism σ A A : QP A → A U such thatσ A A ◦ (Qp P ) = ( A Uλ A ) ◦ ( σ A AU ) .We now want to prove that σ A A : QP A = A U A QP A → A U is a morphism between leftA-module functors which satisfies( A Uλ A ) ◦ ( Aσ A A)= σAA ◦ (A µ Q P A).AU□□

95We have( A Uλ A ) ◦ ( Aσ A A)◦ (AQpP ) defσA A= ( A Uλ A ) ◦ (A A Uλ A ) ◦ ( Aσ A AAU )AUλ A coequ= ( A Uλ A ) ◦ (m AA U) ◦ ( Aσ A AU ) (80)= ( A Uλ A ) ◦ ( σ A AU ) ◦ (A µ Q P A U )defσA= A σA A ◦ (Qp P ) ◦ (A µ Q P A U ) A µ Q= σAA ◦ (A )µ Q P A ◦ (AQpP )and since A, Q preserve coequalizers, AQp P is an epimorphism, so that we get( A Uλ A ) ◦ ( Aσ A A)= σAA ◦ (A µ Q P A).Hence, by Lemma 3.29, there exists a unique morphism A σA A: AQP A → Id A A suchthatAU A σA A = σA.ANow, note that, by definition of σA A , we haveσA A ◦ (Qp P ) = ( A Uλ A ) ◦ ( σ A AU )so that by applying it to A F we get(σAA AF ) ◦ (Qp P A F ) = ( A Uλ AA F ) ◦ ( σ A AU A F ) .Hence, by Proposition 3.34, we obtain that(σAA AF ) ◦ ( Qµ A P)= mA ◦ ( σ A A ) (80)= σ A ◦ ( Qµ A P).Since Qµ A P is an epimorphism, we deduce that σA AAF = σ A .Proposition 6.16. Let A and B be categories with equalizers and let τ : Q → QP Qbe a regular herd for a formal dual structure M = (A, B, P, Q, σ A , σ B ) where theunderlying functors P : A → B, Q : B → A and A : A → A preserve equalizers.Let• C = ( C, ∆ C , ε C) be the comonad on the category A constructed in Proposition6.1;• ( Q, C ρ Q)be the left C-comodule functor constructed in Proposition 6.1;• A Q : B → A A be the functor defined in Lemma 3.29;• Φ : AC → CA be the mixed distributive law between the comonad C and themonad A constructed in Proposition 6.14;• ˜C be the lifting of C on the category A A constructed in Theorem 5.7.Then there exists a functorial morphism eC ρ A Q : A Q → ˜C A Q such thatAU eC ρ A Q = C ρ Q .( )Moreover, AQ, eC ρ A Q is a left ˜C-comodule functor.Proof. Since τ : Q → QP Q is a regular herd for M = (A, B, P, Q, σ A , σ B ), byProposition 6.14, the mixed distributive law Φ : AC → CA is uniquely defined by(iA) ◦ Φ = ( QP σ A) ◦ (τP ) ◦ (A µ Q P ) ◦ (Ai) .Now we prove that C ρ Q yields a functorial morphism eC ρ A Q. In fact we have(iQ) ◦ ( C A µ Q)◦ (ΦQ) ◦(A C ρ Q) i= ( QP A µ Q)◦ (iAQ) ◦ (ΦQ) ◦(A C ρ Q)□

96defΦ= ( QP A µ Q)◦(QP σ A Q ) ◦ (τP Q) ◦ (A µ Q P Q ) ◦ (AiQ) ◦ ( A C ρ Q)(61),(82)= ( QP µ B Q)◦(QP QσB ) ◦ (τP Q) ◦ (A µ Q P Q ) ◦ (Aτ)τ= ( QP µ B Q)◦ (τB) ◦(QσB ) ◦ (A µ Q P Q ) ◦ (Aτ)A µ Q=(QP µBQ)◦ (τB) ◦( Aµ Q B ) ◦ ( AQσ B) ◦ (Aτ)(70)= ( QP µ B Q)◦ (τB) ◦( Aµ Q B ) ◦ (AQu B )A µ Q=(QP µBQ)◦ (τB) ◦ (QuB ) ◦ A µ Q τ = ( QP µ B Q)◦ (QP QuB ) ◦ τ ◦ A µ QQmodfun= τ ◦ A µ Q(61)= (iQ) ◦ C ρ Q ◦ A µ Qand since by construction iQ is a monomorphism we get that(C A µ Q)◦ (ΦQ) ◦(A C ρ Q)= C ρ Q ◦ A µ Q .By Lemma 5.3 we know that( )C A µ Q ◦ (ΦQ) = A µ CQso that we getA µ CQ ◦ ( A C ρ Q)=(C A µ Q)◦ (ΦQ) ◦(A C ρ Q)= C ρ Q ◦ A µ Q .Hence there exists a morphism eC ρ A Q : A Q → ˜C A Q such thatAU eC ρ A Q = C ρ Q .By the coassociativity and counitality(properties)of C ρ Q , we deduce that eC ρ A Q isalso coassociative and counital so that AQ, eC ρ A Q is a left ˜C-comodule functor. □Lemma 6.17. Let M = (A, B, P, Q, σ A , σ B ) be a formal dual structure where theunderlying functors are A : A → A, B : B → B, P : A → B and Q : B → A.Assume that both categories A and B have coequalizers and the functors A, QBpreserve them. Assume thatand• C = ( C, ∆ C , ε C) is a comonad on the category A such that C preservescoequalizers• ˜C(= ˜C, ∆C e, ε e )Cis a lifting of the comonad of C to the category A A( )• AQ, eC ρ A Q is a left ˜C-comodule functor where A U eC ρ A Q = C ρ Q .Consider the functorial morphismscan 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CA( ) ( )Acan A := ˜CA σAA eC◦ ρ A QP A : A QP A → ˜CThen can 1 is an isomorphism if and only if A can A is an isomorphism.

Proof. Note that, since( )AQ, eC ρ A Q is a left ˜C-comodule functor, then ( )Q, C ρ Q isa left C-comodule functor where C ρ Q = A U eC ρ A Q. Let (P A , p P ) be the coequalizerdefined in (6). Now, by Lemma 6.15, σ A induces a morphism σ A A : QP A → A U suchthatσ A A ◦ (Qp P ) = ( A Uλ A ) ◦ ( σ A AU ) . Then, we can consider the morphism(87) can A := ( Cσ A A)◦( Cρ Q P A): QPA = A U A QP A → C A U = A U ˜C.Then, by using the naturality of C ρ Q and the definition of σA A , we obtain(88) can A ◦ (Qp P ) = (C A Uλ A ) ◦ (can 1A U) .Moreover, by Lemma 6.15, there exists a morphism A σA A : AQP A → Id A A suchthat A U A σA A = σA A . Since ˜C is a lifting of the comonad C, we know that CσA A =C A U A σA A = AU ˜C A σA A . Let us set( ) ( )Acan A := ˜CA σAA eC◦ ρ A QP A : A QP A → ˜Cso that we get(89) AU A can A =(AU ˜C))A σAA ◦(AU eC ρ A QP A = ( (CσA) A ◦ C)ρ Q P A = canA .By using the naturality of C ρ Q , we calculate(can 1A U) ◦ ( Qµ A P AU ) = ( Cσ A AU ) ◦ (C ρ Q P A U ) ◦ ( Qµ A P AU )so that we get= ( Cσ A AU ) ◦ ( CQµ A P AU ) ◦ (C ρ Q P A A U )(80)= (Cm AA U) ◦ ( Cσ A A A U ) ◦ (C ρ Q P A A U ) = (Cm AA U) ◦ (can 1 A A U)(90) (can 1A U) ◦ ( Qµ A P AU ) = (Cm AA U) ◦ (can 1 A A U) .Let us consider the following diagramQP A A Ucan 1 A A UCAA A UQµ A P AUQP A Uλ A Cm AA UCA A Uλ Acan 1A UQP A UQp PQP Acan A CA A U C AUλ AC A U = A U ˜C = CA A 0.Now, since can 1 : QP → CA is a functorial morphism and by formula (90), theleft square serially commutes. By formula (88) also the right square commutes.Moreover, by definition, p P and A Uλ A are coequalizers. Since Q and C preservecoequalizers, both the rows are coequalizers .Assume now that can 1 is a functorial isomorphism. Then both can 1 A A U andcan 1A U are isomorphism and we deduce that also can A is an isomorphism. SinceAU A can A = can A and A U reflects isomorphisms, we get that also A can A is an isomorphism.Conversely, assume that A can A is an isomorphism. Then also can A = A U A can Ais an isomorphism. Then, by using (89), (87), Lemma 6.15 and (15), we obtain097

98AU A can AA F = can AA F = ( CσAAF ) A ◦ (C ρ Q P AA F ) = ( Cσ A) ◦ (C ρ Q P ) = can 1 sothat also can 1 is an isomorphism.□6.6. The tame case.Definition 6.18. A formal dual structure M = (A, B, P, Q, σ A , σ B ) is called aMorita context on the categories A and B if it satisfies also the balanced conditions(91) σ A ◦ ( µ B QP ) = σ A ◦ ( Q B µ P)and σ B ◦ ( P A µ Q)= σ B ◦ ( µ A P Q ) .Lemma 6.19. Let M = (A, B, P, Q, σ A , σ B ) be a Morita context on the categories Aand B and assume that A, B, P, Q preserve coequalizers. Hence, there exist functorialmorphisms• AB σ A BA : AQ BB P A → Id A A such that(92) AU AB σ A BA = B σ A BAwhere B σ A BA is uniquely determined by Bσ A BA ◦(Q Bp B P ) = ( A Uλ A )◦ ( Bσ A BAU )and(93) Bσ A B ◦ (p QB P ) = σ A• BA σ B AB : BP AA Q B → Id B B such that(94) BU BA σ B AB = A σ B ABwhere A σ B AB is uniquely determined by Aσ B AB ◦ (P Ap A Q) = ( B Uλ B ) ◦ ( Aσ B A BU )and(95) Aσ B A ◦ (p P A Q) = σ B .Moreover we have that(96) Bσ A BAAF = B σ A B and Aσ B ABBF = A σ B A.Proof. By definition, (Q BB P , p QB P ) = Coequ Fun(µBQ P, Q B µ P)and by assumptionσ A is balanced, so that, by the universal property of the coequalizer, there exists aunique functorial morphism B σ A B : Q BBP → A such that B σ A B ◦ (p QBP ) = σ A . Now,let us consider the following diagramQ BB P A A UQ B µ A B P AU Q B B P A Uλ AQ BB P A UQ B p B PQ BB P ABσ A B A AUBσ A B AUBσ A BAA A Um AA UA A Uλ A A A UAUλ A AUNote that, by naturality of p Q and definition of B σB A we have(Bσ BAU ) A ◦ ( Q B µ A BP AU ) ◦ (p QB P A A U) = ( BσBAU ) A ◦ (p QB P A U) ◦ ( Q B Uµ A BP AU )= ( σ A AU ) ◦ ( Qµ A P AU ) (80)= (m AA U) ◦ ( σ A A A U ) = (m AA U) ◦ ( Bσ A BA A U ) ◦ (p QB P A A U)

and since p QB P A A U is an epimorphism, we get that ( Bσ A BAU ) ◦ ( Q B µ A BP AU ) =(m AA U) ◦ ( Bσ A B A AU ) . Moreover, by using naturality of p Q , definition of B σ A B , naturalityof σ A we have(Bσ A BAU ) ◦ (Q BB P A Uλ A ) ◦ (p QB P A A U) = ( Bσ A BAU ) ◦ (p QB P A U) ◦ (Q B U B P A Uλ A )= ( σ A AU ) ◦ (Q B U B P A Uλ A ) = (A A Uλ A ) ◦ ( σ A AU A F A U )= (A A Uλ A ) ◦ ( Bσ A BA A U ) ◦ (p QB P A A U)and since p QB P A A U is an epimorphism, we get that ( Bσ A BAU ) ◦ (Q BB P A Uλ A ) =(A A Uλ A ) ◦ ( Bσ A B A AU ) so that the left square serially commutes. Since B, P, Q preservecoequalizers, by Corollary <strong>2.</strong>12, also Q BB P = Coequ Fun(µBQ BU B P, Q B Uλ BB P )preserves them so that both the rows are coequalizers. Hence, there exists a uniquefunctorial morphism B σ A BA : Q BBP A → A U such that(97) Bσ A BA ◦ (Q B p B P ) = ( A Uλ A ) ◦ ( Bσ A BAU ) .Now, by using naturality of A µ QB , definition of B σ A BA , definition of Bσ A B , coequalizingproperty of A Uλ A , we computeBσ A BA ◦ (A µ QB BP A)◦ (AQB p B P ) ◦ (Ap QB P A U)= B σBA A ◦ (Q B p B P ) ◦ (A µ QB BP A U ) ◦ (Ap QB P A U)(7)= ( A Uλ A ) ◦ ( BσBAU ) A ◦ (p QB P A U) ◦ (A µ QB U B P A U )= ( A Uλ A ) ◦ ( σ A AU ) ◦ (A µ QB U B P A U )(80)= ( A Uλ A ) ◦ (m AA U) ◦ ( Aσ A AU ) = ( A Uλ A ) ◦ (A A Uλ A ) ◦ ( Aσ A AU )= ( A Uλ A ) ◦ (A A Uλ A ) ◦ ( A B σ A BAU ) ◦ (Ap QB P A U)= ( A Uλ A ) ◦ ( A B σ A BA)◦ (AQB p B P ) ◦ (Ap QB P A U)and since (AQ B p B P ) ◦ (Ap QB P A U) is an epimorphism, we get B σBA A ◦ (A )µ QB BP A =( A Uλ A )◦ ( A B σBA) A so that B σBA A induces a functorial morphism ABσBA A : AQ BB P A →Id A A such that A U AB σBA A = BσBA A . Similarly, one can prove that there exists a uniquefunctorial morphism A σA B : P AAQ → B such that A σA B ◦ (p P AQ) = σ B and it inducesa unique functorial morphism BA σAB B : BP AA Q B → Id B B such that B U BA σAB B = AσABBwhere A σAB B is uniquely determined by AσAB B ◦ (P Ap A Q) = ( B Uλ B ) ◦ ( AσA B BU ) .Finally we compute(Bσ BAAF ) A ◦ (Q B p B P AF ) ◦ (p QB P A) (97)= ( A Uλ AA F ) ◦ ( BσBAU A A F ) ◦ (p QB P A)= m A ◦ ( BσBA ) A ◦ (p QB P A) (93)= m A ◦ ( σ A A ) (80)= σ A ◦ ( )Qµ A P(93)= B σB A ◦ (p QB P ) ◦ ( )Qµ A (11)P = B σB A ◦ (p QB P ) ◦ ( )Q B Uµ A BPp Q=B σ A B ◦ ( Q B µ A BP)◦ (pQB P A) (13),(14)= B σ A B ◦ (Q B p B P AF ) ◦ (p QB P A) .Since B and Q preserve coequalizers, by Corollary <strong>2.</strong>12, also Q B preserves them sothat ( Q B µ A BP)◦ (pQB P A) is epi and we deduce thatBσ A BAAF = B σ A B.99

100Similarly, one can prove the statement for A σ B ABBF = A σ B A .Definitions 6.20. Let M = (A, B, P, Q, σ A , σ B ) be a Morita context. We will saythat M is tame if the lifted functorial morphisms AB σ A BA : AQ BB P A → Id A A andBAσ B AB : BP AA Q B → Id B B are isomorphisms so that the lifted functors A Q B : B B →AA and B P A : A A → B B yield a category equivalence. In this case, if τ : Q → QP Qis a herd for M, we will say that τ is a tame herd.Proposition 6.2<strong>1.</strong> Let M = ( A, B, P, Q, σ A , σ B) be a tame Morita context. Thenunit and counit of the adjunction ( A Q B , B P A ) are given byη (A Q B , B P A ) = ( ) () )BAσABBP B AA Q B ◦ BP A(ABσ BAA −1AQ B ◦ ( BAσAB) B −1and ɛ(A Q B , B P A ) =ABσBA A so thatη (A Q,P A ) = ( BU BA σABBP B AA Q BB F ) () (◦(BU B P A ABσBA) A −1AQ BB F ◦ BU ( )BAσAB) B −1BF ◦u B and ɛ (A Q,P A ) = AB σBA A ◦ ( AQ B λ BB P A ).Proof. It is a well-known fact that, given the two functorial isomorphisms σ : Id →RL and ɛ : LR → Id associated to an equivalence of categories, the unit of an adjunctionis given by η = (σ −1 RL) ◦ (Rɛ −1 L) ◦ σ and the counit is ɛ. Hence, since the isomorphismsare ɛ = AB σBA A : AQ BB P A → Id A A and σ −1 = BA σAB B : BP AA Q B → Id B Bthe unit is η (A Q B , B P A ) = ( ) () )BAσABBP B AA Q B ◦ BP A(ABσ BAA −1AQ B ◦ ( )BAσABB −1and the counit is ɛ (A Q B , B P A ) = AB σBA A . Note that, by Proposition 6.11, ( AQ, P A ) =( A Q BB F , B U B P A ) and ( B P , Q B ) = ( B P AA F , A U A Q B ) are adjunctions. Hence, theunit of the adjunction ( A Q, P A ) is η (A Q,P A ) = ( BUη (A Q B , B P A )BF ) ◦ η (B F , B U) and thusη (A Q,P A ) = ( BU BA σABBP B AA Q BB F ) () (◦(BU B P A ABσBA) A −1AQ BB F ◦ BU ( )BAσAB) B −1BF ◦u B . The counit of the adjunction ( A Q BB F , B U B P A ) = ( A Q, P A ) is given by ɛ (A Q,P A ) =ɛ (A Q B , B P A ) ◦ ( )AQ B ɛ (B F , B U)BP A = AB σBA A ◦ ( AQ B λ BB P A ). A similar result holds forthe other adjunction.□Corollary 6.2<strong>2.</strong> Let M = (A, B, P, Q, σ A , σ B ) be a tame Morita context. Assumethat the functors A, B, P, Q preserve coequalizers. Then the counits of the adjunctions( A Q, P A ) and ( B P , Q B ) are given by ɛ (A Q,P A ) = A σ A A and ɛ ( B P ,Q B ) = B σ B B .Proof. By Proposition 6.11 ( A Q, P A ) and ( B P , Q B ) are adjunctions. Let us considerthe functorial morphism A σA A : AQP A → Id A A constructed in Lemma 6.15 satisfyingAU A σAAF A = σAAF A = σ A . By using naturality of µ B Q , definition of σA A , the balancedproperty of σ A , we computeσA A ◦ ( )µ B QP A ◦ (QBpP ) = σA A ◦ (Qp P ) ◦ ( µ B QP A U )= ( A Uλ A ) ◦ ( σ A AU ) ◦ ( µ B QP A U ) = ( A Uλ A ) ◦ ( σ A AU ) ◦ ( Q B µ P A U )= σ A A ◦ (Qp P ) ◦ ( Q B µ P A U ) (7)= σ A A ◦ ( Q B µ PA)◦ (QBpP )and since QBp P is an epimorphism, we get thatσ A A ◦ ( µ B QP A)= σAA ◦ ( Q B µ PA)i.e.( ) ) ( ) ( )AU A σAA ◦(AUµ B AQP A = AU A σAA ◦ AU A Q B µ PA .□

Since A U reflects and ( A Q BB P A , p A QBP A ) = Coequ Fun(µBQ P A , A Q B µ PA), there existsa unique functorial morphism AB σ A BA : A Q BB P A → Id A A such thatABσ A BA ◦ (p A QBP A ) = A σ A A.Using definition of σ A A , Bσ A B and Bσ A BA , naturality of λ B we compute(AU AB σ A BA)◦ (A Up A QBP A ) ◦ (Qp P ) = A U [ ABσ A BA ◦ (p A QBP A ) ] ◦ (Qp P )= ( AU A σ A A)◦ (QpP ) = ( A Uλ A ) ◦ ( σ A AU ) = ( A Uλ A ) ◦ ( Bσ A BAU ) ◦ (p QB P A U)(14)= ( A Uλ A ) ◦ ( Bσ A BAU ) ◦ (Q B λ BB P A U)= B σ A BA ◦ (Q B p B P ) ◦ (Q B λ BB P A U) = B σ A BA ◦ (Q B λ BB P A ) ◦ (Q BB F B Up B P )= B σ A BA ◦ (Q B λ BB P A ) ◦ (Qp P ) = ( AU AB σ A BA)◦ (A U A Q B λ BB P A ) ◦ (Qp P )and since Qp P is an epimorphism and A U reflects and by definition of AB σ A BA we getABσ A BA ◦ ( A Q B λ BB P A ) = AB σ A BA ◦ (p A QBP A ) = A σ A Aso that ɛ (A Q,P A ) = AB σ A BA ◦ ( AQ B λ BB P A ) = A σ A A .Lemma 6.23. Let M = (A, B, P, Q, σ A , σ B ) be a formal dual structure where theunderlying functors are A : A → A, B : B → B, P : A → B and Q : B → A.Assume that both categories A and B have coequalizers and the functors A, QBpreserve them. Assume that• C = ( C, ∆ C , ε C) is a comonad on the category A such that C preservescoequalizers• ˜C(= ˜C, ∆C e, ε e )Cis a lifting of the comonad C to the category A A( )• AQ, eC ρ A Q is a left ˜C-comodule functor• M is a tame Morita context.Then can 1 is an isomorphism if and only if A can A is an isomorphism if and onlyif A Q is a left ˜C-Galois functor.Proof. Assume that M is a tame Morita context. Then, by Corollary 6.22, ( A Q, P A )is an adjunction with counit ɛ := A σA A : AQP ( A → Id A ( A. Then, ) A Q is a left ˜C-Galois functor if and only if the morphism ˜CA σA)A eC◦ ρ A QP A = A can A is anisomorphism. By using Lemma 6.17 we deduce that can 1 is an isomorphism if andonly if A can A is an isomorphism if and only if A Q is a left ˜C-Galois functor. □The following Theorem is a formulation, in pure categorical terms, of [BV, Theorem<strong>2.</strong>18].Theorem 6.24. Let M = (A, B, P, Q, σ A , σ B ) be a regular tame Morita context.Assume that• both categories A and B have equalizers and coequalizers,• the functors A and B preserve equalizers,• the functors A, B, P, Q preserve coequalizers.Then the existence of the following structures are equivalent:(a) A herd τ : Q → QP Q for M101□

102(b) A comonad C = ( C, ∆ C , ε C) on the category A such that the functor Cpreserves equalizers and a mixed distributive law Φ : AC → CA such thatAQ is a Galois comodule functor over ˜C (where ˜C is the lifting of C)(c) A comonad D = ( D, ∆ D , ε D) on the category B such that the functor Dpreserves equalizers and an opposite mixed distributive law Ψ : DB → BDsuch that B P is a Galois comodule functor over ˜D (where ˜D is the lifting ofD).Proof. By Proposition 6.11 the pairs ( A Q, P A ) and ( B P , Q B ) are adjunctions andhence P A and Q B preserve equalizers. Since A = A U A F and B = B U B F preserveequalizers, by Lemma 3.22 also A F and B F preserve them so that, in view of (15),we get that P = P AA F and Q = Q BB F preserve equalizers.(a) ⇒ (b) Assume that τ : Q → QP Q is a herd for M = ( A, B, P, Q, σ A , σ B) . ByProposition 6.14 there exists a mixed distributive law Φ : AC → CA such that(iA) ◦ Φ = ( QP σ A) ◦ (τP ) ◦ (A µ Q P ) ◦ (Ai) .Then, by Theorem 5.7, there exists a lifting comonad ˜C =(˜C, ∆e C, ε e C)on the categoryA A. By Proposition 6.16, there(exists)a functorial morphism eC ρ A Q : A Q → ˜C A Qsuch that A U eC ρ A Q = C ρ Q and AQ, eC ρ A Q is a left ˜C-comodule functor. Since byassumption we have a regular formal dual structure, by Theorem 6.6, the functorialmorphism can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CA is an isomorphism and so, byLemma 6.23, A Q is a left ˜C-Galois functor.(b) ⇒ (a) Follows by [BM, Theorem 4.4 (1)] where(T , (N A , R A ), (N B , R B ), C, ξ) = ( A A, ( A F , A U), ( A Q, P A ), C, A U A can A ) noting that apretorsor for a formal dual structure is a herd.□6.7. Copretorsors.Proposition 6.25. Let A and B be categories with coequalizers and let P : A →B, Q : B → A, and C : A → A be functors. Assume that all the functors P, Q andC preserve coequalizers. Let ε C : C → A be a functorial morphism and assume that(A, εC ) = Coequ Fun(Cε C , ε C C ) . Let χ : QP Q → Q be a functorial morphism suchthat(98) χ ◦ (QP χ) = χ ◦ (χP Q)and let δ C : C → QP be a functorial morphism such that(99) χ ◦ (δ C Q) = ε C Q.Let w l = (χP ) ◦ (QP δ C ) and w r = QP ε C : QP C → QP . Set(100) (A, x) = Coequ Fun(w l , w r) .There exists a functorial morphism A µ Q : AQ → Q such that(101)A µ Q ◦ (xQ) = χ.There exist functorial morphisms m A : AA → A and u A : A → A such that A =(A, m A , u A ) is a monad over A that preserves coequalizers. Moreover m A and u A

are uniquely determined by(102) x ◦ (χP ) = m A ◦ (xx)and(103) u A ◦ ε C = x ◦ δ C .Finally ( Q, A µ Q)is a left A-module functor.Proof. We haveHenceχ ◦ ( w l Q ) = χ ◦ (χP Q) ◦ (QP δ C Q) (98)= χ ◦ (QP χ) ◦ (QP δ C Q)By Lemma <strong>2.</strong>9, we have that99= χ ◦ ( QP ε C Q ) = χ ◦ (w r Q) .χ ◦ ( w l Q ) = χ ◦ (w r Q) .(AQ, xQ) = Coequ Fun(w l Q, w r Q )and hence there exists a unique functorial morphism A µ Q : AQ → Q which fulfils(101). We computex ◦ ( A µ Q P ) ◦ (Aw l ) ◦ (xQP C) x = x ◦ ( A µ Q P ) ◦ (xQP ) ◦ ( QP w l)(101)= x ◦ (χP ) ◦ ( QP w l) = x ◦ (χP ) ◦ (QP χP ) ◦ (QP QP δ C )(98)= x ◦ (χP ) ◦ (χP QP ) ◦ (QP QP δ C )χ= x ◦ (χP ) ◦ (QP δ C ) ◦ (χP C) = x ◦ w l ◦ (χP C)xcoequ= x ◦ w r ◦ (χP C) = x ◦ ( QP ε C) ◦ (χP C)χ= x ◦ (χP ) ◦ ( QP QP ε C) (101)= x ◦ ( A µ Q P ) ◦ (xQP ) ◦ ( QP QP ε C)x= x ◦ ( A µ Q P ) ◦ ( AQP ε C) ◦ (xQP C) = x ◦ ( A µ Q P ) ◦ (Aw r ) ◦ (xQP C)so that we getx ◦ ( A µ Q P ) ◦ (Aw l ) ◦ (xQP C) = x ◦ ( A µ Q P ) ◦ (Aw r ) ◦ (xQP C)and since xQP C is an epimorphism we deduce thatx ◦ ( A µ Q P ) ◦ (Aw l ) = x ◦ ( A µ Q P ) ◦ (Aw r ).By Corollary <strong>2.</strong>12, A preserves coequalizers so that we get(AA, Ax) = Coequ Fun(Aw l , Aw r) .Hence there exists a unique functorial morphism m A : AA → A such that(104) m A ◦ (Ax) = x ◦ ( A µ Q P )or equivalentlym A ◦ (xx) = m A ◦ (Ax) ◦ (xQP ) = x ◦ ( A µ Q P ) ◦ (xQP ) (101)= x ◦ (χP ) .103

104We calculateso that we getx ◦ δ C ◦ ( Cε C) δ C= x ◦ ( QP ε C) ◦ (δ C C) = x ◦ w r ◦ (δ C C)xcoequ= x ◦ w l ◦ (δ C C) = x ◦ (χP ) ◦ (QP δ C ) ◦ (δ C C)δ C= x ◦ (χP ) ◦ (δ C QP ) ◦ (Cδ C ) (99)= x ◦ ( ε C QP ) ◦ (Cδ C )ε C = x ◦ δ C ◦ (ε C C)x ◦ δ C ◦ (Cε C ) = x ◦ δ C ◦ (ε C C).Since ( A, ε C) = Coequ Fun(Cε C , ε C C ) there exists a unique functorial morphismu A : A → A such that (103) is fulfilled. Now we want to show that A = (A, m A , u A )is a monad over A that isWe calculatem A ◦ (m A A) = m A ◦ (Am A )m A ◦ (Au A ) = A = m A ◦ (u A A) .m A ◦ (m A A) ◦ (xxx) = m A ◦ (m A A) ◦ (xxA) ◦ (QP QP x)(102)= m A ◦ (xA) ◦ (χP A) ◦ (QP QP x)χ= m A ◦ (xA) ◦ (QP x) ◦ (χP QP ) = m A ◦ (xx) ◦ (χP QP )(102)= x ◦ (χP ) ◦ (χP QP ) (98)= x ◦ (χP ) ◦ (QP χP )(102)= m A ◦ (xx) ◦ (QP χP ) = m A ◦ (xA) ◦ (QP x) ◦ (QP χP )(102)= m A ◦ (xA) ◦ (QP m A ) ◦ (QP xx) x = m A ◦ (Am A ) ◦ (xAA) ◦ (QP xx)Thus we get that= m A ◦ (Am A ) ◦ (xxx) .m A ◦ (m A A) ◦ (xxx) = m A ◦ (Am A ) ◦ (xxx)and since xxx is an epimorphism, we deduce that m A is associative. We computeThus we get thatm A ◦ (Au A ) ◦ ( Aε C) ◦ (xC) (103)= m A ◦ (Ax) ◦ (Aδ C ) ◦ (xC)x= m A ◦ (Ax) ◦ (xQP ) ◦ (QP δ C ) = m A ◦ (xx) ◦ (QP δ C )(102)= x ◦ (χP ) ◦ (QP δ C ) = x ◦ w l= x ◦ w r = x ◦ (QP ε C ) x = ( Aε C) ◦ (xC) .m A ◦ (Au A ) ◦ ( Aε C) ◦ (xC) = ( Aε C) ◦ (xC) .and since ( Aε C) ◦ (xC) is epimorphism we deduce thatm A ◦ (Au A ) = A.

105We computeso that we getm A ◦ (u A A) ◦ ( ε C A ) ◦ (Cx) (103)= m A ◦ (xA) ◦ (δ C A) ◦ (Cx)δ C= m A ◦ (xA) ◦ (QP x) ◦ (δ C QP ) = m A ◦ (xx) ◦ (δ C QP )(102)= x ◦ (χP ) ◦ (δ C QP ) (99)= x ◦ (ε C QP ) εC = ( ε C A ) ◦ (Cx)m A ◦ (u A A) ◦ ( ε C A ) ◦ (Cx) = ( ε C A ) ◦ (Cx)and since ( ε C A ) ◦ (Cx) is an epimorphism we deduce thatm A ◦ (u A A) = A.Therefore we obtain that m A is unital. We computeA µ Q ◦ ( A A µ Q)◦ (AxQ) ◦ (xQP Q)(101)= A µ Q ◦ (Aχ) ◦ (xQP Q) x = A µ Q ◦ (xQ) ◦ (QP χ)(101)= χ ◦ (QP χ) (98)= χ ◦ (χP Q) (101)= A µ Q ◦ (xQ) ◦ (χP Q)(102)= A µ Q ◦ (m A Q) ◦ (xxQ) = A µ Q ◦ (m A Q) ◦ (AxQ) ◦ (xQP Q) .Since (AxQ) ◦ (xQP Q) is an epimorphism we getWe calculateA µ Q ◦ ( A A µ Q)= A µ Q ◦ (m A Q) .A µ Q ◦ (u A Q) ◦ ( ε C Q ) (103)= A µ Q ◦ (xQ) ◦ (δ C Q)(101)= χ ◦ (δ C Q) (99)= ( ε C Q ) .Since ( ε C Q ) is an epimorphism we obtainA µ Q ◦ (u A Q) = Q.Proposition 6.26. Let A and B be categories with coequalizers and let P : A → B,Q : B → A, and D : B → B be functors. Assume that all the functors P, Q and Dpreserve coequalizers. Let ε D : D → B be a functorial morphism and assume that(B, εD ) = Coequ Fun(Dε D , ε D D ) . Let χ : QP Q → Q be a functorial morphism suchthatχ ◦ (QP χ) = χ ◦ (χP Q) .Let δ D : D → P Q be a functorial morphism such that(105) χ ◦ (Qδ D ) = Qε D .Let z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q. Set(106) (B, y) = Coequ Fun(z l , z r) .There exists a functorial morphism µ B Q : QB → Q such that(107) µ B Q ◦ (Qy) = χ.□

106There exist functorial morphisms m B : BB → B and u B : B → B such thatB = (B, m B , u B ) is a monad over B that preserves coequalizers. Moreover m B andu B are uniquely determined by(108) m B ◦ (yB) = y ◦ ( )P µ B Qor equivalently(109) m B ◦ (yy) = y ◦ (P χ)and(110) y ◦ δ D = u B ◦ ε D .Moreover ( Q, µ B Q)is a right B-module functor.Proof. By left-right symmetric argument of those used in proof of Proposition 6.25,one can prove this Proposition.□Definition 6.27. Let A and B be categories. A preformal codual structure is aeightuple Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) where C : A → A, D : B → B, P : A → Band Q : B → A are functors, δ C : C → QP, δ D : D → P Q, ε C : C → A, ε D : D → Bare functorial morphisms. A copretorsor χ for Θ is a functorial morphism χ :QP Q → Q satisfying the following conditions:1) Coassociativity, in the sense that(111) χ ◦ (χP Q) = χ ◦ (QP χ)2) Counitality, in the sense that(112) χ ◦ (δ C Q) = ε C Qand(113) χ ◦ (Qδ D ) = Qε D .Definition 6.28. A preformal codual structure Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D)will be called regular whenever ( A, ε C) (= Coequ Fun Cε C , ε C C ) and ( (B, ε D) =Coequ Fun Dε D , ε D D ) . In this case a copretorsor for Θ will be called a regularcopretorsor.Theorem 6.29. Let A and B be categories with coequalizers and let χ : QP Q →Q be a regular copretorsor for Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) . Assume that theunderlying functors P, Q, C and D preserve coequalizers. Let w l = (χP ) ◦ (QP δ C )and w r = QP ε C : QP C → QP . Set(114) (A, x) = Coequ Fun(w l , w r) .There exists a functorial morphism A µ Q : AQ → Q such that(115)A µ Q ◦ (xQ) = χ.There exist functorial morphisms m A : AA → A and u A : A → A such that A =(A, m A , u A ) is a monad over A that preserves coequalizers. Moreover m A and u Aare uniquely determined byx ◦ (χP ) = m A ◦ (xx) and x ◦ δ C = u A ◦ ε C .

Moreover ( Q, A µ Q)is a left A-module functor.Let z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q. Set(116) (B, y) = Coequ Fun(z l , z r) .There exists a functorial morphism µ B Q : QB → Q such that(117) µ B Q ◦ (Qy) = χ.There exist functorial morphisms m B : BB → B and u B : B → B such thatB = (B, m B , u B ) is a monad over B that preserves coequalizers. Moreover m B andu B are uniquely determined bym B ◦ (yy) = y ◦ (P χ) and y ◦ δ D = u B ◦ ε D .Moreover ( Q, µ B Q)is a right B-module functor.Finally ( Q, A µ Q , µ B Q)is an A-B-module functor.Proof. Within these assumption, we can apply Proposition 6.25 to get the monad Aand the functorial morphism A µ Q : AQ → Q satisfying 115 such that ( )Q, A µ Q is aleft A-module functor and Proposition 6.26 to get the monad B and the functorialmorphism µ B Q : QB → Q satisfying 117 such that ( Q, µ Q) B is a right B-modulefunctor. Let us check the compatibility condition. We calculateA µ Q ◦ ( )Aµ B Q ◦ (AQy) ◦ (xQP Q) = x A µ Q ◦ (xQ) ◦ ( )QP µ B Q ◦ (QP Qy)(115),(117)= χ ◦ (QP χ) (98)= χ ◦ (χP Q)(117),(115)= µ B Q ◦ (Qy) ◦ (A µ Q P Q ) ◦ (xQP Q)A µ Q= µBQ ◦ (A µ Q B ) ◦ (AQy) ◦ (xQP Q) .Since (AQy) ◦ (xQP Q) is an epimorphism we get thatA µ Q ◦ ( )Aµ B Q = µBQ ◦ (A µ Q B ) .Therefore (Q, A µ Q , µ B Q ) is an A-B-bimodule functor.□Theorem 6.30. Let χ : QP Q → Q be a regular copretorsor for a preformal codualstructure Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) on categories A and B such that the underlyingfunctors P, Q, C and D preserve coequalizers. Assume that C and D arecomonads, ( P, D ρ P)is a left D-comodule functor and(P, ρCP)is a right C-comodulefunctor. Moreover assume that the functorial morphism δ C is right C-colinear, thatis ( Qρ C P)◦ δC = (δ C C) ◦ ∆ C and the functorial morphism δ D is left D-colinear thatis (D ρ P Q ) ◦ δ D = (Dδ D ) ◦ ∆ D and that they are compatible in the sense that(118) (δ D P ) ◦ D ρ P = (P δ C ) ◦ ρ C P .Then there exists a monad A = (A, m A , u A ) on the category A together with afunctorial morphism A µ Q : AQ → Q such that ( Q, A µ Q)is a left A-module functorand a monad B = (B, m B , u B ) together with a functorial morphism µ B Q : QB → Qsuch that ( Q, µ B Q)is a right B-module functor. The underlying functors are definedas follows(A, x) = Coequ Fun((χP ) ◦ (QP δC ) , QP ε C) .107

108andsatisfying(B, y) = Coequ Fun((P χ) ◦ (δD P Q) , ε D P Q ) .A µ Q ◦ (xQ) = χ and µ B Q ◦ (Qy) = χ.Furthermore1) The morphisms cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP is an isomorphism.2) The morphism cocan 1 := ( P µ B Q)◦ (δD B) : DB → P Q is an isomorphisms.3)4)5)6)(χP ) ◦ (QP δ C ) = cocan 1 ◦ (xC) and (P χ) ◦ (δ D P Q) = cocan 1 ◦ (Dy)x = ( Aε C) ◦ (cocan 1 ) −1y = ( ε D B ) ◦ (cocan 1 ) −1δ C = (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)δ D = ( P µ B Q)◦ (δD B) ◦ (Du B ) .From the last equalities, we deduce that if ε C A is a regular epimorphism, so is σ Aand if Bε D is a regular epimorphism, so is σ B .Proof. Note that we are in the setting of Theorem 6.29.1) Let us check that cocan 1 is an isomorphism.( The) inverse of the functorial morphism cocan 1 is given by cocan −11 = (xC) ◦QρCP : QP → AC. Indeed we compute(xC) ◦ ( Qρ C P)◦ cocan1 ◦ (xC) = (xC) ◦ ( Qρ C P)◦( Aµ Q P ) ◦ (Aδ C ) ◦ (xC)= (xC) ◦ ( Qρ C P)◦( Aµ Q P ) ◦ (xQP ) ◦ (QP δ C )(115)= (xC) ◦ ( Qρ C P)◦ (χP ) ◦ (QP δC ) = (xC) ◦ (χP C) ◦ ( QP Qρ C P)◦ (QP δC )P rightCcol= (xC) ◦ (χP C) ◦ (QP δ C C) ◦ ( QP ∆ C) = (xC) ◦ ( w l C ) ◦ ( QP ∆ C)x coequ= (xC) ◦ (w r C) ◦ ( QP ∆ C) = (xC) ◦ ( QP ε C C ) ◦ ( QP ∆ C) C comonad= (xC) .Since xC is an epimorphism, we obtain that(xC) ◦ ( Qρ C P)◦ cocan1 = AC.On the other hand, we havecocan 1 ◦ (xC) ◦ ( ) (Qρ C P = Aµ Q P ) ◦ (Aδ C ) ◦ (xC) ◦ ( )Qρ C P= (A µ Q P ) ◦ (xQP ) ◦ (QP δ C ) ◦ ( )Qρ C (115)P = (χP ) ◦ (QP δ C ) ◦ ( )Qρ C P(118)= (χP ) ◦ (Qδ D P ) ◦ ( )Q D ρ P(113)= ( Qε D P ) ◦ ( Q D ρ P) D ρ P counital= QP

so we obtain thatcocan −11 = (xC) ◦ ( Qρ C P).2) Similarly we prove that we prove that cocan 1 := ( P µ B Q)◦ (δD B) : DB → P Q isan isomorphism with (Dy) ◦ (D ρ P Q ) its inverse. In fact we have(Dy) ◦ (D ρ P Q ) ◦ cocan 1 ◦ (Dy) = (Dy) ◦ (D ρ P Q ) ◦ ( P µ B Q)◦ (δD B) ◦ (Dy)D ρ P= (Dy) ◦(DP µBQ)◦( Dρ P QB ) ◦ (δ D B) ◦ (Dy)δ D leftDcol= (Dy) ◦ ( DP µ Q) B ◦ (DδD B) ◦ ( ∆ D B ) ◦ (Dy)∆= D(Dy) ◦ ( )DP µ B Q ◦ (DδD B) ◦ (DDy) ◦ ( ∆ D P Q )δ=D(Dy) ◦ ( DP µ Q) B ◦ (DP Qy) ◦ (DδD P Q) ◦ ( ∆ D P Q )(117)= (Dy) ◦ (DP χ) ◦ (Dδ D P Q) ◦ ( ∆ D P Q )ycoequ= (Dy) ◦ ( Dε D P Q ) ◦ ( ∆ D P Q ) Dcomonad= Dyand since D preserves coequalizers, Dy is an epimorphism, so that we get(Dy) ◦ (D ρ P Q ) ◦ cocan 1 = DB.On the other hand we havecocan 1 ◦ (Dy) ◦ (D ρ P Q ) = ( )P µ B Q ◦ (δD B) ◦ (Dy) ◦ (D ρ P Q )δ= ( DP µ Q) B ◦ (P Qy) ◦ (δD P Q) ◦ (D ρ P Q ) (117)= (P χ) ◦ (δ D P Q) ◦ (D ρ P Q )so that we get3) We haveso that(118)= (P χ) ◦ (P δ C Q) ◦ ( ρ C P Q ) (112)= ( P ε C Q ) ◦ ( ρ C P Q )P com= P Qcocan 1 ◦ (Dy) ◦ (D ρ P Q ) = P Q.(χP ) ◦ (QP δ C ) (115)= (A µ Q P ) ◦ (xQP ) ◦ (QP δ C )x= (A µ Q P ) ◦ (Aδ C ) ◦ (xC) defcocan 1= cocan 1 ◦ (xC)(119) (χP ) ◦ (QP δ C ) = cocan 1 ◦ (xC) .Similarly we haveso that(P χ) ◦ (δ D P Q) (117)= ( P µ B Q)◦ (P Qy) ◦ (δD P Q)δ= ( DP µ Q) B ◦ (δD B) ◦ (Dy) defcocan 1= cocan 1 ◦ (Dy)(120) (P χ) ◦ (δ D P Q) = cocan 1 ◦ (Dy) .109

1104) With notations of Theorem 6.29, we havex ◦ cocan 1 ◦ (xC) (119)= x ◦ (χP ) ◦ (QP δ C )= x ◦ w l xcoequ= x ◦ w r = x ◦ ( QP ε C) x = ( Aε C) ◦ (xC) .Since xC is an epimorphism, we deduce thatand henceSimilarly, we havex ◦ cocan 1 = Aε Cx = ( Aε C) ◦ (cocan 1 ) −1 .y ◦ cocan 1 ◦ (Dy) (120)= y ◦ (P χ) ◦ (δ D P Q)= y ◦ z l ycoequ= y ◦ z r = y ◦ ( ε D P Q ) ε D = ( ε D B ) ◦ (Dy) .Since Dy is an epimorphism, we deduce thatand hence5) We have thatso thaty ◦ cocan 1 = ε D By = ( ε D B ) ◦ (cocan 1 ) −1 .δ C = (A µ Q P ) ◦ (u A QP ) ◦ δ Cu A= (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)(121) δ C = (A µ Q P ) ◦ (Aδ C ) ◦ (u A C) .Since (A µ Q P ) ◦ (Aδ C ) = cocan 1 is an isomorphism, we will prove that if u A C isa regular monomorphism, so is δ C . In fact, let (C, u A C) = Equ Fun (υ, ϖ) whereυ, ϖ : AC → T. We know that (A µ Q P ) ◦ (Aδ C ) = cocan 1 is an isomorphism withinverse (xC) ◦ ( )Qρ C P so that we haveso thati.e.υ ◦ (cocan 1 ) −1 ◦ δ C(121)= υ ◦ (cocan 1 ) −1 ◦ (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)cocan 1 iso= υ ◦ (u A C) u ACequ= ϖ ◦ (u A C)cocan 1 iso= ϖ ◦ (cocan 1 ) −1 ◦ (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)(121)= ϖ ◦ (cocan 1 ) −1 ◦ δ Cυ ◦ (cocan 1 ) −1 ◦ δ C = ϖ ◦ (cocan 1 ) −1 ◦ δ Cυ ◦ (xC) ◦ ( Qρ C P)◦ δC = ϖ ◦ (xC) ◦ ( Qρ C P)◦ δC .Moreover, for every ξ : X → QP such thatυ ◦ (xC) ◦ ( Qρ C P)◦ ξ = ϖ ◦ (xC) ◦(QρCP)◦ ξ,

since (C, u A C) = Equ Fun (υ, ϖ) , there exists a unique functorial morphism ξ : X →C such that(u A C) ◦ ξ = (xC) ◦ ( Qρ C P)◦ ξ.Then, by composing to the left with the isomorphism cocan 1 := (A µ Q P ) ◦ (Aδ C ) weget ( Aµ Q P ) ◦ (Aδ C ) ◦ (u A C) ◦ ξ = (A µ Q P ) ◦ (Aδ C ) ◦ (xC) ◦ ( )Qρ C P ◦ ξi.e. by (121) we have(122) δ C ◦ ξ = ξ.Then ξ : X → C is the unique morphism satisfying condition (122) so that(C, δ C ) = Equ Fun(υ ◦ (xC) ◦(QρCP), ϖ ◦ (xC) ◦(QρCP))i.e. also δ C is a regular monomorphism.6) We have thatso thatδ D = ( P µ B Q)◦ (P QuB ) ◦ δ Du B= ( P µ B Q)◦ (δD B) ◦ (Du B )(123) δ D = ( P µ B Q)◦ (δD B) ◦ (Du B ) .Since ( P µ Q) B ◦ (δD B) = cocan 1 is an isomorphism, we will prove that if Du B isa regular monomorphism, so is δ D . In fact, let (D, Du B ) = Equ Fun (ζ, θ) whereζ, θ : DB → L. We know that ( P µ Q) B ◦ (δD B) = cocan 1 is an isomorphism withinverse (Dy) ◦ (D ρ P Q ) so that we haveζ ◦ (cocan 1 ) −1 ◦ δ D(123)= ζ ◦ (cocan 1 ) −1 ◦ ( P µ B Q)◦ (δD B) ◦ (Du B )cocan 1 iso= ζ ◦ (Du B ) Du Bequ= θ ◦ (Du B )cocan 1 iso= θ ◦ (cocan 1 ) −1 ◦ ( P µ B Q)◦ (δD B) ◦ (Du B )(123)= θ ◦ (cocan 1 ) −1 ◦ δ Dso thatζ ◦ (cocan 1 ) −1 ◦ δ D = θ ◦ (cocan 1 ) −1 ◦ δ Di.e.ζ ◦ (Dy) ◦ (D ρ P Q ) ◦ δ D = θ ◦ (Dy) ◦ (D ρ P Q ) ◦ δ D .Moreover, for every ν : Y → P Q such thatζ ◦ (Dy) ◦ (D ρ P Q ) ◦ ν = θ ◦ (Dy) ◦ (D ρ P Q ) ◦ ν,since (D, Du B ) = Equ Fun (ζ, θ) , there exists a unique functorial morphism ν : Y →D such that(Du B ) ◦ ν = (Dy) ◦ (D ρ P Q ) ◦ ν.Then, by composing to the left with the isomorphism cocan 1 := we get( )P µBQ ◦ (δD B) ◦ (Du B ) ◦ ν = ( )P µ B Q ◦ (δD B) ◦ (Dy) ◦ (D ρ P Q ) ◦ νi.e. by (123) we have(124) δ D ◦ ν = ν.111

112Then ν : Y → D is the unique morphism satisfying condition (124) so that(D, δ D ) = Equ Fun(ζ ◦ (Dy) ◦( Dρ P Q ) , θ ◦ (Dy) ◦ (D ρ P Q ))i.e. also δ D is a regular monomorphism.6.8. Coherds. Following [BV], by formally dualizing definitions of formal dualstructure and herd, the notions of formal codual structure and of coherd are introduced.Definition 6.3<strong>1.</strong> A formal codual structure on two categories A and B is a sextupleX = (C, D, P, Q, δ C , δ D ) where C = ( C, ∆ C , ε C) and D = ( D, ∆ D , ε D) arecomonads on on A and B respectively and ( C, D, P, Q, δ C , δ D , ε C , ε D) is a preformalcodual structure. Moreover ( (P : A → B, D ρ P : P → DP, ρ C P : P → P C) andQ : B → A, C ρ Q : Q → CQ, ρ D Q : Q → QD) are bicomodule functors; δ C : C →QP, δ D : D → P Q are subject to the following conditions: δ C is C-bicolinear andδ D is D-bicolinear(125) ( C ρ Q P ) ◦ δ C = (Cδ C ) ◦ ∆ C and ( )Qρ C P ◦ δC = (δ C C) ◦ ∆ C( )(126) P ρDQ ◦ δD = (δ D D) ◦ ∆ D and (D ρ P Q ) ◦ δ D = (Dδ D ) ◦ ∆ Dand the coassociative conditions hold(127) (δ C Q) ◦ C ρ Q = (Qδ D ) ◦ ρ D Q and (δ D P ) ◦ D ρ P = (P δ C ) ◦ ρ C P .Definition 6.3<strong>2.</strong> Consider a formal codual structure X = (C, D, P, Q, δ C , δ D ) inthe sense of the previous definition. A coherd for X is a copretorsor χ : QP Q → Qi.e.(128) χ ◦ (χP Q) = χ ◦ (QP χ)(129) χ ◦ (δ C Q) = ε C Qand(130) χ ◦ (Qδ D ) = Qε D .Definition 6.33. A formal codual structure X = (C, D, P, Q, δ C , δ D ) will be calledregular whenever ( C, D, P, Q, δ C , δ D , ε C , ε D) is a regular preformal codual structure.In this case a coherd for X will be called a regular coherd.Lemma 6.34. Let X = (C, D, P, Q, δ C , δ D ) be a formal codual structure and let χ :QP Q → QQ be a coherd for X. Assume that the underlying functors C and D reflectcoequalizers. Then χ is a regular coherd.Proof. Since C and D are comonads, we have ( Cε C) ◦ ∆ C = Id C and ( Dε D) ◦ ∆ D =Id D . Thus, Cε C and Dε D are split epimorphisms and thus epimorphisms. Since Cand D reflect coequalizers, we deduce that also ε C and ε D are epimorphisms andthus ( A, ε C) (= Coequ Fun ε C C, Cε C) and ( B, ε D) (= Coequ Fun ε D D, Dε D) , i.e. χ isa regular coherd.□Proposition 6.35. Let X = (C, D, P, Q, δ C , δ D ) be a formal codual structure suchthat the lifted functors C Q D : D B → C A and D P C : C A → D B determine an equivalenceof categories. Then ( Q D , D P ) and ( P C , C Q ) are adjunctions.□

113Proof. Since (C U, ) and (D U, D F ) are adjunctions, (C U C Q D , D P CC F ) = ( Q D , D P )and (D U D P C , C Q DD F ) = ( P C , C Q ) are also adjunctions.□6.9. Coherds and Monads. In this subsection we prove that in the case whenthere exist coequalizers in the base categories and all functors occurring in a formalcodual structure preserve them, we establish an equivalence between coherds onone hand, and monads on the other hand, together with two natural isomorphismgenerating a Galois map.Theorem 6.36. Let A and B be categories in both of which the coequalizer of anypair of parallel morphisms exists. Let X = (C, D, P, Q, δ C , δ D ) be a formal codualstructure on A and B. Then we have(1) If A = (A, m A , u A ) is a monad on the category A and ( Q, A µ Q : AQ → Q )is a left A-module functor such that(i) the functorial morphism cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP is anisomorphism(ii) the functorial morphism cocan 2 := (A µ Q D ) ◦ ( AρQ) D : AQ → QD is anisomorphismthen χ := A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) : QP Q → Q is a copretorsor andthus a coherd.(2) If B = (B, m B , u B ) is a monad on the category B and ( Q, µ B Q : QB → Q) isa right B-module functor such that(i) the functorial morphism cocan 1 := ( P µ Q) B ◦ (δD B) : DB → P Q is anisomorphism(ii) the functorial morphism cocan 2 := ( (Cµ Q) B ◦ Cρ Q B ) : QB → CQ is anisomorphismthen χ := µ B Q ◦ ( Qε D B ) ◦ (Qcocan −1 1 ) : QP Q → Q is a copretorsor andthus a coherd.Proof. Let us prove 1), the other is similar. We have to prove thatχ := A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q )satisfies (111) , (112) , (113) . We compute( A µ Q ass)=cocan 1 ◦ (m A C) = (A µ Q P ) ◦ (Aδ C ) ◦ (m A C)= (A µ Q P ) ◦ (m A QP ) ◦ (AAδ C )( Aµ Q P ) ◦ ( A A µ Q P ) ◦ (AAδ C ) = (A µ Q P ) ◦ (Acocan 1 )so we obtaincocan 1 ◦ (m A C) = (A µ Q P ) ◦ (Acocan 1 )i.e.(131) (m A C) ◦ Acocan −11 = ( ) (cocan −11 ◦ Aµ Q P ) .We computeχ ◦ (QP χ)= A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (QP A µ Q )◦ ( QP Aε C Q ) ◦ ( QP cocan −11 Q )

114= A µ Q ◦ ( Aε C Q ) ◦ (AC A µ Q )◦ ( ACAε C Q ) ◦ ( ACcocan −11 Q ) ◦ ( cocan −11 QP Q )= A µ Q ◦ (A A µ Q )◦ ( Aε C AQ ) ◦ ( ACAε C Q ) ◦ ( ACcocan −11 Q ) ◦ ( cocan −11 QP Q )= A µ Q ◦ (m A Q)◦ ( Aε C AQ ) ◦ ( ACAε C Q ) ◦ ( ACcocan −11 Q ) ◦ ( cocan −11 QP Q )= A µ Q ◦ (m A Q)◦ ( AAε C Q ) ◦ ( Aε C CAQ ) ◦ ( ACcocan −11 Q ) ◦ ( cocan −11 QP Q )= A µ Q ◦ ( Aε C Q ) ◦ (m A CQ) ◦ ( Acocan −11 Q ) ◦ ( Aε C QP Q ) ◦ ( cocan −11 QP Q )(131)= A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ ( A µ Q P Q)◦ ( Aε C QP Q ) ◦ ( cocan −11 QP Q )Note that we haveso we have= χ ◦ (χP Q) .(cocan 1 ) ◦ (u A C) = (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)= (A µ Q P ) ◦ (u A QP ) ◦ δ C A µ Q unital= δ C(132) (cocan 1 ) ◦ (u A C) = δ C .Now we computeand so we getWe haveχ ◦ (δ C Q) = A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (δ C Q)(132)= A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (cocan 1 Q) ◦ (u A CQ)= A µ Q ◦ ( Aε C Q ) ◦ (u A CQ) = A µ Q ◦ (u A Q) ◦ ( ε C Q ) ( A µ Q unital)=χ ◦ (δ C Q) = ( ε C Q ) .(Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) ◦ cocan 2= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) ◦ (A µ Q D ) ◦ ( Aρ D Q= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (A µ Q P Q ) ◦ (AQδ D ) ◦ ( Aρ D Q(127)= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (A µ Q P Q ) ◦ (Aδ C Q) ◦ ( A C ρ Q)= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (cocan 1 Q) ◦ ( A C ρ Q)= ( Aε C Q ) ◦ ( A C ρ Q) C ρ Q counital= AQSince cocan 2 is an isomorphism, we have that(133) cocan 2 ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) = QD.Finally we computeρ D Q counital=χ ◦ (Qδ D ) = A µ Q ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D )))(ε C Q )A µ Q ◦ ( AQε D) ◦ ( (AρQ) D ◦ Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D )= ( Qε D) ◦ ( A µ Q D) ◦ ( ) (Aρ D Q ◦ Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D )

so that we obtain= ( Qε D) ◦ cocan 2◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) (133)= ( Qε D)χ ◦ (Qδ D ) = ( Qε D) .Theorem 6.37. Let A and B be categories in both of which the coequalizer of anypair of parallel morphisms exist. Let X = (C, D, P, Q, δ C , δ D ) be a regular formal codualstructure on A and B such that the underlying functors C, D, P and Q preservecoequalizers, then the existence of the following structures are equivalent:(a) A coherd χ : QP Q → Q in X.(b) A monad A = (A : A → A, m A : AA → A, u A : A → A) , such that the functorA preserves coequalizers, together with a left action A µ Q : AQ → Q,subject to the following conditions:(i) The natural transformation cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP isan isomorphism.(ii) The natural transformation cocan 2 := (A µ Q D ) ◦ ( AρQ) D : AQ → QD isan isomorphism.(c) A monad B = (B : B → B, m B : BB → B, u B : B → B) , such that the functorB preserves coequalizers, together with a right action µ B Q : QB → Q,subject to the following conditions:(i) The natural transformation cocan 1 := ( P µ Q) B ◦ (δD B) : DB → P Q isan isomorphism.(ii) The natural transformation cocan 2 := ( (Cµ Q) B ◦ Cρ Q B ) : QB → CQ isan isomorphism.Proof. (a) ⇒ (b) Under weaker assumptions, the monad A = (A, m A , u A ) and theaction A µ Q have been constructed in Proposition 6.25. Moreover, by Theorem 6.301) we already proved that cocan 1 is an isomorphism with (xC) ◦ ( Qρ C P)its inverse.Now we prove that ( Aε C Q ) ◦ ( cocan −11 Q ) ◦(Qδ D ) is the inverse of cocan 2 . We computeρ C P counital=cocan 2 ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D )= (A µ Q D ) ◦ ( ) (Aρ D Q ◦ Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D )= (A µ Q D ) ◦ ( Aρ D Q)◦(Aε C Q ) ◦ (xCQ) ◦ ( Qρ C P Q ) ◦ (Qδ D )= (A µ Q D ) ◦ ( (AρQ) D ◦ (xQ) ◦ QP ε C Q ) ◦ ( Qρ C P Q ) ◦ (Qδ D )( Aµ Q D ) ◦ ( )Aρ D Q ◦ (xQ) ◦ (QδD ) = (A µ Q D ) ◦ (xQD) ◦ ( )QP ρ D Q ◦ (QδD )(101)= (χD) ◦ ( )QP ρ D Q ◦ (QδD ) (126)= (χD) ◦ (Qδ D D) ◦ ( Q∆ D)(113)= ( Qε D D ) ◦ ( Q∆ D) (D comonad)= QDso we obtaincocan 2 ◦ ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) = QD.On the other hand, we have(Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) ◦ cocan 2115□

116= ( Aε C Q ) ◦ ( cocan1 −1 Q ) ◦ (Qδ D ) ◦ (A µ Q D ) ◦ ( )Aρ D Q= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (A µ Q P Q ) ◦ (AQδ D ) ◦ ( )Aρ D Q(127)= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (A µ Q P Q ) ◦ (Aδ C Q) ◦ ( A C ρ Q)= ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (cocan 1 Q) ◦ ( A C ρ Q)= ( Aε C Q ) ◦ ( A C ρ Q) C ρ Q counital= AQso we obtaincocan −12 = ( Aε C Q ) ◦ ( cocan −11 Q ) ◦ (Qδ D ) .(b) ⇒ (a) Follows from Theorem 6.36. (a) ⇔ (c) follows by similar computations.□6.10. Coherds and distributive laws. The following result is a reformulation ofTheorem <strong>2.</strong>16 in [BV] in our categorical setting.Proposition 6.38. Let A and B be categories with equalizers and let χ : QP Q →Q be a regular coherd for X = (C, D, P, Q, δ C , δ D ) where the underlying functorsP : A → B, Q : B → A, C : A → A and D : B → B preserve coequalizers.Let A = (A, m A , u A ) and B = (B, m B , u B ) be the associated monads constructed inProposition 6.25 and in Proposition 6.26. Then1) There exists a mixed distributive law between the monad A and the comonadC, Λ : AC → CA such thatΛ ◦ (xC) = λ = (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) .2) There exists an opposite mixed distributive law between the monad B and thecomonad D, Γ : DB → BD such thatΓ ◦ (Dy) = γ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .Proof. 1) Consider the functorial morphism given byQP C QP δ C−→ QP QP −→ χPQP C ρ Q P−→ CQP −→ CxCAλ = (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Recall that w l = (χP ) ◦ (QP δ C ) and w r = QP ε C : QP C → QP and let us provethatλ ◦ ( w l C ) ? = λ ◦ (w r C)that isLet us compute(Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (χP C) ◦ (QP δ C C)?= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ ( QP ε C C ) .(Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (χP C) ◦ (QP δ C C)χ= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (χP QP ) ◦ (QP QP δ C ) ◦ (QP δ C C)δ C ,(111)= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP χP ) ◦ (QP δ C QP ) ◦ (QP Cδ C )

117(112)= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ ( QP ε C QP ) ◦ (QP Cδ C )ε C = (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ ( QP ε C C )Since (AC, xC) = Coequ Fun(w l C, w r C ) ,by the universal property of coequalizers,there exists a unique functorial morphism Λ : AC → CA such thatΛ ◦ (xC) = λ = (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) .We want to prove that Λ is a mixed distributive law. We compute(Cm A ) ◦ (ΛA) ◦ (AΛ) ◦ (xxC) x = (Cm A ) ◦ (ΛA) ◦ (AΛ) ◦ (xAC) ◦ (QP xC)x= (Cm A ) ◦ (ΛA) ◦ (xCA) ◦ (QP Λ) ◦ (QP xC)defλ= (Cm A ) ◦ (CxA) ◦ (C ρ Q P A ) ◦ (χP A) ◦ (QP δ C A) ◦ (QP Cx)◦ ( QP C ρ Q P ) ◦ (QP χP ) ◦ (QP QP δ C )δ C= (Cm A ) ◦ (CxA) ◦ (C ρ Q P A ) ◦ (χP A) ◦ (QP QP x) ◦ (QP δ C QP )◦ ( QP C ρ Q P ) ◦ (QP χP ) ◦ (QP QP δ C )χ= (Cm A ) ◦ (CxA) ◦ (C ρ Q P A ) ◦ (QP x) ◦ (χP QP ) ◦ (QP δ C QP )◦ ( QP C ρ Q P ) ◦ (QP χP ) ◦ (QP QP δ C )C ρ Q= (CmA ) ◦ (CxA) ◦ (CQP x) ◦ (C ρ Q P QP ) ◦ (χP QP ) ◦ (QP δ C QP )◦ ( QP C ρ Q P ) ◦ (QP χP ) ◦ (QP QP δ C )x,(127)= (Cm A ) ◦ (Cxx) ◦ (C ρ Q P QP ) ◦ (χP QP ) ◦ (QP Qδ D P )◦ ( QP ρ D QP ) ◦ (QP χP ) ◦ (QP QP δ C )(102)= (Cx) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (χP QP ) ◦ (QP Qδ D P )◦ ( QP ρ D QP ) ◦ (QP χP ) ◦ (QP QP δ C )χ= (Cx) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (Qδ D P ) ◦ (χDP ) ◦ ( QP ρ D QP ) ◦ (QP χP )◦ (QP QP δ C )C ρ Q= (Cx) ◦ (CχP ) ◦ (CQδD P ) ◦ (C ρ Q DP ) ◦ (χDP ) ◦ ( QP ρ D QP ) ◦ (QP χP )◦ (QP QP δ C )(113)= (Cx) ◦ ( CQε D P ) ◦ (C ρ Q DP ) ◦ (χDP ) ◦ ( QP ρ D QP ) ◦ (QP χP )◦ (QP QP δ C )C ρ Q= (Cx) ◦( Cρ Q P ) ◦ ( Qε D P ) ◦ (χDP ) ◦ ( QP ρ D QP ) ◦ (QP χP ) ◦ (QP QP δ C )χ= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ ( QP Qε D P ) ◦ ( QP ρ D QP ) ◦ (QP χP ) ◦ (QP QP δ C )Qcomfun= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP χP ) ◦ (QP QP δ C )(111)= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (χP QP ) ◦ (QP QP δ C )

118so that we getχ= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (χP C)defλ= Λ ◦ (xC) ◦ (χP C) (102)= Λ ◦ (m A C) ◦ (xxC)(Cm A ) ◦ (ΛA) ◦ (AΛ) ◦ (xxC) = Λ ◦ (m A C) ◦ (xxC)and since xxC is an epimorphism we obtainLet now compute(Cm A ) ◦ (ΛA) ◦ (AΛ) = Λ ◦ (m A C) .(CΛ) ◦ (ΛC) ◦ ( A∆ C) ◦ (xC) x = (CΛ) ◦ (ΛC) ◦ (xCC) ◦ ( QP ∆ C)defλ= (CΛ) ◦ (CxC) ◦ (C ρ Q P C ) ◦ (χP C) ◦ (QP δ C C) ◦ ( QP ∆ C)defλ= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (CQP δ C ) ◦ (C ρ Q P C ) ◦ (χP C) ◦ (QP δ C C)◦ ( QP ∆ C)C ρ Q ,(125)= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (QP δ C ) ◦ (χP C)◦ ( QP Qρ C P)◦ (QP δC )χ= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (QP δ C ) ◦ ( Qρ C P)◦ (χP ) ◦ (QP δC )C ρ Q= (CCx) ◦(C C ρ Q P ) ◦ (CχP ) ◦ (CQP δ C ) ◦ ( CQρ C P)◦( Cρ Q P ) ◦ (χP ) ◦ (QP δ C )(127)= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (CQδ D P ) ◦ ( CQ D ρ P)◦( Cρ Q P ) ◦ (χP )◦ (QP δ C )(113)= (CCx) ◦ ( C C ρ Q P ) ◦ ( CQε D P ) ◦ ( CQ D ρ P)◦( Cρ Q P ) ◦ (χP ) ◦ (QP δ C )so that we getQcomfun= (CCx) ◦ ( C C ρ Q P ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Qcomfun= (CCx) ◦ ( ∆ C QP ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )∆ C= ( ∆ C A ) ◦ (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) defλ= ( ∆ C A ) ◦ Λ ◦ (xC)(CΛ) ◦ (ΛC) ◦ ( A∆ C) ◦ (xC) = ( ∆ C A ) ◦ Λ ◦ (xC)and since xC is an epimorphism we deduce thatNow we compute(CΛ) ◦ (ΛC) ◦ ( A∆ C) = ( ∆ C A ) ◦ Λ.Λ ◦ (u A C) ◦ ( ε C C ) (103)= Λ ◦ (xC) ◦ (δ C C)= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (δ C C)δ=C(Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (δ C QP ) ◦ (Cδ C )(112)= (Cx) ◦ (C ρ Q P ) ◦ ( ε C QP ) ◦ (Cδ C ) = (Cx) ◦ (C ρ Q P ) ◦ δ C ◦ ( ε C C )

(112)= (Cx) ◦ (Cδ C ) ◦ ∆ C ◦ ( ε C C ) (103)= (Cu A ) ◦ ( Cε C) ◦ ∆ C ◦ ( ε C C )Ccomonad= (Cu A ) ◦ ( ε C C )and since ε C C is an epimorphism we get thatΛ ◦ (u A C) = (Cu A ) .Finally we compute(ε C A ) ◦ Λ ◦ (xC) defλ= ( ε C A ) ◦ (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )119ε C = x ◦ ( ε C QP ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Qcomfun= x ◦ (χP ) ◦ (QP δ C ) (102)= m A ◦ (xx) ◦ (QP δ C )x= m A ◦ (xA) ◦ (QP x) ◦ (QP δ C ) (103)= m A ◦ (xA) ◦ (QP u A ) ◦ ( QP ε C)x= m A ◦ (Au A ) ◦ x ◦ ( QP ε C) Acomonad,x=and since xC is an epimorphism we get that(ε C A ) ◦ Λ = Aε C .2) Consider the functorial morphism given by(AεC ) ◦ (xC)DP Q δ DP Q−→ P QP Q P χ−→ P Q P ρD Q−→ P QD yD−→ BDγ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .Recall that z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q and let us computethat isLet us computeγ ◦ ( Dz l) ? = γ ◦ (Dz r )(yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) ◦ (DP χ) ◦ (Dδ D P Q)?= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) ◦ ( Dε D P Q ) .(yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q) ◦ (DP χ) ◦ (Dδ D P Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (P QP χ) ◦ (δD P QP Q) ◦ (Dδ D P Q)δ D ,(111)= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (P χP Q) ◦ (P QδD P Q) ◦ (δ D DP Q)(113)= (yD) ◦ ( (P ρQ) D ◦ (P χ) ◦ P Qε D P Q ) ◦ (δ D DP Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) ◦ ( Dε D P Q ) .Since (DB, Dy) = Coequ Fun(Dz l , Dz r) , there exists a functorial morphism Γ :DB → BD such thatΓ ◦ (Dy) = γ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .

120We want to prove that Γ is an opposite mixed distributive law. We compute(m B D) ◦ (BΓ) ◦ (ΓB) ◦ (Dyy) y = (m B D) ◦ (BΓ) ◦ (ΓB) ◦ (DyB) ◦ (DP Qy)defγ= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P χB) ◦ (δ D P QB) ◦ (DP Qy)δ D= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P χB) ◦ (P QP Qy) ◦ (δ D P QP Q)χ= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P Qy) ◦ (P χP Q) ◦ (δ D P QP Q)ρ D Q= (m B D) ◦ (BΓ) ◦ (yDB) ◦ (P QDy) ◦ ( P ρ D QP Q ) ◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (BΓ) ◦ (BDy) ◦ (yDP Q) ◦ ( P ρ D QP Q ) ◦ (P χP Q) ◦ (δ D P QP Q)defγ= (m B D) ◦ (ByD) ◦ ( )BP ρ D Q ◦ (BP χ) ◦ (BδD P Q) ◦ (yDP Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (yDP Q) ◦ (P QyD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (yyD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)(109)= (yD) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)(127)= (yD) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P δC QP Q) ◦ ( P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)δ=C(yD) ◦ (P χD) ◦ (P δ C QD) ◦ ( (P CρQ) D ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)(112)= (yD) ◦ ( P ε C QD ) ◦ ( ) (P Cρ D Q ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)ε= C (yD) ◦ ( ) (P ρ D Q ◦ (P χ) ◦ P ε C QP Q ) ◦ ( P C ρ Q P Q ) ◦ (P χP Q) ◦ (δ D P QP Q)Qcomfun= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (P χP Q) ◦ (δD P QP Q)(111)= (yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (P QP χ) ◦ (δD P QP Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) ◦ (DP χ)defγ= Γ ◦ (Dy) ◦ (DP χ) (109)= Γ ◦ (Dm B ) ◦ (Dyy)and since Dyy is an epimorphism we deduce thatLet us compute(m B D) ◦ (BΓ) ◦ (ΓB) = Γ ◦ (Dm B ) .(ΓD) ◦ (DΓ) ◦ ( ∆ D B ) ◦ (Dy) ∆D= (ΓD) ◦ (DΓ) ◦ (DDy) ◦ ( ∆ D P Q )

defγ= (ΓD) ◦ (DyD) ◦ ( )DP ρ D Q ◦ (DP χ) ◦ (DδD P Q) ◦ ( ∆ D P Q )defγ= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ (δ D P QD) ◦ ( DP ρ D Q)◦ (DP χ)◦ (Dδ D P Q) ◦ ( ∆ D P Q )121δ=D(yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q)◦ (δ D DP Q) ◦ ( ∆ D P Q )(126)= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( P QP ρ D Q)◦ (P QP χ) ◦ (P QδD P Q)◦ ( P ρ D QP Q ) ◦ (δ D P Q)(127)= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( P QP ρ D Q)◦ (P QP χ) ◦ (P δC QP Q)◦ ( P C ρ Q P Q ) ◦ (δ D P Q)δ C= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ (P δ C QD) ◦ ( P Cρ D Q)◦ (P Cχ) ◦(P C ρ Q P Q )◦ (δ D P Q)(112)= (yDD) ◦ ( P ρ D QD ) ◦ ( P ε C QD ) ◦ ( ) (P Cρ D Q ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (δ D P Q)ε= C (yDD) ◦ ( P ρ D QD ) ◦ ( ) (P ρ D Q ◦ (P χ) ◦ P ε C QP Q ) ◦ ( P C ρ Q P Q ) ◦ (δ D P Q)Qcomfun= (yDD) ◦ ( P ρ D QD ) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q)Qcomfun= (yDD) ◦ ( P Q∆ D) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q)y= ( B∆ D) ◦ (yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) defγ= ( B∆ D) ◦ Γ ◦ (Dy)and since Dy is an epimorphism we get thatNow we compute(ΓD) ◦ (DΓ) ◦ ( ∆ D B ) = ( B∆ D) ◦ Γ.Γ ◦ (Du B ) ◦ ( Dε D) (110)= Γ ◦ (Dy) ◦ (Dδ D )defγ= (yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q) ◦ (Dδ D )δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (P QδD ) ◦ (δ D D)(113)= (yD) ◦ ( P ρ D Q)◦(P QεD ) ◦ (δ D D)ρ D Q= (yD) ◦ ( P QDε D) ◦ ( P ρ D QD ) ◦ (δ D D)(126)= (yD) ◦ ( P QDε D) ◦ (δ D DD) ◦ ( ∆ D D )δ D= (yD) ◦ (δ D D) ◦ ( DDε D) ◦ ( ∆ D D ) ∆ D= (yD) ◦ (δ D D) ◦ ∆ D ◦ ( Dε D)(110)= (u B D) ◦ ( ε D D ) ◦ ∆ D ◦ ( Dε D) Dcomonad= (u B D) ◦ ( Dε D)

122and since Dε D is an epimorphism we deduce thatFinally we compute(BεD ) ◦ Γ ◦ (Dy) defγΓ ◦ (Du B ) = u B D.= ( Bε D) ◦ (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q)y= y ◦ ( P Qε D) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) Qcomfun= y ◦ (P χ) ◦ (δ D P Q)(106)= y ◦ ( ε D P Q ) ε D = ( ε D B ) ◦ (Dy)and since Dy is an epimorphism we get that(BεD ) ◦ Γ = ε D B.6.1<strong>1.</strong> Coherds and coGalois functors. We keep the details of this subsectionbecause the coGalois case is not as common as the Galois notion in the literature.Lemma 6.39. Let X = (C, D, P, Q, δ C , δ D ) be a formal codual structure where Q :B → A, P : A → B and C = ( C, ∆ C , ε C) is a comonad on the category A, D =(D, ∆ D , ε D) is a comonad on B. Assume that both A and B have equalizers and thatC, QD preserve them. Then, δ C : C → QP induces a morphism δ C : C U → QP C inC A so that there exists a morphism C δ C C : C A → C QP C such that(134)C U C δ C C = δ C C .Moreover δCCCF= δ C : C U C F = C → QP C C F = QP.Proof. Let us consider the following diagram with notations of Proposition 4.29C Uδ C CC Uγ CC C Uδ C C UQP C Qι P QP C U∆ C C UC C Uγ CQρ C P C UQP C Uγ CCC C Uδ C C C U QP C C USince QD preserves equalizers, by Lemma 4.18, also the functor Q preserves equalizers.Since ( δC CU) ◦ (C Uγ C) (equalizes the pair Qρ C P C U, QP C Uγ C) and ( QP C , Qι ) P =Equ Fun(Qρ C P C U, QP C Uγ C) , by the universal property of the equalizer, there existsa unique morphism δC C : C U → QP C such that((135)) QιP◦ δC C = ( δ C C U ) ◦ (C Uγ C) .We now want to prove that δ C C : C U → QP C = C U C QP C is a morphism betweenleft C-comodule functors which satisfies(CδCC)◦( CUγ C) = ( C ρ Q P C ) ◦ δ C C .□We have(CQιP ) ◦ ( Cδ C C)◦( CUγ C) (135)= ( Cδ C C U ) ◦ ( C C Uγ C) ◦ (C Uγ C)C Uγ C equ= ( Cδ C C U ) ◦ ( ∆ C C U ) ◦ (C Uγ C) (125)= ( C ρ Q P C U) ◦ ( δ C C U ) ◦ (C Uγ C)

123(135)= ( C ρ Q P C U) ◦ ( Qι P ) ◦ δ C CC ρ Q=(CQιP ) ◦ ( C ρ Q P C ) ◦ δ C Cand since C, Q preserve equalizers, CQι P is a monomorphism, so that we get(CδCC)◦( CUγ C) = ( C ρ Q P C ) ◦ δ C C .Hence, by Lemma 4.28, there exists a unique morphism there exists a unique morphismC δ C C : C A → C QP C such thatC U C δ C C = δ C C .Moreover, note that by definition of δC C we have( ) QιP◦ δC C = ( δ C C U ) ◦ (C Uγ C)so that by applying it to C F we get(QιP C F ) ◦ ( δ C C C F ) = ( δ C C U C F ) ◦ (C Uγ C C F ) .Hence, by Proposition 4.32, we obtain that(QρDQ)◦(δCC C F ) = (δ C C) ◦ ∆ C (125)= ( Qρ C P)◦ δC .Since Qρ D Q is a monomorphism, we deduce that δC C C F = δ C .Proposition 6.40. Let A and B be categories with coequalizers and let χ : QP Q →Q be a regular coherd for a formal codual structure X = (C, D, P, Q, δ C , δ D ) where theunderlying functors P : A → B, Q : B → A and C : A → A preserve coequalizers.Let• A = (A, m A , u A ) be the monad on the category A constructed in Proposition6.25;• ( Q, A µ Q)be the left A-module functor constructed in Proposition 6.25;• C Q : B → C A be the functor defined in Lemma 4.28;• Λ : AC → CA be the mixed distributive law between the comonad C and themonad A constructed in Proposition 6.38;• Ã be the lifting of A on the category C A constructed in Theorem 5.7.Then there exists a functorial morphism eA µC Q : ÃC Q → C Q such that□Moreover,C U eA µC Q = A µ Q .( )C Q, eA µC Q is a left Ã-module functor.Proof. Since χ : QP Q → Q is a regular coherd for X = (C, D, P, Q, δ C , δ D ), byProposition 6.38 the mixed distributive law Λ : AC → CA is uniquely defined byΛ ◦ (xC) = (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) .Now we prove that A µ Q yields a functorial morphism eA µC Q. In fact we have(C A µ Q)◦ (ΛQ) ◦(A C ρ Q)◦ (xQ) x = ( C A µ Q)◦ (ΛQ) ◦ (xCQ) ◦(QP C ρ Q)defΛ= ( C A µ Q)◦ (CxQ) ◦( Cρ Q P Q ) ◦ (χP Q) ◦ (QP δ C Q) ◦ ( QP C ρ Q)(101),(127)= (Cχ) ◦ (C ρ Q P Q ) ◦ (χP Q) ◦ (QP Qδ D ) ◦ ( QP ρ D Q)

124χ= (Cχ) ◦ (C ρ Q P Q ) ◦ (Qδ D ) ◦ (χD) ◦ ( )QP ρ D Q= (Cχ) ◦ (CQδD ) ◦ (C ρ Q D ) ◦ (χD) ◦ ( )QP ρ D QC ρ Q(130)= ( CQε D) ◦ (C ρ Q D ) ◦ (χD) ◦ ( QP ρ D QC ρ Q= C ρ Q ◦ ( Qε D) ◦ (χD) ◦ ( QP ρ D Q) χ= C ρ Q ◦ χ ◦ ( QP Qε D) ◦ ( QP ρ D QQcomfun= C ρ Q ◦ χ (101)= C ρ Q ◦ A µ Q ◦ (xQ)and since by construction xQ is an epimorphism we get that(C A µ Q)◦ (ΛQ) ◦(A C ρ Q)= C ρ Q ◦ A µ Q .By Lemma 5.5 we know that(ΛQ) ◦ ( A C ρ Q)= C ρ AQ))so that (C A µ Q)◦ C ρ AQ = ( C A µ Q)◦ (ΛQ) ◦(A C ρ Q)= C ρ Q ◦ A µ Q .Hence there exists a morphism eA µC Q : ÃC Q → C Q such thatC U eA µC Q = A µ Q .By the associativity and unitality(properties)of A µ Q , we deduce that eA µC Q is alsoassociative and unital so thatC Q, eA µC Q is a left Ã-module functor. □Lemma 6.4<strong>1.</strong> Let X = (C, D, P, Q, δ C , δ D ) be a formal codual structure with underlyingfunctors P : A → B, Q : B → A, C : A → A and D : B → B. Assume that Aand B are categories with equalizers and C, QD preserves them. Assume that• A = (A, m A , u A ) is a monad on the category A such that A preserves equalizers• ( Q, A µ Q)is a left A-module functorandis a lifting of the monad of A to the category C A( )•C Q, eA µC Q is a left Ã-module functor where C U eA µC Q = A µ Q .• Ã = (Ã, m e A, u e A)Consider the functorial morphismscocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP( ) (ÃC )C cocan C eA:= µC QP C ◦ δCC : Ã → C QP C .Then cocan 1 is an isomorphism if and only if C cocan C is an isomorphism.Proof. Let us consider cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP. Let ( Q D , ι Q) bethe equalizer described in Proposition 4.29. Since A µ Q is a functorial morphism, wehave that (QιP ) ◦ (A µ Q P C) = (A µ Q P C U ) ◦ ( AQι P ) .Now, by Lemma 6.39, δ C induces a morphism δ C C : C U → QP C such that(QιP ) ◦ δ C C = ( δ C C U ) ◦ (C Uγ C) .

Then, we can consider the morphism(136) cocan C := (A µ Q P C) ◦ ( Aδ C C): A C U = C Uà → QP C = C U C QP C .Then we have( ) QιP◦ cocan C = ( Qι ) P ◦ (A µ Q P C) ◦ ( )AδC C A µ Q(=Aµ Q P C U ) ◦ ( AQι ) P ◦ ( )AδCC(135)= (A µ Q P C U ) ◦ ( Aδ C C U ) ◦ ( A C Uγ C) = ( cocan C 1 U ) ◦ ( A C Uγ C)i.e.(137)Now, by assumption we have(QιP ) ◦ cocan C = ( cocan 1 C U ) ◦ ( A C Uγ C) .C U eA µC Q = A µ Q125so thatC U eA µC QP C = A µ Q P C .Moreover, by Lemma 6.39, there exists a morphism C δ C : C A → C QP C such thatC U C δ C C = δ C C .Since à is a lifting of the monad A, by Theorem 5.7 we have a mixed distributivelaw Φ : AC → CA so that we can apply Proposition 5.6 and we get thatAδ C C = A C U C δ C C = C UÃC δ C Cwhere ÃC δ C : à → ÃC QP C is a functorial morphism. Then we can consider themorphism( ) (ÃC )C cocan C eA:= µC QP C ◦ δCC : à → C QP Cand we get that(C ) ( )C U C cocan C = U eA µC QP C ◦C UÃC δCC= (A µ Q P C) ◦ ( A C U C δ C C)=( Aµ Q P C) ◦ ( Aδ C C)= cocan C .We compute(cocan1 C C U ) ◦ ( A∆ C C U ) = (A µ Q P C C U ) ◦ ( Aδ C C C U ) ◦ ( A∆ C C U )(125)= (A µ Q P C C U ) ◦ ( AQρ C P C U ) ◦ ( Aδ C C U )A µ Q=(QρCP C U ) ◦ (A µ Q P C U ) ◦ ( Aδ C C U ) = ( Qρ C P C U ) ◦ ( cocan 1 C U )so that we get((138) cocan1 C C U ) ◦ ( A∆ C C U ) = ( Qρ C P C U ) ◦ ( cocan C 1 U ) .Let us consider the following commutative diagram0 A C U AC Uγ C AC C Ucocan C cocan 1 C U0 QP C QιP QP C UA∆ C C UAC C Uγ CQρ C P C UQP C Uγ CACC C U QP C C Ucocan 1 C C U

126Now, since cocan 1 : AC → QP is a functorial morphism and by formula (138) , theright square serially commutes. By formula (137) also the left square commutes.Moreover, by definition, ι P and C Uγ C are monomorphisms. Since QD preservesequalizers, by Lemma 4.18 also Q preserves equalizers. Since C, Q preserve equalizers,Qι P and A C Uγ C are also monomorphisms. Then, if cocan 1A U is an isomorphism,also cocan C is an isomorphism. Since C U C cocan C = cocan C , by 4.17, also C cocan Cis an isomorphism.Conversely, assume that C cocan C is an isomorphism. Then alsococan C = C U C cocan C is an isomorphism. Then we haveC U C cocan C C F = cocan C C F = (A µ Q P C C F ) ◦ ( Aδ C C C F )Pro4.32,Lem6.39=( Aµ Q P ) ◦ (Aδ C ) = cocan 1so that also cocan 1 is an isomorphism.□6.1<strong>2.</strong> The cotame case. The following subsection is presented without proofs,which can be obtained as the dual versions of <strong>results</strong> of the tame case (see Subsection6.6).Definition 6.4<strong>2.</strong> A formal codual structure X = (C, D, P, Q, δ C , δ D ) is called acoMorita context on the categories A and B if it satisfies also the balanced conditions((139) ρDQ P ) ◦ δ C = ( )Q D ρ P ◦ δC and ( ρ C P Q ) ◦ δ D = (P C ρ Q ) ◦ δ D .Lemma 6.43. Let X = (C, D, P, Q, δ C , δ D ) be a coMorita context on the categories Aand B and assume that C, D, P, Q preserve equalizers. Hence, there exist functorialmorphisms(140)(141)(142)(143)• CD δ DCC: IdC A → C Q DD P C such thatC U CD δCDC = D δCDC( )where D δC DC is uniquely determined by Q D ι D P◦ D δC DC = (D δC DC U ) ◦ (C Uγ C)and(ι QD P ) ◦ D δC D = δ C• DC δ CDD: IdD B → D P C C Q D such thatD U DC δDCD = C δDCD( )where C δD CD is uniquely determined by P C ι C Q◦ C δD CD = (C δD C D U ) ◦ (D Uγ D)and(ιP C Q ) ◦ C δD C = δ DDefinitions 6.44. Let X = (C, D, P, Q, δ C , δ D ) be a coMorita context. We will saythat X is cotame if the lifted functorial morphisms CD δCDC : IdC A → C Q DD P C andDC δDCD : IdD B → D P C C Q D are isomorphisms so that the lifted functors C Q D : D B →C A and D P C : C A → D B yield a category equivalence. In this case, if χ : QP Q → Qis a coherd for X, we will say that χ is a cotame coherd.

Proposition 6.45. Let X = (C, D, P, Q, δ C , δ D ) be a cotame coMorita context.Then unit and counit of the adjunction (D P C , C Q D) are given byη ( D P C , C Q D ) = CD δC DC and ɛ ( D P C , C Q D ) = ( ) (DC δDCD −1◦ D P ( ) )C CD δCDC −1 CQ D ◦( DCδDCDDP C C Q D) so that(C )(η (P C , C Q) = Q D γ DD P C ◦ CD δC DC and ɛ (P C , C Q) = ε D ◦D U ( ) )DC δDCD −1 DF ◦(D U D P ( ) )C CD δCDC −1 CQ DD F ◦ (D U DC δD CDDP C C Q DD F ) .Corollary 6.46. Let X = (C, D, P, Q, δ C , δ D ) be a cotame coMorita context. Assumethat the functors A, B, P, Q preserve equalizers. Then the units of the adjunctions( P C , C Q ) and ( Q D , D P ) are given by ɛ (P C , C Q) = C δ C C and ɛ (Q D , D P ) = D δ D D .Lemma 6.47. Let X = (C, D, P, Q, δ C , δ D ) be a formal codual structure where theunderlying functors are C : A → A, D : B → B, P : A → B and Q : B →A. Assume that both categories A and B have equalizers and the functors C, QDpreserve them. Assume that• A = (A, m A , u A ) is a monad on the category A such that A preserves equalizers)•(Ã, à = mA e, u A e is a lifting of the monad A to the category C A( )•C Q, eA µC Q is a left Ã-module functor• X is a cotame coMorita context.Then cocan 1 is an isomorphism if and only if C cocan C is an isomorphism if andonly if C Q is a left Ã-coGalois functor.The following Theorem is a formulation, in pure categorical terms, for the coherdversion of [BV, Theorem <strong>2.</strong>18].Theorem 6.48. Let X = (C, D, P, Q, δ C , δ D ) be a regular cotame coMorita context.Assume that• both categories A and B have equalizers and coequalizers,• the functors C and D preserve coequalizers,• the functors C, D, P, Q preserve equalizers.Then the existence of the following structures are equivalent:(a) A coherd χ : QP Q → Q for X(b) A monad A = (A, m A , u A ) on the category A such that the functor A preservescoequalizers and a mixed distributive law Λ : AC → CA such that C Qis a coGalois module functor over à (where à is the lifting of A)(c) A monad B = (B, m B , u B ) on the category B such that the functor B preservescoequalizers and an opposite mixed distributive law Γ : DB → BDsuch that D P is a coGalois module functor over ˜B (where ˜B is the lifting ofB).127

1287. Herds and Coherds7.<strong>1.</strong> Constructing the functor Q. Our next task is to construct a D-C-bicomodulefunctor Q. Such a functor appears in [BM, Section 5], but we give here new notations.For the details of the proofs, see the dual <strong>results</strong> in the following.Proposition 7.<strong>1.</strong> In the setting of Theorem 6.5, we define functors Q : A → B viathe equalizerQ q P C (θl P )◦(P i)(θ r P )◦(P i) BP QPThen there exists a unique functorial morphism κ ′ 0 : Q → DP such that(144) (P i) ◦ q = (jP ) ◦ κ ′ 0Moreover(Q,κ′0)= EquFun((P ωl ) ◦ (jP ) , (P ω r ) ◦ (jP ) ) .( The functor)Q can be equipped with the structure of a D-C-bicomodule functorQ, D ρ Q , ρ C where ρ C and D ρQQ Q are uniquely determined by(145) (qC) ◦ ρ C Q = ( P ∆ C) ◦ qand(146) (Dκ ′ 0) ◦ D ρ Q = ( ∆ D P ) ◦ κ ′ 0.Proposition 7.<strong>2.</strong> In the setting of Theorem 6.5 and Proposition 7.1, there existtwo functorial morphisms δ C : C → QQ and δ D : D → QQ where δ C is C-bicolinearand δ D is D-bicolinear and they fulfill(147) (Qq) ◦ δ C = (iC) ◦ ∆ Cand(148) (κ ′ 0Q) ◦ δ D = (Dj) ◦ ∆ D .Moreover the coassociative conditions hold, that is(δ C Q) ◦ C ρ Q = (Qδ D ) ◦ ρ D Q and ( δ D Q ) ◦ D ρ Q = ( Qδ C)◦ ρCQ .7.<strong>2.</strong> From herds to coherds.7.3. Given an herd τ : Q → QP Q in a formal dual structure M = (A, B, P, Q, σ A , σ B ),our purpose is to build the formal codual structure X = (C, D, Q, Q, δ C , δ D ) and thena coherd χ : QQQ → Q in X.Theorem 7.4. Let A and B be categories with equalizers and let P : A → B,Q : B → A, A : A → A and B : B → B be functors. Assume that all thefunctors P, Q, A and B preserve equalizers. Let u A : A → A and u B : B → Bbe functorial monomorphisms and assume that (A, u A ) = Equ Fun (u A A, Au A ) and(B, u B ) = Equ Fun (u B B, Bu B ).Let τ : Q → QP Q be a functorial morphism such that(QP τ) ◦ τ = (τP Q) ◦ τ

129Let σ B : P Q → B be a functorial morphism such that(Qσ B ) ◦ τ = Qu Band let σ A : QP → A be a functorial morphism such that(σ A Q) ◦ τ = u A Q.Then there is a formal codual structure X = (C, D, Q, Q, δ C , δ D ).Proof. In view of Theorem 6.5 and Propositions 7.1, 7.2 a formal codual structureX = (C, D, Q, Q, δ C , δ D ) has been constructed.□Theorem 7.5. Let A and B be categories with equalizers and letM = (A, B, P, Q, σ A , σ B ) be a regular formal dual structure where P : A → B,Q : B → A, A : A → A and B : B → B are functors that preserve equalizers.Let τ : Q → QP Q be a pretorsor. Then there is a formal codual structure X =(C, D, Q, Q, δ C , δ D ). Define χ : QQQ → Q by settingχ := µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) .Then χ is a coherd in X.Proof. By Theorem 7.4 X = (C, D, Q, Q, δ C , δ D ) is a formal codual structure. Toshow that χ is a coherd in X, we have to prove that it satisfies the following conditions.1) Coassociativity, in the sense that χ ◦ ( χQQ ) = χ ◦ ( QQχ ) . Let us computeχ ◦ ( QQχ )= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) ◦ ( QQχ )q= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QP Cχ) ◦ ( QqQQQ )i= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP QP χ) ◦ ( QP iQQQ ) ◦ ( QqQQQ )σ A = µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ (AQP χ) ◦ ( σ A QP QQQ ) ◦ ( QP iQQQ ) ◦ ( QqQQQ )A µ Q= µBQ ◦ ( Qσ B) ◦ (QP χ) ◦ (A µ Q P QQQ ) ◦ ( σ A QP QQQ ) ◦ ( QP iQQQ ) ◦ ( QqQQQ )(82)= A µ Q ◦ ( σ A Q ) ◦ (QP χ) ◦ (A µ Q P QQQ ) ◦ ( σ A QP QQQ ) ◦ ( QP iQQQ ) ◦ ( QqQQQ )σ A = A µ Q ◦ (Aχ) ◦ ( σ A QQQ ) ◦ (A µ Q P QQQ ) ◦ ( σ A QP QQQ ) ◦ ( QP iQQQ ) ◦ ( QqQQQ ) .We haveA µ Q ◦ (Aχ) ◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )= A µ Q ◦ ( Aµ B Q)◦(A A µ Q B ) ◦ ( AAQσ B) ◦ ( Aσ A QP Q ) ◦ (AQP iQ) ◦ (AQqQ)◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )Qbimod,σ A=A µ Q ◦ ( A A µ Q)◦(AAµBQ)◦(Aσ A QB ) ◦ ( AQP Qσ B) ◦ (AQP iQ) ◦ (AQqQ)◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )A µ Q ass= A µ Q ◦ (m A Q) ◦ ( AAµ B Q)◦(Aσ A QB ) ◦ ( AQP Qσ B) ◦ (AQP iQ) ◦ (AQqQ)

130◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )m=A A µ Q ◦ ( )Aµ B Q ◦ (mA QB) ◦ [( Aσ A QB ) ◦ ( AQP Qσ B) ◦ (AQP iQ) ◦ (AQqQ) ]◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )Qis a bim= µ B Q ◦ (A µ Q B ) ◦ (m A QB) ◦ [( Aσ A QB ) ◦ ( AQP Qσ B) ◦ (AQP iQ) ◦ (AQqQ) ]◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )A µ Q ass= µ B Q ◦ (A µ Q B ) ◦ [( A A µ Q B ) ◦ ( Aσ A QB ) ◦ ( AQP Qσ B) ◦ (AQP iQ) ◦ (AQqQ) ]◦ ( σ A QQQ ) ◦ (A µ Q P QQQ )(80)= µ B Q ◦ (A µ Q B ) ◦ [( A A µ Q B ) ◦ ( Aσ A QB ) ◦ ( AQP Qσ B) ◦ (AQP iQ) ◦ (AQqQ) ]◦ ( m A QQQ ) ◦ ( Aσ A QQQ )m A= µ B Q ◦ (A µ Q B ) ◦ (m A QB) ◦ [ ( AA A µ Q B ) ◦ ( AAσ A QB ) ◦ ( AAQP Qσ B) ◦ (AAQP iQ)◦ (AAQqQ)] ◦ ( Aσ A QQQ )A µ Q ass= µ B Q ◦ (A µ Q B ) ◦ ( A A µ Q B ) ◦ [ ( AA A µ Q B ) ◦ ( AAσ A QB ) ◦ ( AAQP Qσ B)◦ (AAQP iQ) ◦ (AAQqQ)] ◦ ( Aσ A QQQ )Qis a bim= A µ Q ◦ ( Aµ B Q)◦(A A µ Q B ) ◦ ( AA A µ Q B ) ◦ ( AAσ A QB ) ◦ ( AAQP Qσ B)◦ (AAQP iQ) ◦ (AAQqQ) ◦ ( Aσ A QQQ )Qis a bim= A µ Q ◦ ( A A µ Q)◦(AAµBQ)◦(AA A µ Q B ) ◦ ( AAσ A QB ) ◦ ( AAQP Qσ B)◦ (AAQP iQ) ◦ (AAQqQ) ◦ ( Aσ A QQQ )(82)= A µ Q ◦ ( A A µ Q)◦(AAµBQ)◦(AAµBQ B ) ◦ ( AAQσ B B ) ◦ ( AAQP Qσ B)◦ (AAQP iQ) ◦ (AAQqQ) ◦ ( Aσ A QQQ )σ A = A µ Q ◦ ( A A µ Q)◦(Aσ A Q ) ◦ ( AQP µ B Q)◦(AQP µBQ B ) ◦ ( AQP Qσ B B )◦ ( AQP QP Qσ B) ◦ (AQP QP iQ) ◦ (AQP QqQ)(82)= A µ Q ◦ ( (Aµ Q) ) B ◦ AQσB◦ ( ) (AQP µ B Q ◦ AQP µBQ B ) ◦ ( AQP Qσ B B )◦ ( AQP QP Qσ B) ◦ (AQP QP iQ) ◦ (AQP QqQ)Qis a bim= µ B Q◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( ) (AQP µ B Q ◦ AQP µBQ B ) ◦ ( AQP Qσ B B )◦ ( AQP QP Qσ B) ◦ (AQP QP iQ) ◦ (AQP QqQ)µ B Q= assµ B Q◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( ) [AQP µ B Q ◦ (AQP QmB ) ◦ ( AQP Qσ B B )]◦ ( AQP QP Qσ B) ◦ (AQP QP iQ) ◦ (AQP QqQ)(81)= µ B Q◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( (AQP µ Q) ) B ◦ AQP QσB◦ ( )AQP QP µ B Q◦ ( AQP QP Qσ B) ◦ (AQP QP iQ) ◦ (AQP QqQ)

131A µ Q= µBQ ◦ [( Qσ B) ◦ ( (QP µ Q)] ) B ◦ QP QσB◦ ( ) (QP QP µ )B Q ◦ QP QP QσB◦ (QP QP iQ) ◦ (QP QqQ) ◦ (A µ Q P QQQ )(81)= µ B Q ◦ (Qm B ) ◦ ( Qσ B B ) ◦ ( QP Qσ B) ◦ ( QP QP µ B Q◦ (QP QP iQ) ◦ (QP QqQ) ◦ (A µ Q P QQQ )σ B = µ B Q ◦ (Qm B ) ◦ ( QBσ B) ◦ ( QBP µ B Q)◦(QP QP QσB ))◦(QBP QσB ) ◦ (QBP iQ) ◦ (QBqQ)◦ ( Qσ B QQ ) ◦ (A µ Q P QQQ )µ B Q ass= µ B Q ◦ ( µ B QB ) ◦ ( QBσ B) ◦ ( QBP µ B Q)◦(QBP QσB ) ◦ (QBP iQ) ◦ (QBqQ)◦ ( Qσ B QQ ) ◦ (A µ Q P QQQ )µ B Q= µ B Q ◦ ( Qσ B) ◦ ( ) (QP µ ) B Q ◦ QP QσB◦ (QP iQ) ◦ (QqQ) ◦ ( µ B QQQ )◦ ( Qσ B QQ ) ◦ (A µ Q P QQQ )A µ Q= µBQ ◦ ( Qσ B) ◦ ( ) (QP µ ) B Q ◦ QP QσB◦ (QP iQ) ◦ (QqQ) ◦ ( µ B QQQ )◦ (A µ Q BQQ ) ◦ ( AQσ B QQ ))◦(QP QσB ) ◦ (QP iQ) ◦ (QqQ) ◦ ( µ B QQQ )(82)= A µ Q ◦ ( σ A Q ) ◦ ( QP µ B Q◦ (A µ Q BQQ ) ◦ ( AQσ B QQ )σ A = A µ Q ◦ ( Aµ B Q)◦(AQσB ) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) ◦ ( µ B QQQ )◦ (A µ Q BQQ ) ◦ ( AQσ B QQ )Qis a bim= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) ◦ ( µ B QQQ )◦ (A µ Q BQQ ) ◦ ( AQσ B QQ )and thus we getχ ◦ ( QQχ ) == A µ Q ◦ (Aχ) ◦ ( σ A QQQ ) ◦ (A µ Q P QQQ ) ◦ ( σ A QP QQQ ) ◦ ( QP iQQQ )◦ ( QqQQQ )= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) ◦ ( µ B QQQ )◦ (A µ Q BQQ ) ◦ ( AQσ B QQ ) ◦ ( σ A QP QQQ ) ◦ ( QP iQQQ ) ◦ ( QqQQQ )= χ ◦ ( χQQ )2) Counitality, in the sense that χ ◦ (Qδ D ) = Qε D and χ ◦ (δ C Q) = ε C Q. We haveχ ◦ (Qδ D ) == µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) ◦ (Qδ D )(144)= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QjP Q) ◦ (Qκ ′ 0Q) ◦ (Qδ D )(148)= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QjP Q) ◦ (QDj) ◦ ( Q∆ D)

132= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (Qjj) ◦ ( Q∆ D)(67)= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP τ) ◦ (Qj)We computeσ A = µ B Q ◦ (A µ Q B ) ◦ ( σ A QB ) ◦ ( QP Qσ B) ◦ (QP τ) ◦ (Qj)(70)= µ B Q ◦ (A µ Q B ) ◦ ( σ A QB ) ◦ (QP Qu B ) ◦ (Qj)σ A = µ B Q ◦ (A µ Q B ) ◦ (AQu B ) ◦ ( σ A Q ) ◦ (Qj)Qis a bim= A µ Q ◦ ( Aµ B Q)◦ (AQuB ) ◦ ( σ A Q ) ◦ (Qj)Qis a mod= A µ Q ◦ ( σ A Q ) ◦ (Qj) (82)= µ B Q ◦ ( Qσ B) ◦ (Qj)(67)= µ B Q ◦ (Qu B ) ◦ ( Qε D) Qis a mod= Qε D .χ ◦ (δ C Q) =µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (QqQ) ◦ (δ C Q)(147)= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (QP iQ) ◦ (iCQ) ◦ ( ∆ C Q )= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (iiQ) ◦ ( ∆ C Q )(63)= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ (τP Q) ◦ (iQ)(69)= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ (u A QP Q) ◦ (iQ)u A= µ B Q ◦ (A µ Q B ) ◦ (u A QB) ◦ ( Qσ B) ◦ (iQ)Qis a bim= A µ Q ◦ ( Aµ B Q)◦ (uA QB) ◦ ( Qσ B) ◦ (iQ)u A= A µ Q ◦ (u A Q) ◦ µ B Q ◦ ( Qσ B) ◦ (iQ)Qis a mod= µ B Q ◦ ( Qσ B) ◦ (iQ) (82)= A µ Q ◦ ( σ A Q ) ◦ (iQ)(63)= A µ Q ◦ (u A Q) ◦ ( ε C Q ) Qis a mod= ε C Q.□7.3. Constructing the functor ̂Q. Our next task is to construct a B-A-bimodulefunctor ̂Q.Proposition 7.6. Within the assumptions and notations of Theorem 6.29, definea functor ̂Q via the coequalizer DP QP (P x)◦(zl P )P A l ˆQ(P x)◦(z r P )Then there exists a unique functorial morphism ν ′ 0 : BP → ̂Q such that(149) ν ′ 0 ◦ (yP ) = l ◦ (P x) .

133Moreover ( )( ( ̂Q, ν′0 = Coequ ) Fun (yP ) ◦ P wl, (yP ) ◦ (P w r ) )The functor ̂Q can be equipped with the structure of a B-A-bimodule functor(̂Q, µ A b Q, B µb Q)where µ A b Qand B µb Qare uniquely defined by(150) µ A b Q◦ (lA) = l ◦ (P m A )and(151)Proof. By construction we haveBy Lemma <strong>2.</strong>9, we haveB µb Q◦ (Bν ′ 0) = ν ′ 0 ◦ (m B P ) .l ◦ (P x) ◦ ( z l P ) = l ◦ (P x) ◦ (z r P ) .(BP, yP ) = Coequ Fun(z l P, z r P ) .By the universality of coequalizers, there exists a unique functorial morphism ν ′ 0 :BP → ̂Q which fulfils (149). Let us prove that(̂Q, ν′0)= Coequ Fun((yP ) ◦(P wl ) , (yP ) ◦ (P w r ) ) . We haveν ′ 0 ◦ (yP ) ◦ ( P w l) (149)= l ◦ (P x) ◦ ( P w l) defx= l ◦ (P x) ◦ (P w r )(149)= ν ′ 0 ◦ (yP ) ◦ (P w r ) .Let now ξ : BP → X be a morphism such that ξ ◦ (yP ) ◦ ( P w l) = ξ ◦ (yP ) ◦ (P w r ).Since P preserves coequalizers, we have(P A, P x) = Coequ Fun(P w l , P w r) .By universality of coequalizers there exists a unique functorial morphism ν 0 : P A →X such that(152) ν 0 ◦ (P x) = ξ ◦ (yP ) .We computeν 0 ◦ (P x) ◦ ( z l P ) (152)= ξ ◦ (yP ) ◦ ( z l P )= ξ ◦ (yP ) ◦ (z r P ) (152)= ν 0 ◦ (P x) ◦ (z r P ) .( )By the universality of the coequalizer ̂Q, l , there exists a unique functorial morphismν : ̂Q → X such thatWe computeSince yP is epi, we getν ◦ l = ν 0 .ν ◦ ν ′ 0 ◦ (yP ) (149)= ν ◦ l ◦ (P x) = ν 0 ◦ (P x) (152)= ξ ◦ (yP ) .ξ = ν ◦ ν ′ 0.

134Assume now that there is another morphism t : ̂Q → X such that ξ = t ◦ ν ′ 0. Thenwe havet ◦ l ◦ (P x) (149)= t ◦ ν ′ 0 ◦ (yP ) = ξ ◦ (yP ) = ν ◦ ν ′ 0 ◦ (yP ) (149)= ν ◦ l ◦ (P x) .Since l ◦ (P x) is an epimorphism, we deduce that t = ν.(2) We want to equip ̂Q with the structure of a B-A-bimodule functor. To beginwith, let us prove a number of equalities. Let us calculateso thatχ ◦ ( Qz l) = χ ◦ (QP χ) ◦ (Qδ D P Q) (98)= χ ◦ (χP Q) ◦ (Qδ D P Q)(105)= χ ◦ ( Qε D P Q ) = χ ◦ (Qz r )(153) χ ◦ ( Qz l) = χ ◦ (Qz r ) .Let(154) b = m A ◦ (xA) .Thenso thatx ◦ (χP ) (102)= m A ◦ (xx) = m A ◦ (xA) ◦ (QP x)(155) x ◦ (χP ) = b ◦ (QP x) .We haveand hence(P χ) ◦ ( z l P Q ) = (P χ) ◦ (P χP Q) ◦ (δ D P QP Q)(98)= (P χ) ◦ (P QP χ) ◦ (δ D P QP Q)δ D= (P χ) ◦ (δ D P Q) ◦ (DP χ) = z l ◦ (DP χ)y ◦ (P χ) ◦ ( z l P Q ) = y ◦ z l ◦ (DP χ) ycoequ= y ◦ z r ◦ (DP χ) = y ◦ ( ε D P Q ) ◦ (DP χ)so that we getε D = y ◦ (P χ) ◦ ( ε D P QP Q ) = y ◦ (P χ) ◦ (z r P Q)(156) y ◦ (P χ) ◦ ( z l P Q ) = y ◦ (P χ) ◦ (z r P Q) .From previous equalities, it follows thatl ◦ (P b) ◦ ( z l P A ) ◦ (DP QP x) = zll ◦ (P b) ◦ (P QP x) ◦ ( z l P QP )(155)= l ◦ (P x) ◦ (P χP ) ◦ ( z l P QP ) (149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ ( z l P QP )(156)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (z r P QP ) (149)= l ◦ (P x) ◦ (P χP ) ◦ (z r P QP )(155)= l ◦ (P b) ◦ (P QP x) ◦ (z r P QP ) zr= l ◦ (P b) ◦ (z r P A) ◦ (DP QP x) .Since DP QP x is an epimorphism, we obtain(157) l ◦ (P b) ◦ ( z l P A ) = l ◦ (P b) ◦ (z r P A)

135that isl ◦ (P m A ) ◦ (P xA) ◦ ( z l P A ) = l ◦ (P m A ) ◦ (P xA) ◦ (z r P A) .From <strong>2.</strong>9 we have that(̂QA, lA)= Coequ Fun((P xA) ◦(z l P A ) , (P xA) ◦ (z r P A) ) .Hence there exists a unique functorial morphism µ A Q b : ̂QA → ̂Q which satisfies (150).( )Now we want to prove that ̂Q, µ A Q b is a right A-module functor. First let us provethat µ A b Qis associative that isµ A b Q◦( ) ( )µ A Q bA= µ A Q b ◦ ̂QmA .We computeµ A b Q◦( )µ A Q bA◦ (lAA) (150)= µ A Q b ◦ (lA) ◦ (P m A A)(150)= l ◦ (P m A ) ◦ (P m A A) m Aass= l ◦ (P m A ) ◦ (P Am A )(150)= µ A b Q◦ (lA) ◦ (P Am A ) l = µ A b Q◦( )̂QmA ◦ (lAA) .Since lAA is an epimorphism, we get that µ A b Qis associative. Let us prove that µ A b Qis unital that isin fact( )µ A Q b ◦ ̂QuA( )µ A Q b ◦ ̂QuA = ̂Q◦ l l = µ A b Q◦ (lA) ◦ (P Au A ) (150)= l ◦ (P m A ) ◦ (P Au A ) Amonad= land since l is an epimorphism we conclude. We want to prove a series of equalities.First of all, let us prove that(158)(159)In fact, we haveand(χP ) ◦ ( QP w l) = w l ◦ (χP C)(χP ) ◦ (QP w r ) = w r ◦ (χP C)(χP ) ◦ ( QP w l) = (χP ) ◦ (QP χP ) ◦ (QP QP δ C )(98)= (χP ) ◦ (χP QP ) ◦ (QP QP δ C )χ= (χP ) ◦ (QP δ C ) ◦ (χP C) = w l ◦ (χP C)(χP ) ◦ (QP w r ) = (χP ) ◦ ( QP QP ε C) χ=(QP εC ) ◦ (χP ) = w r ◦ (χP C) .From (158) we deduce thatx ◦ (χP ) ◦ ( QP w l) (158)= x ◦ w l ◦ (χP C)xcoequ= x ◦ w r ◦ (χP C) (159)= x ◦ (χP ) ◦ (QP w r )

136and hence we get(160) x ◦ (χP ) ◦ ( QP w l) = x ◦ (χP ) ◦ (QP w r ) .We observe thatν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ ( BP w l) ◦ (yP QP C)= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (yP QP ) ◦ ( P QP w l)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ ( P QP w l)(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ ( P QP w l)(149)= l ◦ (P x) ◦ (P χP ) ◦ ( P QP w l)(160)= l ◦ (P x) ◦ (P χP ) ◦ (P QP w r )(149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P QP w r )(109)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P QP w r )= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (yP QP ) ◦ (P QP w r )= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (BP w r ) ◦ (yP QP C)since yP QP C is an epimorphism, we obtain thatν 0 ′ ◦ (m B P ) ◦ (ByP ) ◦ ( BP w l) = ν 0 ′ ◦ (m B P ) ◦ (ByP ) ◦ (BP w r ) .(Since B preserves coequalizers, we have that B ̂Q,)Bν 0′ = Coequ Fun ((ByP ) ◦( ) BP wl, (ByP ) ◦ (BP w r )) so that there exists a unique functorial morphism B µb Q:( )̂Q, B µb QB ̂Q → ̂Q which satisfies (151).Now we want to show thatB-module functor. First let us prove that B µb Qis associative that is) ( )(B B µb Qm B ̂Q .B µb Q◦= B µb Q◦is a leftWe have)B µb Q◦(B B µb Q◦ (BBν ′ 0) (151)= B µb Q◦ (Bν ′ 0) ◦ (Bm B P )(151)= ν ′ 0 ◦ (m B P ) ◦ (Bm B P ) m Bass(151)= B µb Q◦ (Bν ′ 0) ◦ (m B BP ) m B= B µb Q◦= ν 0 ′ ◦ (m B P ) ◦ (m B BP )( )m B ̂Q ◦ (BBν 0) ′ .Since BBν 0 ′ is an epimorphism, we get that B µb Qis associative. Let us prove thatB µb Qis unital that is( )B µb Q◦ u B ̂Q = ̂Q.We calculate( )B µb Q◦ u B ̂Q◦ ν ′ 0 = B µb Q◦ (Bν ′ 0) ◦ (u B BP ) (151)= ν ′ 0 ◦ (m B P ) ◦ (u B BP ) = ν ′ 0.

Since ν 0 ′ is an epimorphism, we get that B µb Qis unital. Finally we have to prove thecompatibility condition) ( )(Bµ A Q b = µ A Q b ◦B µb QA .B µb Q◦137We haveB µb Q◦) (Bµ A Q b ◦ (BlA) ◦ (BP xx) ◦ (yP QP QP )(150)= B µb Q◦ (Bl) ◦ (BP m A ) ◦ (BP xx) ◦ (yP QP QP )(102)= B µb Q◦ (Bl) ◦ (BP x) ◦ (BP χP ) ◦ (yP QP QP )(149)= B µb Q◦ (Bν ′ 0) ◦ (ByP ) ◦ (BP χP ) ◦ (yP QP QP )y= B µb Q◦ (Bν ′ 0) ◦ (ByP ) ◦ (yP QP ) ◦ (P QP χP )= B µb Q◦ (Bν ′ 0) ◦ (yyP ) ◦ (P QP χP )(151)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P QP χP )(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P QP χP )(98)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P χP QP )(149)= l ◦ (P x) ◦ (P χP ) ◦ (P χP QP )(102)= l ◦ (P m A ) ◦ (P xx) ◦ (P χP QP )= l ◦ (P m A ) ◦ (P xA) ◦ (P QP x) ◦ (P χP QP )χ= l ◦ (P m A ) ◦ (P xA) ◦ (P χP A) ◦ (P QP QP x)(150)= µ A b Q◦ (lA) ◦ (P xA) ◦ (P χP A) ◦ (P QP QP x)(149)= µ A b Q◦ (ν ′ 0A) ◦ (yP A) ◦ (P χP A) ◦ (P QP QP x)(109)= µ A b Q◦ (ν ′ 0A) ◦ (m B P A) ◦ (yyP A) ◦ (P QP QP x)= µ A b Q◦ (ν ′ 0A) ◦ (m B P A) ◦ (ByP A) ◦ (yP QP A) ◦ (P QP QP x)y= µ A Q b ◦ (ν 0A) ′ ◦ (m B P A) ◦ (ByP A) ◦ (BP QP x) ◦ (yP QP QP )( )(151)= µ A Q b ◦B µb QA ◦ (Bν 0A) ′ ◦ (ByP A) ◦ (BP QP x) ◦ (yP QP QP )( )(149)= µ A Q b ◦B µb QA ◦ (BlA) ◦ (BP xA) ◦ (BP QP x) ◦ (yP QP QP )( )= µ A Q b ◦B µb QA ◦ (BlA) ◦ (BP xx) ◦ (yP QP QP )Since (BlA) ◦ (BP xx) ◦ (yP QP QP ) is an epimorphism, we conclude. Then(̂Q, B µb Q, µ A b Q)is a B-A-bimodule functor.□

138Proposition 7.7. In the setting of Theorem 6.29 and Proposition 7.6, there existtwo functorial morphisms σ A : Q ̂Q → A and σ B : ̂QQ → B where σ A is A-bilinearand σ B is B-bilinear and they fulfill(161) σ A ◦ (Ql) = m A ◦ (xA)and(162) σ B ◦ (ν ′ 0Q) = m B ◦ (By) .Moreover the associative conditions hold, that isA µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B) and B µb Q◦Proof. First we want to prove thatIn fact we have( ) (σ B ̂Q = µ A Q b ◦ ̂QσA).m A ◦ (xA) ◦ (QP x) ◦ ( Qz l P ) = m A ◦ (xA) ◦ (QP x) ◦ (Qz r P ) .m A ◦ (xA) ◦ (QP x) ◦ ( Qz l P ) (102)= x ◦ (χP ) ◦ ( Qz l P )= x ◦ (χP ) ◦ (QP χP ) ◦ (Qδ D P QP ) (128)= x ◦ (χP ) ◦ (χP QP ) ◦ (Qδ D P QP )(130)= x ◦ (χP ) ◦ ( Qε D P QP ) = x ◦ (χP ) ◦ (Qz r P )(102)= m A ◦ (xx) ◦ (Qz r P ) = m A ◦ (xA) ◦ (QP x) ◦ (Qz r P ) .Since Q preserves coequalizers we have(Q ̂Q, Ql) = Coequ Fun((QP x) ◦(Qz l P ) , (QP x) ◦ (Qz r P ) )so that there exists a functorial morphism σ A : Q ̂Q → A which satisfies (161). Nowwe want to show that σ A is A-bilinear that is the following equalities hold( )σ A ◦A µ Q ̂Q = m A ◦ ( Aσ A))σ A ◦(Qµ A Q b = m A ◦ ( σ A A ) .We computem A ◦ ( Aσ A) ◦ (AQl) (161)= m A ◦ (Am A ) ◦ (AxA)m A ass= m A ◦ (m A A) ◦ (AxA) (104)= m A ◦ (xA) ◦ (A µ Q P A )(161)= σ A ◦ (Ql) ◦ (A µ Q P A ) A µ Q= σ A ◦Since AQl is an epimorphism, we get that σ A ◦(A µ Q ̂Q( )A µ Q ̂Q)◦ (AQl) .m A ◦ ( σ A A ) ◦ (QlA) (161)= m A ◦ (m A A) ◦ (xAA)= m A ◦ (Am A ) ◦ (xAA) = x m A ◦ (xA) ◦ (QP m A ))(161)= σ A ◦ (Ql) ◦ (QP m A ) (150)= σ A ◦(Qµ A Q b ◦ (QlA)= m A ◦ ( Aσ A) . We compute

)Since QlA is an epimorphism, we obtain that σ A ◦(Qµ A Q b = m A ◦ ( σ A A ) . Symmetrically,wewant to define σ B . We prove thatIn fact, we havem B ◦ (By) ◦ (yP Q) ◦ ( P w l Q ) = m B ◦ (By) ◦ (yP Q) ◦ (P w r Q) .m B ◦ (By) ◦ (yP Q) ◦ ( P w l Q ) = m B ◦ (yy) ◦ ( P w l Q )(109)= y ◦ (P χ) ◦ ( P w l Q ) = y ◦ (P χ) ◦ (P χP Q) ◦ (P QP δ C Q)(128)= y ◦ (P χ) ◦ (P QP χ) ◦ (P QP δ C Q)(129)= y ◦ (P χ) ◦ ( P QP ε C Q ) = y ◦ (P χ) ◦ (P w r Q)(109)= m B ◦ (yy) ◦ (P w r Q) = m B ◦ (By) ◦ (yP Q) ◦ (P w r Q) .By Lemma <strong>2.</strong>9, we have(̂QQ, ν′0 Q)= Coequ Fun((yP Q) ◦(P w l Q ) , (yP Q) ◦ (P w r Q) )139so that there exists a functorial morphism σ B : ̂QQ → B which satisfies (162). Nowwe want to show that σ B is B-bicolinear that is the following equalities hold( )σ B ◦B µb QQ = m B ◦ ( Bσ B)m B ◦ ( σ B B ) ( )= σ B ◦ ̂QµBQWe calculatem B ◦ ( Bσ B) ◦ (Bν ′ 0Q) (162)= m B ◦ (Bm B ) ◦ (BBy)m B ass= m B ◦ (m B B) ◦ (BBy) m =Bm B ◦ (By) ◦ (m B P Q)( )(162)= σ B ◦ (ν 0Q) ′ ◦ (m B P Q) (151)= σ B ◦B µb QQ ◦ (Bν 0Q) ′ .( )Since Bν 0Q ′ is an epimorphism, we deduce that σ B ◦B µb QQ = m B ◦ ( Bσ B) . Wecomputem B ◦ ( σ B B ) ◦ (ν ′ 0QB) (162)= m B ◦ (m B B) ◦ (ByB)m B ass= m B ◦ (Bm B ) ◦ (ByB) (108)(162)= σ B ◦ (ν ′ 0Q) ◦ ( BP µ B Q= m B ◦ (By) ◦ ( )BP µ B Q( )̂QµBQ ◦ (ν 0QB) ′ .) ν ′0= σ B ◦Since ν 0QB ′ is an epimorphism, we get that m B ◦ ( σ B B ) = σ B ◦have to prove the associative conditionsA µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B)( ) (σ B ̂Q = µ A Q b ◦ ̂QσA).B µb Q◦(̂QµBQ). Finally we

140We calculateA µ Q ◦ ( σ A Q ) ◦ (QlQ) ◦ (QP xQ) (161)= A µ Q ◦ (m A Q) ◦ (xAQ) ◦ (QP xQ)= A µ Q ◦ (m A Q) ◦ (xxQ) (102)= A µ Q ◦ (xQ) ◦ (χP Q)(101)= χ ◦ (χP Q) (128)= χ ◦ (QP χ) (107)= µ B Q ◦ (Qy) ◦ (QP χ)(109)= µ B Q ◦ (Qm B ) ◦ (Qyy) = µ B Q ◦ (Qm B ) ◦ (QBy) ◦ (QyP Q)(162)= µ B Q ◦ ( Qσ B) ◦ (Qν ′ 0Q) ◦ (QyP Q)(149)= µ B Q ◦ ( Qσ B) ◦ (QlQ) ◦ (QP xQ) .Since (QlQ)◦(QP xQ) is an epimorphism, we deduce that A µ Q ◦ ( σ A Q ) = µ B Q ◦( Qσ B) .We compute( (µ A Q b ◦ ̂QσA)◦ ν 0Q ′ ̂Q) (◦ yP Q ̂Q)◦ (P QP Ql) ◦ (P QP QP x)( ((149)= µ A Q b ◦ ̂QσA)◦ lQ ̂Q) (◦ P xQ ̂Q)◦ (P QP Ql) ◦ (P QP QP x)l= µ A Q b ◦ (lA) ◦ ( P Aσ A) (◦ P xQ ̂Q)◦ (P QP Ql) ◦ (P QP QP x)(150)= l ◦ (P m A ) ◦ ( P Aσ A) (◦ P xQ ̂Q)◦ (P QP Ql) ◦ (P QP QP x)x= l ◦ (P m A ) ◦ (P xA) ◦ ( P QP σ A) ◦ (P QP Ql) ◦ (P QP QP x)(161)= l ◦ (P m A ) ◦ (P xA) ◦ (P QP m A ) ◦ (P QP xA) ◦ (P QP QP x)= l ◦ (P m A ) ◦ (P xA) ◦ (P QP m A ) ◦ (P QP xx)(102)= l ◦ (P m A ) ◦ (P xA) ◦ (P QP x) ◦ (P QP χP )(151)= l ◦ (P m A ) ◦ (P xx) ◦ (P QP χP )(102)= l ◦ (P x) ◦ (P χP ) ◦ (P QP χP )(128)= l ◦ (P x) ◦ (P χP ) ◦ (P χP QP )(149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P χP QP )(109)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P χP QP )= B µb Q◦ (Bν ′ 0) ◦ (yBP ) ◦ (P QyP ) ◦ (P χP QP )y= B µb Q◦(149)= B µb Q◦χ= B µb Q◦(109)= B µb Q◦( )y ̂Q ◦ (P Qν 0) ′ ◦ (P QyP ) ◦ (P χP QP )( )y ̂Q ◦ (P Ql) ◦ (P QP x) ◦ (P χP QP )( ) ( )y ̂Q ◦ P χ ̂Q ◦ (P QP Ql) ◦ (P QP QP x)( ) (m B ̂Q ◦ yy ̂Q)◦ (P QP Ql) ◦ (P QP QP x)

( ) (= B µb Q◦ m B ̂Q ◦ By ̂Q) (◦ yP Q ̂Q)◦ (P QP Ql) ◦ (P QP QP x)( ) ((162)= B µb Q◦ σ B ̂Q ◦ ν 0Q ′ ̂Q) (◦ yP Q ̂Q)◦ (P QP Ql) ◦ (P QP QP x)(Since ν 0Q ′ ̂Q) (◦ yP Q ̂Q)◦ (P QP Ql) ◦ (P QP QP x) is an epimorphism, we get that( ) ( )B µb Q◦ σ B ̂Q = µ A Q b ◦ ̂QσA. □7.4. From coherds to herds.7.8. Given a coherd χ : QP Q → Q in a formal codual structureX = (C, D, P, Q, δ C , δ D ), our purpose is to build the formal dual structure M =(A, B, ̂Q, Q, σ A , σ B ) and then an herd τ : Q → Q ̂QQ in M.Theorem 7.9. Let A and B be categories with coequalizers and let P : A → B,Q : B → A, C : A → A and D : B → B be functors. Assume that all thefunctors P, Q, C and D preserve coequalizers. Let ε C : C → A and ε D : D → Bbe functorial epimorphisms and assume that ( A, ε C) (= Coequ Fun Cε C , ε C C ) ( ) (andB, εD= Coequ Fun Dε D , ε D D ) . Let χ : QP Q → Q be a functorial morphism suchthatχ ◦ (QP χ) = χ ◦ (χP Q) .Let δ C : C → QP be a functorial morphism such thatχ ◦ (δ C Q) = (ε C Q)and let δ D : D → P Q be a functorial morphism such thatχ ◦ (Qδ D ) = (Qε D ).Then there is a formal dual structure M = (A, B, ̂Q, Q, σ A , σ B ).Proof. In view of Theorem 6.29 and Propositions 7.6, 7.7 a formal dual structureM = (A, B, ̂Q, Q, σ A , σ B ) has been constructed.□Theorem 7.10. Let A and B be categories with coequalizers and letX = (C, D, P, Q, δ C , δ D ) be a regular formal codual structure where P : A → B,Q : B → A, C : A → A and D : B → B are functors that preserve coequalizers.Let χ : QP Q → Q be a copretorsor. Then there is a formal dual structure M =(A, B, ̂Q, Q, σ A , σ B ). Define τ : Q → Q ̂QQ by settingτ := (QlQ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q.Then τ is an herd in M.Proof. By Theorem 7.9, M = (A, B, ̂Q, Q, σ A , σ B ) is a formal dual structure. To showthat τ is an herd in M, we have to prove ( that it satisfies the following conditions.1) Associativity, in the sense that Q ̂Qτ) (◦ τ = τ ̂QQ)◦ τ. We have(Q ̂Qτ)◦ τ(= Q ̂Qτ)◦ (QlQ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q141

142x=and=δ C=(l= QlQ ̂QQ)◦ (QP Aτ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(QlQ ̂QQ) (◦ QP xQ ̂QQ)◦ (QP QP τ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(δ=CQlQ ̂QQ) (◦ QP xQ ̂QQ) (◦ δ C QP Q ̂QQ)◦ (CQP τ) ◦ (CQδ D )δ C=(125)=◦ (C ρ Q D ) ◦ ρ D QC ρ Q=(QlQ ̂QQ) (◦ QP xQ ̂QQ) (◦ δ C QP Q ̂QQ) (◦C ρ Q P Q ̂QQ)◦ (QP τ)(127)=(QlQ ̂QQ) (◦ QP xQ ̂QQ(QlQ ̂QQ) (◦ QP xQ ̂QQ)◦(C ρ Q P Q ̂QQ)◦ (Qδ D ) ◦ ρ D Q)◦(δ C QP Q ̂QQ)◦◦ (δ C Q) ◦ C ρ Q(δ C QP Q ̂QQ)◦◦ (Cτ) ◦ C ρ Q(δ C Q ̂QQ)◦ (Cτ) ◦ C ρ Q(C ρ Q P Q ̂QQ)◦ (QP τ)(C ρ Q P Q ̂QQ) (◦ δ C Q ̂QQ)◦(C ρ Q P Q ̂QQ) (◦ δ C Q ̂QQ)◦ (CQlQ) ◦ (CQP xQ) ◦ (Cδ C QP Q) ◦ (CCQδ D )◦ ( C C ρ Q D ) ◦ ( CρQ) D ◦ C ρ Q(C ρ Q P Q ̂QQ) (◦ δ C Q ̂QQ)◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D ) ◦ (Cδ C QD)◦ ( C C ρ Q D ) ◦ ( CρQ) D ◦ C ρ Q(Cδ C Q ̂QQ) (◦ ∆ C Q ̂QQ)◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D ) ◦ (Cδ C QD)Since◦ ( C C ρ Q D ) ◦ ( Cρ D Q)◦ C ρ Q(C C ρ Q D ) ◦ ( )Cρ D Q ◦ C ρ QQis a bicom=( ) ( )CCρDQ ◦ C C ρ Q ◦ C Qis a comρ Q = ( (CCρQ) D ◦ ∆ C Q ) ◦ C ρ Q)◦ C Qis a bicom (ρ Q = ∆ C QD ) ◦ (C ρ Q D ) ◦ ρ D Q∆ C= ( ∆ C QD ) ◦ ( Cρ D QQis a com= ( C C ρ Q D ) ◦ (C ρ Q D ) ◦ ρ D Qwe obtain(Cδ C Q ̂QQ) (◦ ∆ C Q ̂QQ)◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D ) ◦ (Cδ C QD)◦ ( C C ρ Q D ) ◦ ( CρQ) D ◦ C ρ Q(= Cδ C Q ̂QQ) (◦ ∆ C Q ̂QQ)◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D ) ◦ (Cδ C QD)◦ ( C C ρ Q D ) ◦ (C ρ Q D ) ◦ ρ D Q

(127)=(Cδ C Q ̂QQ) (◦ ∆ C Q ̂QQ)◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D )143∆ C=Qis a com=Qis a bicom=◦ (CQδ D D) ◦ ( Cρ D QD ) ◦ (C ρ Q D ) ◦ ρ D Q(Cδ C Q ̂QQ)◦ (CCQlQ) ◦ (CCQP xQ) ◦ (CCQP Qδ D ) ◦ (CCQδ D D)◦ ( CCρ D QD ) ◦ ( ∆ C QD ) ◦ (C ρ Q D ) ◦ ρ D Q(Cδ C Q ̂QQ)◦ (CCQlQ) ◦ (CCQP xQ) ◦ (CCQP Qδ D ) ◦ (CCQδ D D)◦ ( CCρ D QD ) ◦ ( C C ρ Q D ) ◦ (C ρ Q D ) ◦ ρ D Q(Cδ C Q ̂QQ)◦ (CCQlQ) ◦ (CCQP xQ) ◦ (CCQP Qδ D ) ◦ (CCQδ D D)◦ ( C C ρ Q DD ) ◦ ( Cρ D QD ) ◦ (C ρ Q D ) ◦ ρ D QC ρ Q=(Cδ C Q ̂QQ))◦(C C ρ Q ̂QQ ◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D ) ◦ (CQδ D D)Qis a bicom=◦ ( Cρ D QD ) ◦ (C ρ Q D ) ◦ ρ D Q(Cδ C Q ̂QQ))◦(C C ρ Q ̂QQ ◦ (CQlQ) ◦ (CQP xQ) ◦ (CQP Qδ D )◦ (CQδ D D) ◦ (C ρ Q DD ) ◦ ( ρ D QD ) ◦ ρ D QC ρ Q=(Cδ C Q ̂QQ))◦(C C ρ Q ̂QQ (C )◦ ρ Q ̂QQ ◦ (QlQ) ◦ (QP xQ) ◦ (QP Qδ D )(127)=(127)=Qis a bicom=δ C=◦ (Qδ D D) ◦ ( ρ D QD ) ◦ ρ D Q) ((CQδ D ̂QQ ◦ Cρ D ̂QQ) (C )Q ◦ ρ Q ̂QQ ◦ (QlQ) ◦ (QP xQ) ◦ (QP Qδ D )◦ (Qδ D D) ◦ ( ρ D QD ) ◦ ρ D Q) ((CQδ D ̂QQ ◦ Cρ D ̂QQ) (C )Q ◦ ρ Q ̂QQ ◦ (QlQ) ◦ (QP xQ) ◦ (QP Qδ D )◦ (δ C QD) ◦ (C ρ Q D ) ◦ ρ D Q) ((CQδ D ̂QQ ◦C ρ Q D ̂QQ) (◦ ρ D ̂QQ)Q ◦ (QlQ) ◦ (QP xQ) ◦ (QP Qδ D )◦ (δ C QD) ◦ (C ρ Q D ) ◦ ρ D Q) ((CQδ D ̂QQ ◦C ρ Q D ̂QQ) (◦ ρ D ̂QQ)Q ◦ (QlQ) ◦ (QP xQ) ◦ (δ C QP Q)Hence we obtain◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(Q ̂Qτ)◦ τ(= QlQ ̂QQ) (◦ QP xQ ̂QQ) (◦ δ C QP Q ̂QQ) (◦C ρ Q P Q ̂QQ) (◦ δ C Q ̂QQ)=(QlQ ̂QQ) (◦ QP xQ ̂QQ)◦◦ (Cτ) ◦ C ρ Q(δ C QP Q ̂QQ))◦(CQδ D ̂QQ ◦(C ρ Q D ̂QQ)

144(◦ ρ D ̂QQ)Q◦ (QlQ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(= τ ̂QQ)◦ τ.2) Counitality, in the sense that ( Qσ B) ◦ τ = Qu B and ( σ A Q ) ◦ τ = u A Q. Let usprove that (QσB ) ◦ τ = Qu B .In fact, we have(QσB ) ◦ τ= ( Qσ B) ◦ (QlQ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(149)= ( Qσ B) ◦ (Qν ′ 0Q) ◦ (QyP Q) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(162)= (Qm B ) ◦ (QBy) ◦ (QyP Q) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D QLet us prove that= (Qm B ) ◦ (Qyy) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(109)= (Qy) ◦ (QP χ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Qδ C= (Qy) ◦ (QP χ) ◦ (QP Qδ D ) ◦ (δ C QD) ◦ (C ρ Q D ) ◦ ρ D Q(130)= (Qy) ◦ ( QP Qε D) ◦ (δ C QD) ◦ (C ρ Q D ) ◦ ρ D Qδ C= (Qy) ◦ (δ C Q) ◦ ( CQε D) ◦ (C ρ Q D ) ◦ ρ D QQis a bicom= (Qy) ◦ (δ C Q) ◦ ( CQε D) ◦ (Cρ D Q) ◦ C ρ QQis a com= (Qy) ◦ (δ C Q) ◦ C ρ Q(127)= (Qy) ◦ (Qδ D ) ◦ ρ D Q(110)= (Qu B ) ◦ ( Qε D) ◦ ρ D Qis a comQ = Qu B .(σ A Q ) ◦ τ = (u A Q).We calculate(σ A Q ) ◦ τ= ( σ A Q ) ◦ (QlQ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(161)= (m A Q) ◦ (xAQ) ◦ (QP xQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q= (m A Q) ◦ (xxQ) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(102)= (xQ) ◦ (χP Q) ◦ (δ C QP Q) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Q(129)= (xQ) ◦ ( ε C QP Q ) ◦ (CQδ D ) ◦ (C ρ Q D ) ◦ ρ D Qε C = (xQ) ◦ (Qδ D ) ◦ ( ε C QD ) ◦ (C ρ Q D ) ◦ ρ D QQis a com= (xQ) ◦ (Qδ D ) ◦ ρ D (127)Q = (xQ) ◦ (δ C Q) ◦ C ρ Q(103)= (u A Q) ◦ ( ε C Q ) ◦ C ρ QQis a com= u A Q.

7.5. Herd - Coherd - Herd.7.1<strong>1.</strong> Let τ : Q → QP Q be a herd for a regular formal dual structure M =(A, B, P, Q, σ A , σ B , ) where P : A → B, Q : B → A, A : A → A and B : B → Bare functors that preserve equalizers. Then, by Propositions 6.1 and 6.2, we canconstruct comonads C = ( C, ∆ C , ε C) and D = ( D, ∆ D , ε D) and functorial morphismsC ρ Q : Q → CQ and ρ D Q : Q → QD such that (Q, C ρ Q , ρ D Q ) is a C-Dbicomodule) functor (see Theorem 6.5). Let Q as defined in Proposition 7.<strong>1.</strong> Then(Q, D ρ Q , ρ C is a D-C-bicomodule functor. By Theorem 7.5, we construct a coherdQX = ( C, D, Q, Q, δ C , δ D , χ ) where χ := µ B Q ◦(A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦(QP iQ)◦(QqQ). Then we can construct monads A ′ = (A ′ , m A ′, u A ′) and B ′ = (B ′ , m B ′, u B ′)following respectively Proposition 6.25 and Proposition 6.26 as the coequalizers QQC (χQ)◦(QQδC )QQ x′ A ′QQε C DQQ (Qχ)◦(δD QQ)SQ y′ B ′ε D QQThis means that the following holdm A ′ ◦ (x ′ x ′ ) = x ′ ◦ ( χQ ) and u A ′ ◦ ε C = x ′ ◦ δ C(163) m B ′ ◦ (y ′ y ′ ) = y ′ ◦ ( Qχ ) and u B ′ ◦ ε D = y ′ ◦ δ D .Notation 7.1<strong>2.</strong> With notations of Theorem 6.5 and Proposition 7.1, let h : Q → Pbe defined by settingh = B µ P ◦ ( σ B P ) ◦ (P i) ◦ q.The following theorem reformulates Theorem 3.5 in [BV] in our categorical setting.Theorem 7.13. Let M = (A, B, P, Q, σ A , σ B ) be a tame Morita context and letτ : Q → QP Q be a herd for M such that A and B reflect equalizers and coequalizers.We denote by A ′ and B ′ the monads constructed in Claim 7.1<strong>1.</strong> Then1) There are functorial morphisms ν A : A ′ → A and ν B : B ′ → B such that ν Aand ν B are morphisms of monads.2) If the functorial morphism hQ : QQ → P Q, where h = B µ P ◦ ( σ B P ) ◦(P i)◦q,is an isomorphism, then ν A and ν B are isomorphisms.3) If P C = Q ≃ Q ′ = DP then hQ is an isomorphism and hence ν A and ν Bare isomorphisms of monads.Proof. Note that, since A and B reflect equalizers, by Lemma 6.10, we have aregular herd, i.e. the assumptions (A, u A ) = Equ Fun (u A A, Au A ) and (B, u B ) =Equ Fun (u B B, Bu B ) are fulfilled. We will prove only the statement for the monadB ′ , for A ′ the proof is similar.1) Consider the functorial morphismσ B : QQ → B145□

146given byWe computeσ B = m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ)(144)= m B ◦ ( σ B σ B) ◦ (jP Q) ◦ (κ ′ 0Q) .(σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ ) ◦ ( δ D QQ )q= ( σ B P Q ) ◦ (P iQ) ◦ (P Cχ) ◦ ( qQQQ ) ◦ ( δ D QQ )i= ( σ B P Q ) ◦ (P QP χ) ◦ ( P iQQQ ) ◦ ( qQQQ ) ◦ ( δ D QQ )(144)= ( σ B P Q ) ◦ (P QP χ) ◦ ( jP QQQ ) ◦ ( κ ′ 0QQQ ) ◦ ( δ D QQ )(148)= ( σ B P Q ) ◦ (P QP χ) ◦ ( jP QQQ ) ◦ ( DjQQ ) ◦ ( ∆ D QQ )(67)= ( σ B P Q ) ◦ (P QP χ) ◦ ( P τQQ ) ◦ ( jQQ )= ( σ B P Q ) ◦ ( ) (P QP µ B Q ◦ P QP A µ Q B ) ◦ ( P QP AQσ B) ◦ ( P QP σ A QP Q )◦ (P QP QP iQ) ◦ (P QP QqQ) ◦ ( P τQQ ) ◦ ( jQQ )σ= ( A σ B P Q ) ◦ ( (P QP µ Q) B ◦ P QP A µ Q B ) ◦ ( P QP σ A QB ) ◦ ( P QP QP Qσ B)◦ (P QP QP iQ) ◦ (P QP QqQ) ◦ ( P τQQ ) ◦ ( jQQ )(82)= ( σ B P Q ) ◦ ( ) (P QP µ B Q ◦ P QP µBQ B ) ◦ ( P QP Qσ B B )◦ ( P QP QP Qσ B) ◦ (P QP QP iQ) ◦ (P QP QqQ) ◦ ( P τQQ ) ◦ ( jQQ )σ= ( ) ( BBP µ B Q ◦ BP µBQ B ) ◦ ( BP Qσ B B ) ◦ ( BP QP Qσ B) ◦ (BP QP iQ)◦ (BP QqQ) ◦ ( σ B P QQQ ) ◦ ( P τQQ ) ◦ ( jQQ )defD= ( BP µ B Q)◦(BP µBQ B ) ◦ ( BP Qσ B B ) ◦ ( BP QP Qσ B) ◦ (BP QP iQ)◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ )and thus we obtain (σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ ) ◦ ( δ D QQ )(164)= ( BP µ B Q)◦(BP µBQ B ) ◦ ( BP Qσ B B ) ◦ ( BP QP Qσ B) ◦ (BP QP iQ)Let us compute◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ ) .σ B ◦ ( Qχ ) ◦ ( δ D QQ )= m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ ) ◦ ( δ D QQ )= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ ) ◦ ( δ D QQ )(164)= m B ◦ ( Bσ B) ◦ ( ) (BP µ B Q ◦ BP µBQ B ) ◦ ( BP Qσ B B )◦ ( BP QP Qσ B) ◦ (BP QP iQ) ◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ )u B= m B ◦ (u B B) ◦ σ B ◦ ( P µ B Q)◦(P µBQ B ) ◦ ( P Qσ B B ) ◦ ( P QP Qσ B)

◦ (P QP iQ) ◦ ( jQQ )Bmonad= σ B ◦ ( P µ B Q)◦(P µBQ B ) ◦ ( P Qσ B B ) ◦ ( P QP Qσ B) ◦ (P QP iQ)◦ ( jQQ )(82)= σ B ◦ ( (P µ Q) B ◦ P A µ Q B ) ◦ ( P σ A QB ) ◦ ( P QP Qσ B) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )147σ= A σ B ◦ ( ) (P µ B Q ◦ P A µ Q B ) ◦ ( P AQσ B) ◦ ( P σ A QP Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )(81)= m B ◦ ( σ B B ) ◦ ( P A µ Q B ) ◦ ( P AQσ B) ◦ ( P σ A QP Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )A µ Q= mB ◦ ( σ B B ) ◦ ( P Qσ B) ◦ ( P A µ Q P Q ) ◦ ( P σ A QP Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )(82)= m B ◦ ( σ B B ) ◦ ( P Qσ B) ◦ ( P µ B QP Q ) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )σ B = m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ ( P µ B QP Q ) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )(81)= m B ◦ ( Bσ B) ◦ (m B P Q) ◦ ( σ B BP Q ) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )σ B = m B ◦ ( Bσ B) ◦ (m B P Q) ◦ ( Bσ B P Q ) ◦ (BP iQ) ◦ (BqQ) ◦ ( σ B QQ ) ◦ ( jQQ )(67)= m B ◦ ( Bσ B) ◦ (m B P Q) ◦ ( Bσ B P Q ) ◦ (BP iQ) ◦ (BqQ) ◦ ( u B QQ )◦ ( ε D QQ )u B= m B ◦ ( Bσ B) ◦ (m B P Q) ◦ (u B BP Q) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( ε D QQ )so that we getBmonad= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( ε D QQ )= σ B ◦ ( ε D QQ )(165) σ B ◦ ( Qχ ) ◦ ( δ D QQ ) = σ B ◦ ( ε D QQ )and since (B ′ , y ′ ) = Coequ Fun((Qχ)◦(δD QQ ) , ε D QQ ) there exists a unique functorialmorphism ν B : B ′ → B such that(166) ν B ◦ y ′ = σ B = m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) .Now we want to prove that ν B is a morphism of monads. Let us computem B ◦ (ν B ν B ) ◦ (y ′ y ′ ) = m B ◦ (ν B B) ◦ (B ′ ν B ) ◦ (y ′ B ′ ) ◦ ( QQy ′)

148andy= ′m B ◦ (ν B B) ◦ (y ′ B) ◦ ( ) (QQν )B ◦ QQy′(166)= m B ◦ (m B B) ◦ ( σ B σ B B ) ◦ (P iQB) ◦ (qQB) ◦ ( QQm B)◦(QQσ B σ B)◦ ( QQP iQ ) ◦ ( QQqQ )(144)= m B ◦ (m B B) ◦ ( σ B σ B B ) ◦ (jP QB) ◦ (κ ′ 0QB) ◦ ( QQm B)◦(QQσ B σ B)◦ ( QQjP Q ) ◦ ( QQκ ′ 0Q )= m B ◦ (m B B) ◦ ( Bσ B B ) ◦ ( σ B P QB ) ◦ (jP QB) ◦ (κ ′ 0QB) ◦ ( QQm B)◦ ( QQBσ B) ◦ ( QQσ B P Q ) ◦ ( QQjP Q ) ◦ ( QQκ ′ 0Q )(67)= m B ◦ (m B B) ◦ ( Bσ B B ) ◦ (u B P QB) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQm B)◦ ( QQBσ B) ◦ ( QQu B P Q ) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )u B= m B ◦ (m B B) ◦ (u B BB) ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQm B)◦ ( QQu B B ) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )Bmonad= m B ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )ν B ◦ m B ′ ◦ (y ′ y ′ ) (163)= ν B ◦ y ′ ◦ ( Qχ ) = m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ )(144)= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) ◦ (κ ′ 0Q) ◦ ( Qχ )(67)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qχ )u B= m B ◦ (u B B) ◦ σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qχ )Bmonad= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qχ )defχ= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ B Q ◦ Q A µ Q B ) ◦ ( QAQσ B) ◦ ( Qσ A QP Q )◦ ( QQP iQ ) ◦ ( QQqQ )A µ Q= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ ) B Q ◦ QQσB◦ ( Q A µ Q P Q ) ◦ ( Qσ A QP Q )◦ ( QQP iQ ) ◦ ( QQqQ )(82)= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ ) B Q ◦ QQσB◦ ( Qµ B QP Q )◦ ( QQσ B P Q ) ◦ ( QQP iQ ) ◦ ( QQqQ )(144)= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ ) B Q ◦ QQσB◦ ( Qµ B QP Q )◦ ( QQσ B P Q ) ◦ ( QQjP Q ) ◦ ( QQκ ′ 0Q )(67)= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qµ B Q◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )Qmodulefunctor= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qµ B Q)◦(QQσB ) ◦ ( Qµ B QP Q ) ◦ ( QQu B P Q ))◦(QQσB ) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )

κ ′ 0= σ B ◦ ( ε D P Q ) ◦ ( )DP µ B Q ◦ (κ′0 QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )ε= D σ B ◦ ( (P µ Q) B ◦ ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )(81)= m B ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )so that we obtainm B ◦ (ν B ν B ) ◦ (y ′ y ′ ) = ν B ◦ m B ′ ◦ (y ′ y ′ )and since y ′ is an epimorphism we deduce thatNow, let us calculatem B ◦ (ν B ν B ) = ν B ◦ m B ′.ν B ◦ u B ′ ◦ ε D (163)= ν B ◦ y ′ ◦ δ D(166)= m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) ◦ δ D= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ δ D(144)= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) ◦ (κ ′ 0Q) ◦ δ D(67),(148)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ (ε D P Q) ◦ (Dj) ◦ ∆ Du B ,ε D= m B ◦ (u B B) ◦ σ B ◦ j ◦ (ε D D) ◦ ∆ DBmonadDcomonad= σ B ◦ j (67)= u B ◦ ε D .Since the tame condition is assumed, BA σ B ABAσ B A(97)= A σ B ABBF (94)= B U BA σ B ABAFis also an isomorphism. By (95) we have thatAσ B A ◦ (p P A Q) = σ Bis a functorial isomorphism and thusand Lemma <strong>2.</strong>6, since p P A Q is a regular epimorphism by construction, we deducethat σ B is also a regular epimorphism. Thus, by Theorem 6.6, so is Bε D . Since Breflects coequalizers we get that ε D is an epimorphism and therefore we obtainHence ν B is a morphism of monads.2) Consider the following diagramν B ◦ u B ′ = u B .149DQQDhQDP Q(Qχ)◦(δ D QQ)(ε D QQ)(P µ S Q )◦(jS)◦(Dσ S) (ε D P Q)QQ y′ hQ P Q π Z ZS ′ ν Zwhere (Z, π Z ) = Coequ Fun((P µBQ)◦ (jB) ◦(DσB ) , ε D P Q ) . Note that(ε D P Q ) ◦ (DhQ) εD = (hQ) ◦ ( ε D QQ ) .

150Now we compute(hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) ? = ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)(hQ) ◦ ( Qχ ) ◦ ( δ D QQ )= (B µ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ ) ◦ ( δ D QQ )(164)= (B µ P Q ) ◦ ( ) (BP µ B Q ◦ BP µBQ B ) ◦ ( BP Qσ B B )◦ ( BP QP Qσ B) ◦ (BP QP iQ) ◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ )Qmodfunctor=( Bµ P Q ) ◦ ( BP µ B Q)◦ (BP QmB ) ◦ ( BP Qσ B B ) ◦ ( BP QP Qσ B)◦ (BP QP iQ) ◦ (BP QqQ) ◦ ( u B P QQQ ) ◦ ( jQQ )u= (B Bµ P Q ) ◦ (u B P Q) ◦ ( P µ Q) B ◦ (P QmB ) ◦ ( P Qσ B B ) ◦ ( P QP Qσ B)◦ (P QP iQ) ◦ (P QqQ) ◦ ( jQQ )Qmodfunctor ( )= P µBQ ◦ (P QmB ) ◦ ( P Qσ B B ) ◦ ( P QP Qσ B) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )σ= ( ) BP µ B Q ◦ (P QmB ) ◦ ( P QBσ B) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )(81)= ( P µ B Q)◦(P QσB ) ◦ ( P Q B µ P Q ) ◦ ( P Qσ B P Q ) ◦ (P QP iQ)◦ (P QqQ) ◦ ( jQQ )j= ( P µ B Q)◦ (jB) ◦(DσB ) ◦ ( D B µ P Q ) ◦ ( Dσ B P Q ) ◦ (DP iQ) ◦ (DqQ)= ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)Then the diagram above serially commute and hence in particularso thatπ Z ◦ (hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) = π Z ◦ ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)π Z= π Z ◦ ( ε D P Q ) ◦ (DhQ) = π Z ◦ (hQ) ◦ ( ε D QQ )π Z ◦ (hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) = π Z ◦ (hQ) ◦ ( ε D QQ ) .Since (B ′ , y ′ ) = Coequ Fun((Qχ)◦(δD QQ ) , ε D QQ ) , by the universal property ofcoequalizers, there exists a unique functorial morphism ν Z : B ′ → Z such that(167) ν Z ◦ y ′ = π Z ◦ (hQ) .We want to prove that ν Z is an isomorphism. Since hQ is an isomorphism, thereexists (hQ) −1 : P Q → QQ. Note that fromwe deduce that(hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) = ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (DhQ)(hQ) −1 ◦ (hQ) ◦ ( Qχ ) ◦ ( δ D QQ ) ◦ ( D (hQ) −1) == (hQ) −1 ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB◦ (DhQ) ◦ ( D (hQ) −1)

151that is(168)Similarly, fromwe deduce that( ) (Qχ ◦ δD QQ ) ◦ ( D (hQ) −1) = (hQ) −1 ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB.(ε D P Q ) ◦ (DhQ) = (hQ) ◦ ( ε D QQ )(169) (hQ) −1 ◦ ( ε D P Q ) = ( ε D QQ ) ◦ ( D (hQ) −1) .Thus we haveso thaty ′ ◦ (hQ) −1 ◦ ( P µ B Q)◦ (jB) ◦(DσB ) (168)= y ′ ◦ ( Qχ ) ◦ ( δ D QQ ) ◦ ( D (hQ) −1)defy ′= y ′ ◦ ( ε D QQ ) ◦ ( D (hQ) −1) (169)= y ′ ◦ (hQ) −1 ◦ ( ε D P Q )y ′ ◦ (hQ) −1 ◦ ( P µ B Q)◦ (jB) ◦(DσB ) = y ′ ◦ (hQ) −1 ◦ ( ε D P Q ) .(( ) (Since (Z, π Z ) = Coequ ) Fun P µBQ ◦ (jB) ◦ DσB, ε D P Q ) , by the universal propertyof coequalizers, there exists a unique functorial morphism ν Z′ : Z → B′ suchthat(170) ν ′ Z ◦ π Z = y ′ ◦ (hQ) −1 .Now we want to prove that ν ′ Z is the two-sided inverse of ν Z. Let us computeand since y ′ is an epimorphism we getMoreoverν ′ Z ◦ ν Z ◦ y ′ (167)= ν ′ Z ◦ π Z ◦ (hQ)(170)= y ′ ◦ (hQ) −1 ◦ (hQ) = y ′ν ′ Z ◦ ν Z = Id B ′.ν Z ◦ ν ′ Z ◦ π Z(170)= ν Z ◦ y ′ ◦ (hQ) −1 (167)= π Z ◦ (hQ) ◦ (hQ) −1 = π Zand since π Z is an epimorphism we deduce thatν Z ◦ ν ′ Z = Id Z .Thus ν Z is a functorial isomorphism between B ′ and Z with inverse ν Z ′ . Now we wantto construct an isomorphism between B and Z. Consider the parallel pairDP Q(P µ S Q )◦(jS)◦(Dσ S) (ε D P Q) P Qand computeσ B ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB (81),j= m B ◦ ( σ B B ) ◦ ( P Qσ B) ◦ (jP Q)σ= B m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) (67)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ ( ε D P Q )u B= m B ◦ (u B B) ◦ σ B ◦ ( ε D P Q ) Bmonad= σ B ◦ ( ε D P Q ) .

152Thus we obtainσ B ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB= σ B ◦ ( ε D P Q )and since (Z, π Z ) = Coequ Fun((P µBQ)◦ (jB) ◦(DσB ) , ε D P Q ) , by the universalproperty of coequalizers, there exists a unique functorial morphism λ : Z → B suchthat(171) λ ◦ π Z = σ B .Since we already proved in 1) that σ B is a regular epimorphism, in particular wecan write ( B, σ B) = Coequ Fun (ξ, ζ). Let us computeso thatπ Z ◦ ξ ◦ ( ε D P Q ) ε D = π Z ◦ ( ε D P Q ) ◦ (Dξ)defπ Z= πZ ◦ ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (Dξ)σ B coequ= π Z ◦ ( (P µ Q) ) B ◦ (jB) ◦ DσB◦ (Dζ)defπ=ZπZ ◦ ( ε D P Q ) ◦ (Dζ) εD = π Z ◦ ζ ◦ ( ε D P Q )π Z ◦ ξ ◦ ( ε D P Q ) = π Z ◦ ζ ◦ ( ε D P Q ) .Since σ B is a regular epimorphism, by Theorem 6.6, so is Bε D . Since by assumptionB reflects coequalizers, also ε D is an epimorphism so that we get(172) π Z ◦ ξ = π Z ◦ ζ.Since ( B, σ B) = Coequ Fun (ξ, ζ) , by the universal property of coequalizers thereexists a unique functorial morphism λ ′ : B → Z such that(173) λ ′ ◦ σ B = π Z .We prove that λ ′ is the two-sided inverse of λ. In factλ ′ ◦ λ ◦ π Z(171)= λ ′ ◦ σ B (173)= π Z .Since π Z is an epimorphism we deduce thatSimilarlyλ ′ ◦ λ = Id Z .λ ◦ λ ′ ◦ σ B (173)= λ ◦ π Z(171)= σ Band, since also σ B is an epimorphism, we deduce thatWe now want to prove thati.e.We computeλ ◦ λ ′ = Id B .ν B = λ ◦ ν Zλ ′ ◦ ν B = ν Z .λ ′ ◦ ν B ◦ y ′ (109)= λ ′ ◦ m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ)

153(81)= λ ′ ◦ σ B ◦ (B µ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ)(173)= π Z ◦ (B µ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) = π Z ◦ (hQ) (167)= ν Z ◦ y ′ .Since y ′ is an epimorphism, we getλ ′ ◦ ν B = ν Zso that we deduce that ν B : B ′ → B is an isomorphism and thus an isomorphism ofmonads.3) If P preserves equalizers and P C = Q → κ′Q ′ = DP then ν A and ν B areisomorphisms of monads.By assumption we have that q = Id P C and q ′ = Id DP and κ and κ ′ are isomorphisms.Then we can rewrite the initial diagram as followsDP CQDhQDP Q(P Cχ)◦(δD P CQ)(ε D P CQ)(P µ S Q )◦(jS)◦(Dσ S)(ε D P Q)P CQ y′ hQ P Q π Z ZS ′ ν Z((Since (C, i) = Equ ) )Fun QP σA((◦ (τP ) , QP u A , by Lemma <strong>2.</strong>10 we also have(CQ, iQ) = Equ Fun QP σ A Q ) ◦ (τP Q) , QP u A Q ) . We compute(QP σ A Q ) ◦ (τP Q) ◦ τ (68)= ( QP σ A Q ) ◦ (QP τ) ◦ τ (69)= (QP u A Q) ◦ τso that there exists a unique functorial morphism C ρ Q : Q → CQ such that (61)holds i.e.(iQ) ◦ C ρ Q = τas constructed in Proposition 6.<strong>1.</strong> Moreover ( Q, C ρ Q)is a left C-comodule by Proposition6.<strong>1.</strong> We also get(P iQ) ◦ ( P C ρ Q)= P τ.Let us computeand(hQ) ◦ ( P C ρ Q)=( Bµ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ ( P C ρ Q)q=Id P C=( Bµ P Q ) ◦ ( σ B P Q ) ◦ (P iQ) ◦ ( P C ρ Q)(61)= (B µ P Q ) ◦ ( σ B P Q ) ◦ (P τ) (82)= ( µ A P Q ) ◦ ( P σ A Q ) ◦ (P τ)(69)= ( µ A P Q ) ◦ (P u A Q) P module= P Q(P C ρ Q)◦ (hQ) =(P C ρ Q)◦( Bµ P Q ) ◦ ( σ B P Q ) ◦ (P iQ)(82)= ( ) (P C ρ Q ◦ µAP Q ) ◦ ( P σ A Q ) ◦ (P iQ)(63)= ( ) (P C ρ Q ◦ µAP Q ) ◦ (P u A Q) ◦ ( P ε C Q )P module= ( P C ρ Q)◦(P ε C Q ) Qcomodule= P Q.

154Thus hQ is an isomorphism with inverse P C ρ Q and we can conclude by applying2). □8. Equivalence for (co)module categories8.<strong>1.</strong> Equivalence for module categories coming from copretorsor. In thissubsection we prove that, for given categories A and B, under the assumptions ofTheorem 6.29, there exist a monad A on A and a monad B on B such that theircategories of modules are equivalent. We outline that the assumptions quoted aboveare satisfied in the particular case of a regular coherd.First of all we need to define the functors A Q B and B ̂QA which will be used to setthe equivalence between these module categories.Using the functors Q and ̂Q, we construct the lifting functors A Q B : B B → A A andB ̂Q A : A A → B B.Proposition 8.<strong>1.</strong> In the setting of 6.29 there exists a functor A (Q B ) : B B → A Asuch that A U A (Q B ) = Q B where (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B). Moreoverwe have(174) p Q ◦ (A µ Q BU ) = A µ QB ◦ (Ap Q )where A µ QB = A Uλ AA (Q B ) : AQ B → Q B .Proof. In view of Theorem 6.29, we can apply Proposition 3.30.8.<strong>2.</strong> In light of Proposition 8.1, a functor Q : B → A introduced in 6.29 inducesa functor A (Q B ) : B B → A A for the monads A and B. Our next task is to provethat ( the ) B-A-bimodule functor ̂Q, constructed in Proposition 7.6, induces a functor̂QA : A A → B B which yields the inverse of A (Q B ) .BProposition 8.3. Within the assumptions and notations of Theorem 6.29, thereexists a functor B ̂QA : AA → B B such that B U B ̂QA = ̂Q( )A where ̂QA , pb Q=(Coequ Fun µ A Q bAU, ̂Q)A Uλ A . Moreover we have) ( )(175)A µ QB ◦(Bpb Q= pb Q◦B µb Q AUwhere B µb QA= B Uλ BB ̂QA : B ̂Q A → ̂Q A , so thatfunctor.(̂QA , B µb QA)□is an B-left moduleProof. In view of Proposition 7.6, we can apply Proposition 3.30 where Q is ̂Q andwe exchange the role of A and B, A and B.□Now we want to prove the first isomorphism.Within the assumptions and notations of Theorem 6.29, we will construct a functorialisomorphism B ̂QAA Q B∼ = B B.Lemma 8.4. Within the assumptions and notations of Theorem 6.29 the followingequality(176) ν ′ 0 ◦ (u B P ) = l ◦ (P u A )

155holds where ν ′ 0 is defined in (149) .Proof. We computeν ′ 0 ◦ (u B P ) ◦ ( ε D P ) ◦ ( DP ε C) (110)= ν ′ 0 ◦ (yP ) ◦ (δ D P ) ◦ ( DP ε C)δ D= ν ′ 0 ◦ (yP ) ◦ ( P QP ε C) ◦ (δ D P C)(149)= l ◦ (P x) ◦ ( P QP ε C) ◦ (δ D P C) = l ◦ (P x) ◦ (P w r ) ◦ (δ D P C)xcoequ= l ◦ (P x) ◦ ( P w l) ◦ (δ D P C) = l ◦ (P x) ◦ (P χP ) ◦ (P QP δ C ) ◦ (δ D P C)δ D= l ◦ (P x) ◦ (P χP ) ◦ (δ D P QP ) ◦ (DP δ C )(149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (δ D P QP ) ◦ (DP δ C ) = ν ′ 0 ◦ (yP ) ◦ (z l P ) ◦ (DP δ C )ycoequ= ν ′ 0 ◦ (yP ) ◦ (z r P ) ◦ (DP δ C ) = ν ′ 0 ◦ (yP ) ◦ (ε D P QP ) ◦ (DP δ C )(149)= l ◦ (P x) ◦ (ε D P QP ) ◦ (DP δ C ) = l ◦ (P x) ◦ (P δ C ) ◦ (ε D P C)(103)= l ◦ (P u A ) ◦ ( P ε C) ◦ (ε D P C) εD = l ◦ (P u A ) ◦ ( ε D P ) ◦ (DP ε C ).Since ( ε D P ) ◦ ( DP ε C) is an epimorphism (recall that both ε D and ε C are coequalizers),we conclude.□Proposition 8.5. Within the assumptions and notations of Theorem 6.29, thereexists a functorial morphism α : B B U → ̂Q AA Q B such that(̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ) . Moreover for every morphism h : B B U →X such thath ◦ (m BB U) = h ◦ (B B Uλ B )if ĥ : ̂Q AA Q B → X is the unique morphism such that ĥ ◦ α = h, we have that)(177) ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h ◦ (y B U) ◦ ( P A µ Q BU ) .Proof. Let us prove that)(178)(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U))=(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U) .Using Proposition 8.1, we compute) (pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U))u=A(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P u A QP Q B U)(101)=(174)=) (pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ ( P A A µ QB U ) ◦ (P AxQ B U)◦ (P u A QP Q B U)) (pb Q AQ B ◦ (l A U A Q B ) ◦ ( )P A A µ QB ◦ (P AApQ ) ◦ (P AxQ B U)◦ (P u A QP Q B U)

156) ( )l=(pb Q AQ B ◦ ̂QA µ QBp bQ coeq=(150)=def A µ QB( )= pb Q AQ B ◦m=A◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P u A QP Q B U)( )̂QA Uλ AA Q B ◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U)◦ (P u A QP Q B U)) )(pb Q AQ B ◦(µ A Q bQ B ◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P u A QP Q B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P m A Q B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P u A QP Q B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P m A Q B U) ◦ (P u A AQ B U) ◦ (P xQ B U))Amonad=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P xQ B U)Now we want to prove that(pb Q AQ B)◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ ( z l BU ))=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (z r BU) .We have(pb Q AQ B)◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ ( z l BU ))=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U) ◦ (δ D P Q B U))(178)=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (δ D P Q B U))x=(pb Q AQ B ◦ (lQ B ) ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (δ D P Q B U))δ=D(pb Q AQ B ◦ (lQ B ) ◦ (P xQ B ) ◦ (δ D P Q B ) ◦ (DP p Q ))(149)=(pb Q AQ B ◦ (ν 0Q ′ B ) ◦ (yP Q B ) ◦ (δ D P Q B ) ◦ (DP p Q ))(110)=(pb Q AQ B ◦ (ν 0Q ′ B ) ◦ (u B P Q B ) ◦ ( )ε D P Q B ◦ (DP pQ ))(176)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ ( )ε D P Q B ◦ (DP pQ )) (pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ ( ε D P Q B U )ε D =u A=Since, by Lemma <strong>2.</strong>10)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (z r BU)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (z r BU)(B B U, y B U) = Coequ Fun(zlB U, z r BU ) ,there exists a functorial morphism α : B B U → ̂Q AA Q B such that(179) α ◦ (y B U) = (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) .

157Now we want to prove that(̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ).Let us show the fork property for α, that is(180) α ◦ (m BB U) = α ◦ (B B Uλ B ) .We have(179)=α ◦ (B B Uλ B ) ◦ (yy B U) = α ◦ (B B Uλ B ) ◦ (yB B U) ◦ (P Qy B U)y= α ◦ (y B U) ◦ (P Q B Uλ B ) ◦ (P Qy B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P Q B Uλ B ) ◦ (P Qy B U))=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (P Q B Uλ B ) ◦ (P Qy B U)( )defp Q= pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ ( P µ B QBU ) ◦ (P Qy B U))(107)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (P χ B U))u=A(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U)(179)= α ◦ (y B U) ◦ (P χ B U) (109)= α ◦ (m BB U) ◦ (yy B U)and, since yy B U is an epimorphism, we conclude. Now, let us consider a functorialmorphism h : B B U → X such that h ◦ (m BB U) = h ◦ (B B Uλ B ) . We have to showthat there exists a unique functorial morphism ĥ : ̂Q AA Q B → X such thatĥ ◦ α = h.First we will show that there exists a functorial morphism ĥ such that ĥ and h fulfill(177) i.e.)ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h ◦ (y B U) ◦ ( P A µ QB U ) .To do this, we need a series of equalities. First of all, let us show that(181) y ◦ ( P A µ Q)◦(P AµBQ)◦ (P AQy) ◦ (P xQP Q) = mB ◦ (yy) ◦ (P χP Q) .In fact, we haveNow let us prove thaty ◦ ( P A µ Q)◦(P AµBQ)◦ (P AQy) ◦ (P xQP Q)(107)= y ◦ ( P A µ Q)◦ (P Aχ) ◦ (P xQP Q)x= y ◦ ( )P A (101)µ Q ◦ (P xQ) ◦ (P QP χ) = y ◦ (P χ) ◦ (P QP χ)(98)= y ◦ (P χ) ◦ (P χP Q) (109)= m B ◦ (yy) ◦ (P χP Q) .(182) (yy) ◦ (P χP Q) = (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .

158In fact we haveTherefore we deduce that(yy) ◦ (P χP Q) = (yB) ◦ (P Qy) ◦ (P χP Q)χ= (yB) ◦ (P χB) ◦ (P QP Qy)(101)= (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .y ◦ ( ) ( )(183) P A µ Q ◦ P AµBQ ◦ (P AQy) ◦ (P xQP Q)= m B ◦ (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .Now we computeh ◦ (y B U) ◦ ( P A µ QB U ) ◦ ( P Aµ B QBU ) ◦ (P AQy B U) ◦ (P xQP Q B U)(183)= h ◦ (m BB U) ◦ (yB B U) ◦ ( P A µ Q B B U ) ◦ (P xQB B U) ◦ (P QP Qy B U)assumpth= h ◦ (B B Uλ B ) ◦ (yB B U) ◦ ( P A µ Q B B U ) ◦ (P xQB B U) ◦ (P QP Qy B U)y= h ◦ (y B U) ◦ (P Q B Uλ B ) ◦ ( P A µ Q B B U ) ◦ (P xQB B U) ◦ (P QP Qy B U)A µ Q= h ◦ (yB U) ◦ ( P A µ QB U ) ◦ (P AQ B Uλ B ) ◦ (P xQB B U) ◦ (P QP Qy B U)x= h ◦ (y B U) ◦ ( P A µ QB U ) ◦ (P AQ B Uλ B ) ◦ (P AQy B U) ◦ (P xQP Q B U)Since (P AQy B U) ◦ (P xQP Q B U) is an epimorphism, we obtainh ◦ (y B U) ◦ ( P A µ QB U ) ◦ ( P Aµ B QBU ) = h ◦ (y B U) ◦ ( P A µ QB U ) ◦ (P AQ B Uλ B )Since (P AQ B , P Ap Q ) = Coequ Fun(P AµBQ BU, P AQ B Uλ B), there exists a functorialmorphism h 1 : P AQ B → X such that(184) h 1 ◦ (P Ap Q ) = h ◦ (y B U) ◦ ( P A µ Q BU ) .Now we prove that(185) y ◦ ( P A µ Q)◦ (P xQ) ◦(z l P Q ) = y ◦ ( P A µ Q)◦ (P xQ) ◦ (z r P Q)In fact, we havey ◦ ( P A µ Q)◦ (P xQ) ◦(z l P Q ) (101)= y ◦ (P χ) ◦ ( z l P Q )(156)= y ◦ (P χ) ◦ (z r P Q) (101)= y ◦ ( P A µ Q)◦ (P xQ) ◦ (z r P Q) .Using the previous equalities, we obtainh 1 ◦ (P xQ B ) ◦ ( )z l P Q B ◦ (DP QP pQ ) = zlh 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ ( z l P Q B U )x= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ ( z l P Q B U )(184)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ ( z l P Q B U )(185)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ (z r P Q B U)(184)= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (z r P Q B U)x= h 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (z r P Q B U)

z r= h 1 ◦ (P xQ B ) ◦ (z r P Q B ) ◦ (DP QP p Q ) .Since DP QP p Q is an epimorphism, we deduce thath 1 ◦ (P xQ B ) ◦ ( )z l P Q B = h1 ◦ (P xQ B ) ◦ (z r P Q B ) .( )Since ̂QQB lQ B = Coequ Fun ((P xQ B )◦ ( )z l P Q B , (P xQB )◦(z r P Q B )), there existsa functorial morphism h 2 : ̂QQ B → X such that(186) h 2 ◦ (lQ B ) = h 1 .We computeso that we gety ◦ ( P A µ Q)◦ (P xQ) ◦ (P χP Q)(101)= y ◦ (P χ) ◦ (P χP Q)(98)= y ◦ (P χ) ◦ (P QP χ)(101)= y ◦ ( P A µ Q)◦ (P xQ) ◦ (P QP χ)x= y ◦ ( P A µ Q)◦ (P Aχ) ◦ (P xQP Q)(187) y ◦ ( P A µ Q)◦ (P xQ) ◦ (P χP Q) = y ◦(P A µ Q)◦ (P Aχ) ◦ (P xQP Q) .We also have(lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(101)= (lQ B ) ◦ (P Ap Q ) ◦ ( P A A µ Q BU ) ◦ (P AxQ B U) ◦ (P xQP Q B U)(174)= (lQ B ) ◦ ( )P A A µ QB ◦ (P AApQ ) ◦ (P AxQ B U) ◦ (P xQP Q B U)( )l= ̂QA µ QB ◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P xQP Q B U)( )x= ̂QA µ QB ◦ (lAQ B ) ◦ (P AxQ B ) ◦ (P AQP p Q ) ◦ (P xQP Q B U)( )̂QA µ QB ◦ (lAQ B ) ◦ (P AxQ B ) ◦ (P xQP QQ B ) ◦ (P QP QP p Q )( )= ̂QA µ QB ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )x=so that we get(188)=(lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)( )̂QA µ QB ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )and hence we obtain)h 2 ◦(µ A Q bQ B ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )(150)= h 2 ◦ (lQ B ) ◦ (P m A Q B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )(186)= h 1 ◦ (P m A Q B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )(102)= h 1 ◦ (P xQ B ) ◦ (P χP Q B ) ◦ (P QP QP p Q )159

160χ= h 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (P χP Q B U)x= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (P χP Q B U)(184)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ (P χP Q B U)(187)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(184)= h 1 ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(186)= h 2 ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(188)= h 2 ◦ ( ̂Q A µ QB ) ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q ) .Since (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q ) is an epimorphism, we deduce thath 2 ◦ ( ̂Q)A µ QB ) = h 2 ◦(µ A Q bQ B .By Proposition 8.3 we have(̂QAA Q B , pb Q AQ B)= Coequ Fun(µ A b QAU A Q B , ̂Q A Uλ AA Q B)Coequ Fun(µ A b QQ B , ̂Q A µ QB)and hence we infer that there exists a functorial morphismĥ : ̂Q AA Q B → X such that)ĥ ◦(pb Q AQ B = h 2 .Hence we get)ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h 2 ◦ (l A U A Q B ) ◦ (P Ap Q )= h 2 ◦ (lQ B ) ◦ (P Ap Q ) (186)= h 1 ◦ (P Ap Q ) (184)= h ◦ (y B U) ◦ ( P A µ Q BU )and hence equality (177) is proven. Now we have( )ĥ ◦ α ◦ (y B U) (179)= ĥ ◦ pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)(177)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P u A Q B U) A µ Q is unital= h ◦ (y B U)so we getĥ ◦ α = h.Let now ĥ′ : ̂QAA Q B → X be another functorial morphism such that ĥ′ ◦ α = h.Then we haveĥ ◦ (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)(179)= ĥ ◦ α ◦ (y BU) = h ◦ (y B U) = ĥ′ ◦ α ◦ (y B U)(179)= ĥ′ ◦ (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) .)Note that, by (178) , since the second term is an epimorphism,(pb Q AQ B ◦(l A U A Q B )◦(P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U) is epi and so (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)is also ( an epimorphism. Therefore we deduce that ĥ′ = ĥ. Hence we have provedthat ̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ). □=

Theorem 8.6. Within the assumptions and notations of Theorem 6.29, we have afunctorial isomorphism B ̂QAA Q B∼ = B B.Proof. In Proposition 8.5 we have proved the existence of a functorial morphism α :B B U → ̂Q(AA Q B such that ̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B. ) . By Proposition3.13 and Proposition 3.14 also ( B U, ( B Uλ B )) = Coequ Fun (m BB U, B B Uλ B. ). Inview of uniqueness of a coequalizer up to isomorphisms, there exists a functorialisomorphismNow sinceβ : ̂Q AA Q B = B U B ̂QAA Q B → B U such that β ◦ α = B Uλ B .( B Uλ B ) ◦ (m BB U) = ( B Uλ B ) ◦ (B B Uλ B )and since β ◦ α = B Uλ B , by applying (177) where ”ĥ” = β and ”h” = BUλ B , wededuce that)β ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = ( B Uλ B ) ◦ (y B U) ◦ ( P A µ Q BU )equivalentlyi.e.(189) β ◦)β ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U)= ( B Uλ B ) ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U)(101)= ( B Uλ B ) ◦ (y B U) ◦ (P χ B U)(pb Q AQ B)◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U) = ( B Uλ B ) ◦ (y B U) ◦ (P χ B U) .Recall ( that, in view of Proposition 3.13, ( B U, B Uλ B ) is an B-left module functor.Also ̂QAA Q B , B µb QA A B)Q is an B-left module functor (see proof of Proposition3.30 and Lemma 3.17) where B µb QA= B Uλ BB ̂QA : B ̂Q A → ̂Q A .Now we want toprove that β lifts to a functorial morphism B ̂QAA Q B → B B i.e. thatβ :(̂QAA Q B , B µb QA AQ B)→ ( B U, B Uλ B )is a morphism of B-left module functors. Thus we have to proveWe calculate( B Uλ B ) ◦ (Bβ) = β ◦ ( B µb QA AQ B )(y ̂Q)◦ (P Ql) ◦ (P QP x)B µb Q◦ (Bl) ◦ (yP A) ◦ (P QP x) = B µb Q◦((149)= B µb Q◦ y ̂Q)◦ (P Qν 0) ′ ◦ (P QyP ) = y B µb Q◦ (Bν 0) ′ ◦ (yBP ) ◦ (P QyP )(151)= ν ′ 0 ◦ (m B P ) ◦ (yBP ) ◦ (P QyP ) = ν ′ 0 ◦ (m B P ) ◦ (yyP )(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) (149)= l ◦ (P x) ◦ (P χP )161

162so that we obtain:(190)We computeB µb Q◦ (Bl) ◦ (yP A) ◦ (P QP x) = l ◦ (P x) ◦ (P χP ) .β ◦ ( B µb QA AQ B ) ◦(175)= β ◦ (pb Q AQ B ) ◦= β ◦ (pb Q AQ B ) ◦(B µb Q AU A Q B))(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (yP xp Q ) =(B µb Q AU A Q B)◦ (Bl A U A Q B ) ◦ (yP xp Q )◦ (Bl A U A Q B ) ◦ (yP AQ B ) ◦ (P QP xQ B )◦ (P QP QP p Q )(190)= β ◦ (pb Q AQ B ) ◦ (l A U A Q B ) ◦ (P x A U A Q B ) ◦ (P χP A U A Q B )◦ (P QP QP p Q )χ= β ◦ (pb Q AQ B ) ◦ (l A U A Q B ) ◦ (P x A U A Q B ) ◦ (P QP p Q ) ◦ (P χP Q B U)x= β ◦ (pb Q AQ B ) ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (P χP Q B U)(189)= ( B Uλ B ) ◦ (y B U) ◦ (P χ B U) ◦ (P χP Q B U)(98)= ( B Uλ B ) ◦ (y B U) ◦ (P χ B U) ◦ (P QP χ B U)(109)= ( B Uλ B ) ◦ (m BB U) ◦ (yy B U) ◦ (P QP χ B U)BUλ B coequ= ( B Uλ B ) ◦ (B B Uλ B ) ◦ (By B U) ◦ (yP Q B U) ◦ (P QP χ B U)y= ( B Uλ B ) ◦ (B B Uλ B ) ◦ (By B U) ◦ (BP χ B U) ◦ (yP QP Q B U))(189)= ( B Uλ B ) ◦ (Bβ) ◦(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (BP Ap Q ) ◦ (BP xQ B U)◦ (yP QP Q B U))= ( B Uλ B ) ◦ (Bβ) ◦(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (yP xp Q ))Since(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (yP xp Q ) is an epimorphism, we get that ( B Uλ B ) ◦(Bβ) = β ◦ ( B µb QA A Q B). Therefore β is a morphism of B-left module functorsand ̂Qhence, in view of Proposition 3.25, it gives rise to a functorial isomorphismB AA Q B∼ = B B.□Now, we prove the second isomorphism.Within the assumptions and notations of Theorem 6.29, we will construct a functorialisomorphism A Q BB ̂QA ∼ = A A.Lemma 8.7. Within the assumptions and notations of Theorem 6.29, there exists afunctorial morphism Ξ : A A U → Q BB ̂QA uniquely determined by) )(191)(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) = Ξ ◦ (x A U)such that(192) Ξ ◦ (m AA U) = Ξ ◦ (A A U A λ) .

163( )Proof. Since (Q B , p Q ) = Coequ Fun µBQ BU, Q B Uλ B we have that))(p QB ̂QA ) ◦(µ B QBU B ̂QA = (p QB ̂QA ) ◦(Q B Uλ BB ̂QAso that we obtain) )(p QB ̂QA ◦(Qpb Q(107)== (p QB ̂QA ) ◦ (Q B µb QA)(◦ χ ̂Q) )χ=A U(p QB ̂QA ◦(χ ̂Q) )A ◦(QP Qpb Q) (p QB ̂QA ◦(µ B ̂Q)Q A ◦(Qy ̂Q) )A ◦(QP Qpb Q(Qy ̂Q) )A ◦(QP Qpb Q) ((QBpb Q◦ Qy ̂Q)A U) ) ( ) ((p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U= (p QB ̂QA ) ◦ (Q B µb QA) ◦y(175)== (p QB ̂QA ) ◦ (Q B µb QA) ◦and hence we get) ) ((193)(p QB ̂QA ◦(Qpb Q◦ χ ̂Q)A U =) ) ( ) ((p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A Uso that) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U)) ) (χ=(p QB ̂QA ◦(Qpb Q◦ χ ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)) ) ( ) ((193)=(p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)) ) ( )y=(p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ (QBl A U) ◦ (QyP A A U) ◦ (QP QP x A U)) )(190)=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U) .Therefore we deduce that) )(194)(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U)) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U)By using this equality we compute) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ ( w l AU ) ◦ ( QP Cε C AU )) )w=(p lQB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ ( QP ε C AU ) ◦ ( w l C A U )) )(103)=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP δ C A U) ◦ ( w l C A U )) )w=(p lQB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ ( w l QP A U ) ◦ (QP Cδ C A U)) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U) ◦ (QP δ C QP A U)◦ (QP Cδ C A U)

164(194)=) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U) ◦ (QP δ C QP A U)(99)=◦ (QP Cδ C A U)) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ ( QP ε C QP A U )◦ (QP Cδ C A U)) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP δ C A U) ◦ ( QP Cε C AU )) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ ( QP ε C AU ) ◦ ( QP Cε C AU )) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (w r AU) ◦ ( QP Cε C AU )ε C =(103)=since QP Cε C AU is epi we deduce that) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ ( w l AU )) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (w r AU) .(Since (A A U, x A U) = Coequ Fun wl A U, w r AU ) , there exists a functorial morphismΞ : A A U → Q BB ̂QA such that) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) = Ξ ◦ (x A U) .Let us prove thatBy the definition of pb Qwe have that( )µ A Q bAUso thatand hencepb Q◦Ξ ◦ (m AA U) = Ξ ◦ (A A Uλ A ) .◦ (lA A U) = pb Q◦(̂QA U A λpb Q◦ (l A U) ◦ (P m AA U) (150)= pb Q◦p bQ coequ= pb Q◦(̂QA U A λ)◦ (lA A U)( )µ A Q bAU◦ (lA A U))◦ (lA A U)(195) pb Q◦ (l A U) ◦ (P m AA U) = pb Q◦(̂QA U A λWe calculate(191)=Ξ ◦ (m AA U) ◦ (xx A U) ◦ ( QP QP ε C AU ))◦ (lA A U) .(102)= Ξ ◦ (x A U) ◦ (χP A U) ◦ ( QP QP ε C AU )) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (χP A U)◦ ( QP QP ε C AU )(χ ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP u AA U)) )χ=(p QB ̂QA ◦(Qpb Q◦

◦ ( QP QP ε C AU )) ) ( ) ((p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP u AA U)◦ ( QP QP ε C AU )) ) ( )y=(p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ (QBl A U) ◦ (QyP A A U) ◦ (QP QP u AA U)◦ ( QP QP ε C AU )) ) ( )(103)=(p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ (QBl A U) ◦ (QyP A A U) ◦ (QP QP x A U)(193)=(190)=(102)=(195)=( )l,x= p QB ̂QA ◦Amonad=◦ (QP QP δ C A U)) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U)◦ (QP QP δ C A U)) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xx A U)◦ (QP QP δ C A U)) ) ((p QB ̂QA ◦(Qpb Q◦ Q ̂Q)A U A λ ◦ (QlA A U) ◦ (QP xA A U)◦ (QP QP x A U) ◦ (QP QP δ C A U)) ) ((103)=(p QB ̂QA ◦(Qpb Q◦ Q ̂Q)A U A λ ◦ (QlA A U) ◦ (QP xA A U)◦ (QP QP u AA U) ◦ ( QP QP ε C AU )) (Qpb Q◦ (Ql A U) ◦ (QP A A U A λ) ◦ (QP Au AA U) ◦ (QP x A U)◦ ( QP QP ε C AU )) )Amonad=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ ( QP QP ε C AU )) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP u A A A U) ◦ (QP x A U)◦ ( QP QP ε C AU )) ) ((195)=(p QB ̂QA ◦(Qpb Q◦ Q ̂Q)A U A λ ◦ (QlA A U) ◦ (QP u A A A U)◦ (QP x A U) ◦ ( QP QP ε C AU )) )l=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP A A U A λ) ◦ (QP u A A A U) ◦ (QP x A U)◦ ( QP QP ε C AU )u A=) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (QP A U A λ) ◦ (QP x A U)◦ ( QP QP ε C AU )(191)= Ξ ◦ (x A U) ◦ (QP A U A λ) ◦ (QP x A U) ◦ ( QP QP ε C AU )x= Ξ ◦ (A A U A λ) ◦ (xA A U) ◦ (QP x A U) ◦ ( QP QP ε C AU )165

166= Ξ ◦ (A A U A λ) ◦ (xx A U) ◦ ( QP QP ε C AU )Since (xx A U) ◦ ( QP QP ε C AU ) is epi, we deduce (192) . Note that, in particular, wehave) )(196)(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (χP A U) ◦ ( QP QP ε C AU )) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ ( QP QP ε C AU )) )and since the second term is epi, also the first is epi, and hence(p QB ̂QA ◦(Qpb Q◦(Ql A U) ◦ (QP u AA U) is an epimorphism.Proposition 8.8. Within the assumptions and notations of Theorem 6.29, thereexists ( a functorial ) morphism Ξ : A A U → Q BB ̂QA such thatQ BB ̂QA , Ξ = Coequ Fun (m AA U, A A U A λ) . Moreover for every morphism k suchthatk ◦ (m AA U) = k ◦ (A A U A λ)if ̂k : Q BB ̂QA → Y is the unique morphism such that ̂k ◦ Ξ = k, we have that( ) )(197) ̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .Proof. By Lemma 8.7 we already know thatΞ ◦ (m AA U) = Ξ ◦ (A A U A λ) .Now we want to prove that( )Q BB ̂QA , Ξ = Coequ Fun (m AA U, A A U A λ).Let k : A A U → Y be a functorial morphism such that(198) k ◦ (m AA U) = k ◦ (A A U A λ)We have to show that there exists a functorial morphism ̂k : Q BB ̂QA → Y such that̂k ◦ Ξ = k.First we will show that there exists a functorial morphism ̂k such that ̂k and k fulfil(197) i.e.k ◦ (m AA U) ◦ (xA A U) = ̂k) )◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) .We proceed in several steps. First of all we computek ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ ( Qz l P A U )(102)= k ◦ (x A U) ◦ (χP A U) ◦ ( Qz l P A U )= k ◦ (x A U) ◦ (χP A U) ◦ (QP χP A U) ◦ (Qδ D P QP A U)(98)= k ◦ (x A U) ◦ (χP A U) ◦ (χP QP A U) ◦ (Qδ D P QP A U)(105)= k ◦ (x A U) ◦ (χP A U) ◦ ( Qε D P QP A U )(102)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ ( Qε D P QP A U )□

= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ (Qz r P A U) .Since ( Q preserves coequalizers by assumption, by Lemma<strong>2.</strong>9 we haveQ ̂Q)(A U, Ql A U = Coequ Fun (QP xA U ◦ Qz l P A U), (QP x A U ◦ Qz r P A U) ) , so we deducethat there exists a unique functorial morphism k 1 : Q ̂Q A U → Y such that(199) k 1 ◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .Then we havek 1 ◦( )Qµ A Q bAU◦ (QlA A U) (150)= k 1 ◦ (Ql A U) ◦ (QP m AA U)(199)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP m AA U)x= k ◦ (m AA U) ◦ (Am AA U) ◦ (xAA A U) m Aass= k ◦ (m AA U) ◦ (m A A A U) ◦ (xAA A U)m A(198)= k ◦ (A A Uλ A ) ◦ (m A A A U) ◦ (xAA A U)= k ◦ (m AA U) ◦ (AA A U A λ) ◦ (xAA A U) = x k ◦ (m AA U) ◦ (xA A U) ◦ (QP A A U A λ)((199)= k 1 ◦ (Ql A U) ◦ (QP A A U A λ) = l k 1 ◦ Q ̂Q)A U A λ ◦ (QlA A U)( )Since QlA A U is epi, we get that k 1 ◦ Qµ A Q bAU= k 1 ◦(Q ̂Q)A Uλ A . Since Q preservescoequalizers,(Q ̂Q)(A , Qpb Q= Coequ Fun Qµ A Q bAU, Q ̂Q)A Uλ A , then there exists aunique functorial morphism k 2 : Q ̂Q A → Y such that)(200) k 1 = k 2 ◦(Qpb Q.We have) ) (k 2 ◦(µ B QBU B ̂QA ◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U))) )y= k 2 ◦(µ B QBU B ̂QA ◦(Qy B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U)) )(107)= k 2 ◦(χ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U))χ= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U)(200)= k 1 ◦ ◦ (Ql A U) ◦ (QP x A U) ◦ (χP QP A U)(199)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ (χP QP A U)(102)= k ◦ (x A U) ◦ (χP A U) ◦ (χP QP A U)(98)= k ◦ (x A U) ◦ (χP A U) ◦ (QP χP A U)(102)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ (QP χP A U)(199)= k 1 ◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U))(200)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U)167

168)(155)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP b A U) ◦ (QP QP x A U))(154)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xAU A ) ◦ (QP QP xU A ))= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xx A U))(102)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U))(149)= k 2 ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (QyP A U) ◦ (QP χP A U))(109)= k 2 ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (Qm B P A U) ◦ (QyyP A U)) ( )(151)= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0P ′ A U) ◦ (QyyP A U)) ( )= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0P ′ A U) ◦ (QByP A U) ◦ (QyP QP A U)) ( )(149)= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBlP A U) ◦ (QBP x A U) ◦ (QyP QP A U)) ( ) (y= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U ◦ (QP QlP A U) ◦ (QP QP x A U)) ) ((175)= k 2 ◦(Q B µb QA◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)def A µ QB) ) (= k2 ◦(Q B Uλ BB ̂QA ◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)) (Since(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U) is epi, we get))k 2 ◦(µ B QBU B ̂QA = k 2 ◦(Q B Uλ BB ̂QA .)()From(Q BB ̂QA , p QB ̂QA = Coequ Fun µ B QBU B ̂QA , Q B Uλ BB ̂QA we deduce thatthere exists a unique functorial morphism ̂k : Q BB ̂QA → Y such that(201) ̂k ◦(p QB ̂QA)= k 2 .Moreover we have( ) ))̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) = k 2 ◦(Qpb Q◦ (Ql A U)i.e.(202) ̂k ◦(p QB ̂QA)◦Now we compute(200)= k 1 ◦ (Ql A U) (199)= k ◦ (m AA U) ◦ (xA A U)̂k ◦ Ξ ◦ (xA U) (191)= ̂k ◦) (Qpb Q◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)(202)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP u AA U) x = k ◦ (m AA U) ◦ (Au AA U) ◦ (x A U)= k ◦ (x A U) .

Since x A U is epi we get that̂k ◦ Ξ = k.Let now ̂k ′ : Q BB ̂QA → Y be another functorial morphism such that ̂k ′ ◦ Ξ = k.Then we have( ) )̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)(191)= ̂k ◦ Ξ ◦ (x A U) = k ◦ (x A U) = ̂k ′ ◦ Ξ ◦ (x A U)(191)= ̂k) )′ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)) )Since we already observed from (196) that(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)is an epimorphism, we have ̂k ′ = ̂k.( )Hence we have proved that Q BB ̂QA , Ξ =Coequ Fun (m AA U, A A U A λ) and, in view of (202) , we get that (197)holds.Theorem 8.9. Within the assumptions and notations of Theorem 6.29, we have afunctorial isomorphism A A ∼ = A Q BB ̂QA .Proof. In Proposition 8.8 we ( have proved)the existence of a functorial morphismΞ : AU A → Q BB ̂QA such that Q BB ̂QA , Ξ = Coequ Fun (m AA U, A A Uλ A ). By Proposition3.13 and Proposition 3.14 also ( A U, A Uλ A ) = Coequ Fun ((m AA U, A A Uλ A )) , inview of uniqueness of a coequalizer up to isomorphisms, there exists a functorialisomorphismρ : A U A Q BB ̂QA = Q BB ̂QA → A U such that ρ ◦ Ξ = A Uλ A .Now since( A Uλ A ) ◦ (m AA U) = ( A Uλ A ) ◦ (A A Uλ A )and since ρ ◦ Ξ = A Uλ A , by Proposition 8.8, we deduce that) )(203) ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) = ( A Uλ A ) ◦ (m AA U) ◦ (xA A U)Now we want to prove that ρ lifts to a functorial morphism A Q BB ̂QA → A A of A-leftmodule functors. First we observe that ( A U, A Uλ A ) is an A-left module functor inview of Proposition 3.13. Also(Q BB ̂QA , A µ ̂Q)QB B A is an A-left module functor(see proof of Proposition 3.30) where A µ QB = A Uλ AA Q B : AQ B → Q B . To showthat ρ is morphism of A-left module functors we have to prove( A Uλ A ) ◦ (Aρ) = ρ ◦ ( A µ QB B ̂Q A ).We have(A ρ ◦ µ ̂Q))) )QB B A ◦(Ap QB ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U))(174)= ρ ◦(p (A )) )QB ̂QA ◦ µ Q BU B ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U)◦ (QP QP x A U))) )(101)= ρ ◦(p QB ̂QA ◦(χ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U)169□

170) ) (χ= ρ ◦(p QB ̂QA ◦(Qpb Q◦ χ ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)) ) ( ) ((193)= ρ ◦(p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U ◦ (QP Ql A U))y= ρ ◦(p QB ̂QA(149)= ρ ◦◦ (QP QP x A U)) ( )◦(Qpb Q◦ Q B µb Q AU ◦ (QBl A U) ◦ (QBP x A U) ◦ (QyP QP A U)) ) ( )(p QB ̂QA ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0AU) ′ ◦ (QByP A U)◦ (QyP QP A U)) )(151)= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (Qm B P A U) ◦ (QByP A U)◦ (QyP QP A U)) )= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (Qm B P A U) ◦ (QyyP A U)) )(109)= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (QyP A U) ◦ (QP χP A U)) )(149)= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U)) )(102)= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xx A U)) )= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xA A U) ◦ (QP QP x A U)) ) ( )(150)= ρ ◦(p QB ̂QA ◦(Qpb Q◦ Qµ A Q bAU◦ (QlA A U) ◦ (QP xA A U) ◦ (QP QP x A U)defp bQ( ) )= ρ ◦ p QB ̂QA ◦(Qpb Q◦(Q ̂Q)A Uλ A ◦ (QlA A U) ◦ (QP xA A U) ◦ (QP QP x A U)) )l= ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP A A Uλ A ) ◦ (QP xA A U) ◦ (QP QP x A U)(203)= ( A Uλ A ) ◦ (m AA U) ◦ (xA A U) ◦ (QP A A Uλ A ) ◦ (QP xA A U) ◦ (QP QP x A U)x= ( A Uλ A ) ◦ (m AA U) ◦ (AA A Uλ A ) ◦ (xAA A U) ◦ (QP xA A U) ◦ (QP QP x A U)m A= ( A Uλ A ) ◦ (A A Uλ A ) ◦ (m A A A U) ◦ (xAA A U) ◦ (QP xA A U) ◦ (QP QP x A U)AUλ A ass= ( A Uλ A ) ◦ (m AA U) ◦ (m A A A U) ◦ (xAA A U) ◦ (QP xA A U)m A ass= ( A Uλ A ) ◦ (m AA U) ◦ (Am AA U) ◦ (xAA A U) ◦ (QP xA A U)AUλ A ass= ( A Uλ A ) ◦ (A A Uλ A ) ◦ (Am AA U) ◦ (xAA A U) ◦ (QP xA A U)= ( A Uλ A ) ◦ (A A Uλ A ) ◦ (Am AA U) ◦ (AxA A U) ◦ (xQP A A U)) )(203)= ( A Uλ A ) ◦ (Aρ) ◦(Ap QB ̂QA ◦(AQpb Q◦ (AQl A U) ◦ (xQP A A U))x= ( A Uλ A ) ◦ (Aρ) ◦(Ap QB ̂QA ◦(xQ ̂Q) )A ◦(QP Qpb Q◦ (QP Ql A U)

)) )Since(Ap QB ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) is an epimorphism, weget that ρ ◦ ( A µ ̂Q QB B A ) = ( A Uλ A ) ◦ (Aρ). Therefore ρ is a morphism of A-leftmodule functors and hence, in view of Proposition 3.25, it gives rise to a functorialisomorphism A Q BB ̂QA ∼ = A A.□8.<strong>2.</strong> Equivalence for comodule categories coming from pretorsor. The <strong>results</strong>obtained in this subsection can be found in [BM].Given categories A and B, under the assumptions of Theorem 6.5, one can provethat there exist a comonad C on A and a comonad D on B such that their categoriesof comodules are equivalent. We outline that the assumptions quoted above aresatisfied in the particular case of a regular herd.Using the functors Q and Q, we construct the functors C Q D : D B → C A andD Q C : C A → D B which will be used to set the equivalence between these comodulecategories.Proposition 8.10. In the setting of Theorem 6.5 there exists a functor C ( Q D) :D B → C A such that C U C ( Q D) = Q D where ( Q D , ι Q) = Equ Fun(ρDQ D U, Q D Uγ D) .Moreover we have(204)( Cρ Q D U ) ◦ ι Q = ( Cι Q) ◦ C ρ Q Dwhere C ρ Q D = C Uγ C C ( Q D) : Q D → CQ D .Proof. In view of Theorem 6.5, we can apply Proposition 4.29.8.1<strong>1.</strong> In light of Proposition 8.10, a functor Q : B → A introduced in Theorem 6.5induces a functor C ( Q D) : D B → C A for the comonads C and D. Our next task is toprove that the D-C-bicomodule functor Q, constructed in Proposition 7.1, inducesa functor D (Q C) : C A → D B which yields the inverse of C ( Q D) .Proposition 8.1<strong>2.</strong> Within the assumptions and notations of Theorem 6.5, there )exists a functor D Q C : C A → D B such that D U D Q C = Q C where(Q C , ι Q =(Equ Fun ρ C QC U, Q C Uγ C) . Moreover we have)(205)(Dι Q ◦ D ρ C Q= (D ρ C Q U ) ◦ ι Q( )where D ρ C Q= D Uγ DD Q C : Q C → DQ C so that Q C , D ρ C Qis a left D-comodulefunctor.Proof. In view of Proposition 7.1, we can apply Proposition 4.29 where Q is Q andwe exchange the role of A and B, C and D.□Within the assumptions and notations of Theorem 6.5, one can construct functorialisomorphism D Q C C Q D ∼ = D B and C Q DD Q C ∼ = C A. Such a result can be obtainedby dualizing all the ingredients proved in details for the equivalence between modulecategories.171□

172Theorem 8.13. Let A and B be categories with equalizers and let τ : Q → QP Q bea regular pretorsor for Ξ = ( A, B, P, Q, σ A , σ B , u A , u B). Assume that the underlyingfunctors P, Q, A and B preserve equalizers. Then we have functorial isomorphismsD B ∼ = D Q CC Q D and C A ∼ = C Q DD Q C .9. EXAMPLESLemma 9.<strong>1.</strong> Let T Σ R be a bimodule. Let L = − ⊗ T Σ : Mod-T → Mod-R andlet H = Hom R (Σ, −) . Assume that T Σ is faithfully flat. Then the unit η of theadjunction (L, H) is a regular mono.Proof. It is well known (see e.g. [BM, Lemma <strong>2.</strong>3]) that the diagramL −→ LηLHL LηHL⇒ LHLHLLHLηis a contractible (split) equalizer with respect to the functorial morphisms (ɛL, ɛLHL).Since T Σ is faithfully flat we get that the diagramis exact.Id Mod-Rη−→ HL ηHL⇒ HLHLHLηCorollary 9.<strong>2.</strong> Let α : T → A be a ring homomorphism and assume that T A isfaithfully flat. Let γ : A → A ⊗ T A be the map defined by settingThen (T, α) = Ker (γ) .γ (a) = 1 A ⊗ T a − a ⊗ T 1 A .Corollary 9.3. Let α : A → T be a ring homomorphism such that T A is faithfullyflat. Then the unit of the adjunction (− ⊗ A T, Hom T (T, −)) is a regular monomorphism.Proof. Let η : Mod-T → Hom T (T, −) ⊗ A T be the unit of the adjunction. Then,for every M ∈ Mod-T we haveWe haveηM : M → M ⊗ A Tx ↦→ x ⊗ A 1 T .ηM (x) = x ⊗ A 1 T = (M ⊗ A α) (x ⊗ A 1 A ) = (M ⊗ A α) ◦ ( r A M) −1(x) .Then, for every M ∈ Mod-T, we haveKer (M ⊗ A γ) = (M ⊗ A A, M ⊗ A α) ∼ =(M, (M ⊗ A α) ◦ ( ) )rMA −1□= (M, ηM) .Hence, (M, ηM) = Ker (M ⊗ A γ) , i.e. η is a regular monomorphism.□We try here to apply the previous theory to a particular setting in which, first ofall, we compute all the ingredients we need to understand the form of the herd.9.4. Let us consider

• R = associative unital algebra• A = R-ring• C = R-coring• ψ : C ⊗ R A → A ⊗ R C a right entwining• ˜C = A ⊗ R C the induced A-coring• (Σ A , ρ e CΣ ) = right ˜C-comodule = right entwined module.• T = End eC (Σ) .Note that if T ⊆ A is a right C-Galois extension i.e.comodule and the canonical Galois mapcan C : A ⊗ T A → A ⊗ R Ct ⊗ T t ′ ↦→ tρ C A (t ′ ) = tt ′ 0 ⊗ R t ′ 1is an isomorphism, then we can consider the right entwiningψ : C ⊗ R A → A ⊗ R C(c ⊗ R t ↦→ can C can−1C (1 A ⊗ R c) t )173( )AR , ρ C A is a right C-and hence the A-coring A ⊗ R C, which turns out to be a right Galois coring i.e. Ais a right comodule over the A-coring A ⊗ R C via ρ e CA defined byA ∼ = A ⊗ A A A⊗ Aρ C A−→ A ⊗ A A ⊗ R C ∼ = A ⊗ R C,t ↦→ 1 A ⊗ A t 0 ⊗ R t 1 .The coinvariants of A with respect to this coaction is still T and the canonical Galoismap iscan A⊗R C : A ⊗ T A → A ⊗ A A ⊗ R Ct ⊗ T t ′ ↦→ tρ A⊗ RCA(t ′ ) = t ⊗ A t ′ 0 ⊗ R t ′ 1 = 1 ⊗ A tt ′ 0 ⊗ R t ′ 1 = 1 ⊗ A can C (t ⊗ T t ′ )so that can A⊗R C is still an isomorphism. Therefore we can consider this case as aparticular case of the previous one, where(Σ A , ρ e CΣ ) =(A A , ρ e )CA .Let A be a right Galois extension of B over the Hopf algebra H. In thiscase we haveA = Mod-R,B = Mod-B where B = A co(H)A = − ⊗ R A : A = Mod-R −→ A = Mod-RB = − ⊗ B A : B = Mod-B −→ B = Mod-BQ = − ⊗ B A : B = Mod-B → A = Mod-RP = − ⊗ R A B : A = Mod-R → B = Mod-BQP = − ⊗ R A ⊗ B A m A→ − ⊗ R AP Q = − ⊗ B A ⊗ R A m A→ − ⊗ B AC = − ⊗ R H : A = Mod-R −→ A = Mod-R

174l = eC ρ L : L = − ⊗ B A −→ ˜CL = − ⊗ B Σ ⊗ A A ⊗ R H ∼ = − ⊗ B A ⊗ R CA second particular case of this situation we have the one where R = kis a commutative ring, C = H is a Hopf algebra over k and A is a rightGalois extension of T = A co(H) .Now let us setA = Mod-R,B = Mod-T where T = End eC (Σ)A = − ⊗ R A : A = Mod-R −→ A = Mod-RC = − ⊗ R C : A = Mod-R −→ A = Mod-RB = − ⊗ T B : B = Mod-T −→ B = Mod-T where B = Hom A (Σ A , Σ A )Ψ = − ⊗ R ψ : AC = − ⊗ R C ⊗ R A −→ CA = − ⊗ R A ⊗ R CL = − ⊗ T Σ : Mod-T −→ Mod-A ∼ = A AW = Hom A (Σ, −) : Mod-A −→ Mod-T˜C = − ⊗ A A ⊗ R C : Mod-A −→ Mod-Al = eC ρ L : L = − ⊗ T Σ −→ ˜CL = − ⊗ T Σ ⊗ A A ⊗ R C ∼ = − ⊗ T Σ ⊗ R CWhen Σ A is f.g.p., we setA (Σ ∗ ) B= Hom A ( B Σ, A A)and we consider the following formal dual structure M = (A, B, P, Q, σ A , σ B ) on thecategories A and B .• A = (− ⊗ R A, − ⊗ R m A , − ⊗ R u A ) is a monad on A = Mod-R• B = (− ⊗ T B, − ⊗ T m B , − ⊗ T u B ) is a monad on B = Mod-T where T =End eC (Σ)• P = − ⊗ R Σ ∗ T : Mod-R → Mod-T• Q = A U ◦ L = − ⊗ T Σ R : Mod − T → Mod − R − ⊗ T Σ : Mod-T → Mod-R• σ A : QP → A is defined by• σ B : P Q → B is defined byσ A : QP = − ⊗ R Σ ∗ ⊗ T Σ → − ⊗ R A− ⊗ R f ⊗ T x ↦→ − ⊗ R f (x)σ B : P Q = − ⊗ T Σ R ⊗ R Σ ∗ T → B = − ⊗ T B ∼ = − ⊗ T Σ R ⊗ A Σ ∗ T− ⊗ T y ⊗ R γ ↦→ − ⊗ T y · γ () ≃ − ⊗ T yγ (x i ) ⊗ A x ∗ i= − ⊗ T y ⊗ A γ (x i ) x ∗ i = − ⊗ T y ⊗ A γ• ( P : A → B, B µ P : BP → P, µ A P : P A → P ) is a bimodule functor• ( Q : B → A, A µ Q : AQ → Q, µ B Q : QB → Q) is a bimodule functor• σ A : QP → A is A-bilinear• σ B : P Q → B is B-bilinearσ A ◦ (A µ Q P ) = m A ◦ ( Aσ A) and σ A ◦ ( )Qµ A P = mA ◦ ( σ A A )σ B ◦ (B µ P Q ) = m B ◦ ( Bσ B) and σ B ◦ ( )P µ B Q = mB ◦ ( σ B B )

175and the associative conditions holdA µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B) and B µ P ◦ ( σ B P ) = µ A P ◦ ( P σ A) .In fact, we compute[σ A ◦ (A µ Q P )] (− ⊗ R f ⊗ T x ⊗ R a) = σ A (− ⊗ R f ⊗ T xa)= − ⊗ R f (xa) = − ⊗ R f (x) aand[mA ◦ ( Aσ A)] (− ⊗ R f ⊗ T x ⊗ R a) = m A (− ⊗ R f (x) ⊗ R a) = − ⊗ R f (x) aso thatσ A ◦ (A µ Q P ) = m A ◦ ( Aσ A) .We compute[σ A ◦ ( )]Qµ A P (− ⊗R a ⊗ R f ⊗ T x) = σ A (− ⊗ R af ⊗ T x) = − ⊗ R af (x)and[mA ◦ ( σ A A )] (− ⊗ R a ⊗ R f ⊗ T x) = m A (− ⊗ R a ⊗ R f (x)) = − ⊗ R af (x)so that we getσ A ◦ ( Qµ A P)= mA ◦ ( σ A A ) .We compute[σ B ◦ (B µ P Q )] (− ⊗ T x ⊗ R f ⊗ T b) = σ B (− ⊗ T x ⊗ R fb)= σ B (− ⊗ T x ⊗ R f (b ())) = − ⊗ T x · f (b ())[mB ◦ ( Bσ B)] (− ⊗ T x ⊗ R f ⊗ T b) = m B (− ⊗ T x · f () ⊗ T b) = − ⊗ T [(x · f ()) ◦ b]Let us compute, for every y ∈ Σ we have[− ⊗ T x · f (b ())] (y) = − ⊗ T xf (b (y)) =− ⊗ T [(x · f ()) ◦ b] (y) = − ⊗ T (x · f ()) (b (y)) = − ⊗ T xf (b (y))so that we obtainσ B ◦ (B µ P Q ) = m B ◦ ( Bσ B) .Now we compute[σ B ◦ ( )]P µ B Q (− ⊗T b ⊗ T x ⊗ R f) = σ B (− ⊗ T b (x) ⊗ R f) = − ⊗ T b (x) · f ()and[mB ◦ ( σ B B )] (− ⊗ T b ⊗ T x ⊗ R f) = m B (− ⊗ T b ⊗ T x · f ()) = − ⊗ T [b ◦ (x · f ())]so that, for every y ∈ Σ we haveand[− ⊗ T b (x) · f ()] (y) = − ⊗ T b (x) f (y)(− ⊗ T [b ◦ (x · f ())]) (y) = − ⊗ T b (x · f () (y)) = − ⊗ T b (xf (y)) = − ⊗ T b (x) f (y)so that we getσ B ◦ ( P µ B Q)= mB ◦ ( σ B B ) .

and[µBQ ◦ ( Qσ B)] (− ⊗ T x ⊗ R f ⊗ T y) = µ B Q (− ⊗ T x · f () ⊗ T y)176Now we have[ Aµ Q ◦ ( σ A Q )] (− ⊗ T x ⊗ R f ⊗ T y) = A µ Q (− ⊗ T x ⊗ R f (y)) = − ⊗ T xf (y)so that= − ⊗ T x · f () (y) = − ⊗ T xf (y)A µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B) .Finally we compute[ Bµ P ◦ ( σ B P )] (− ⊗ R f ⊗ T x ⊗ R g) = B µ P (− ⊗ R f ⊗ T x · g ()) = − ⊗ R f (x · g ())and[µAP ◦ ( P σ A)] (− ⊗ R f ⊗ T x ⊗ R g) = µ A P (− ⊗ R f (x) ⊗ R g) = − ⊗ R f (x) g ()so that, for every y ∈ Σ we haveand[− ⊗ R f (x · g ())] (y) = − ⊗ R f (xg (y)) = − ⊗ R f (x) g (y)so that we get[− ⊗ R f (x) g ()] (y) = − ⊗ R f (x) g (y)B µ P ◦ ( σ B P ) = µ A P ◦ ( P σ A) .Note that, in the case R A is faithfully flat,by Corollary 9.2,(A, u A ) = (Mod-R, − ⊗ R u A ) = Ker (− ⊗ R γ)= Equ Fun (− ⊗ R u A ⊗ R A, − ⊗ R u A ⊗ R A) .Analogously if T B is faithfully flat we have(B, u B ) = (Mod-T, − ⊗ T u B ) = Equ Fun (− ⊗ T u B ⊗ T B, − ⊗ T u B ⊗ T B).Thus, in the following we will assume that both R A and T B are faithfullyflat so that we have a regular formal dual structure.The counit ɛ of the adjunction (L, W ) is given byɛ M : Hom A (Σ, M) ⊗ T Σ −→ Mf ⊗ T x ↦→ f (x)for each M ∈ Mod-A. Therefore we get that) ( ) eCcan =(˜Cɛ ◦ ρ L W : LW = Hom A (Σ, −) ⊗ T Σ −→ ˜C = − ⊗ A A ⊗ R C Ris defined bycan M : Hom A (Σ, M) ⊗ T Σ −→ M ⊗ A A ⊗ R Cγ ⊗ T x ↦→ γ (x 0 ) ⊗ R x 1)for each M ∈ Mod-A. Hence we deduce that(L, eC ρ L is a left ˜C-Galois functorif and only if (Σ A , ρ e CΣ ) is a right Galois comodule.

By Lemma 3.29, we have A Q = L = − ⊗ T Σ : Mod-T → Mod-A. We have thatP A is a right adjoint of A Q, so that, by the uniqueness of the adjoint, we haveP A = W = − ⊗ A Σ ∗ T : Mod-A → Mod-T.Note that, by Corollary 6.22,, A σ A A : AQP A → A A = Mod-A is the counit ɛ of theadjunction ( A Q,P A ) = (L, W ) , i.e.Now, we can considerAσ A AM = ɛM : A QP A M = M ⊗ A Σ ∗ ⊗ T Σ A → Mm ⊗ A f ⊗ T x ↦→ mf (x) .Acan A =(˜CA σ A A)( ) eC◦ ρ L W : LW → ˜C.For every M ∈ Mod-A, we have( ) ( )Acan A M = ˜CA σAA eC◦ ρ L W M : M ⊗ A Σ ∗ ⊗ T Σ A → M ⊗ R Cm ⊗ A f ⊗ T x ↦→ mf (x 0 ) ⊗ R x 1 .Therefore we deduce that A can A = can. Recall now that, by Lemma 6.17 A can A isan isomorphism if and only ifcan 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CAis an isomorphism. For every M ∈ Mod-R, we havecan 1 M : QP M = M ⊗ R Σ ∗ T ⊗ T Σ R −→ CAM = M ⊗ R A ⊗ R Cm ⊗ R f ⊗ T x ↦→ ( Cσ A M ) (m ⊗ R f ⊗ T x 0 ⊗ R x 1 ) = m ⊗ R f (x 0 ) ⊗ R x 1 .Assume now that (Σ A , ρ e CΣ ) is a right Galois comodule. Thus we deduce that(L, eC ρ L)is a left Galois functor and thus can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CAis an isomorphism. Therefore, we can consider the compositeτ := ( (can 1 ) −1 Q ) ◦ (Cu A Q) ◦ C ρ Q : Q → QP Qand we can apply Theorem 6.24, that implies that the functorial morphism τ is aregular herd. It is defined bywhereτ : M ⊗ T Σ R −→ M ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T Σ Rm ⊗ T x ↦→ m ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1(( ) ) ( )m ⊗ T x 0 ⊗ R x 1 1 x21 ⊗T0 x2 = (can 1 1 1QM) ( )m ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1= [ (can 1 QM) ◦ ( (can 1 ) −1 QM ) ◦ (Cu A QM) ◦ (C ρ Q M )] (m ⊗ T x)For every c ∈ C, we denote by= [ (Cu A QM) ◦ (C ρ Q M )] (m ⊗ T x)= m ⊗ T x 0 ⊗ R 1 A ⊗ R x 1 .− ⊗ R c 1 ⊗ T c 2 = (can 1 ) −1 (− ⊗ R 1 A ⊗ R c)177

178so that− ⊗ R 1 A ⊗ R c = ( can 1 ◦ (can 1 ) −1) (− ⊗ R 1 A ⊗ R c) = can 1(− ⊗R c 1 ⊗ T c 2)= − ⊗ R c (( 1 c 2) (0) ) ⊗R c21i.e.− ⊗ R c (( 1 c 2) (0) ) ⊗R c2 = − ⊗ 1 R 1 A ⊗ R c.Now, starting from a pretorsor, we want to compute the two comonads associated.First of all we compute the comonad E = ( E, ∆ E , ε E) corresponding to the comonadC defined in Proposition 6.<strong>1.</strong> We have((E, i) = Equ Fun ω l , ω r)where ω l = ( QP σ A) ◦ (τP ) and ω r = QP u A : QP → QP A, i.e.ω l : QP = − ⊗ R Σ ∗ T ⊗ T Σ R → QP A = − ⊗ R A ⊗ R Σ ∗ T ⊗ T Σ Rand− ⊗ R f ⊗ T x ↦→ − ⊗ R f ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ↦→ − ⊗ R f (x 0 ) ⊗ R x 1 1 ⊗ T x 2 1We computeω r : QP = − ⊗ R Σ ∗ T ⊗ T Σ R → QP A = − ⊗ R A ⊗ R Σ ∗ T ⊗ T Σ R− ⊗ R f ⊗ T x ↦→ − ⊗ R 1 A ⊗ R f ⊗ T x.(can 1 A) ◦ ω l = (can 1 A) ◦ ( QP σ A) ◦ (τP ) = ( CAσ A) ◦ (can 1 QP ) ◦ (τP )= ( CAσ A) ◦ (can 1 QP ) ◦ ( (can 1 ) −1 QP ) ◦ (Cu A QP ) ◦ (C ρ Q P )= ( CAσ A) ◦ (Cu A QP ) ◦ (C ρ Q P )= (Cu A A) ◦ ( Cσ A) ◦ (C ρ Q P )= (Cu A A) ◦ can 1so that we get(can 1 A) ◦ ω l = (Cu A A) ◦ can 1Moreover(can 1 A) ◦ ω r = (can 1 A) ◦ (QP u A ) = (CAu A ) ◦ can 1i.e.(can 1 A) ◦ ω r = (CAu A ) ◦ can 1 .Assume that the functor C : A = Mod-R −→ A = Mod-R preserves equalizers.Then we know that(C, (Cu A )) = Equ Fun ((Cu A A) , (CAu A ))and thus we have(C, can−11 ◦ (Cu A ) ) = Equ Fun ((Cu A A) ◦ can 1 , (CAu A ) ◦ can 1 )so that we get(C,can−11 ◦ (Cu A ) ) = Equ Fun((can1 A) ◦ ω l , (can 1 A) ◦ ω r)i.e.(C, can−11 ◦ (Cu A ) ) = Equ Fun(ω l , ω r) .

Note that, in view of our assumptions, (B, u B ) = Equ Fun (u B B, Bu B ) and hence wecan apply Proposition 6.<strong>2.</strong> Now, we compute the comonad D = ( D, ∆ D , ε D) definedin Proposition 6.<strong>2.</strong> We have(D, j) = Equ Fun(θ l , θ r)where θ l = ( σ B P Q ) ◦ (P τ) and θ r = u B P Q : P Q → BP Q = W LP Q, i.e.θ l : P Q = − ⊗ T Σ R ⊗ R Σ ∗ T → BP Q = − ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T B− ⊗ T x ⊗ R f ↦→ − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ⊗ R f ↦→ − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 · fθ l : P Q = − ⊗ T Σ R ⊗ R Σ ∗ T → W LP Q = − ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T Σ ⊗ A Σ ∗− ⊗ T x ⊗ R f ↦→ − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ⊗ R f ↦→ − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1f (x i ) ⊗ A x ∗ i= − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ⊗ A f (x i ) x ∗ i = − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ⊗ A fθ r : P Q = − ⊗ T Σ R ⊗ R Σ ∗ T → BP Q = − ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T B− ⊗ T x ⊗ R f ↦→ − ⊗ T x ⊗ R f ⊗ T 1 B .θ r : P Q = − ⊗ T Σ R ⊗ R Σ ∗ T → BP Q = W LP Q = − ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T Σ ⊗ A Σ ∗− ⊗ T x ⊗ R f ↦→ − ⊗ T x ⊗ R f ⊗ T 1 B (x i ) ⊗ A x ∗ i = − ⊗ T x ⊗ R f ⊗ T x i ⊗ A x ∗ i .Note that P A = W = −⊗ A Σ ∗ and by (15) we have P AA F = W A F = −⊗ R A⊗ A Σ ∗ =− ⊗ R Σ ∗ = P. Let us consider( ) ( )Acan AA F = ˜CA σAAFA eC◦ ρ L W A F( ) ( )= ˜CA σAAFA eC◦ ρ L P AA F( ) ( )(15)= ˜CA σAAF A eC◦ ρ L PwhereAcan AA F : LW A F = LP −→ ˜C A Fis thus defined by settingor simplyWe havei.e.Acan AA F : − ⊗ R A ⊗ A Σ ∗ ⊗ T Σ −→ − ⊗ R A ⊗ A A ⊗ R C ∼ = − ⊗ R A ⊗ R C− ⊗ R a ⊗ A f ⊗ T x ↦→ − ⊗ R a ⊗ A f (x 0 ) ⊗ R x 1 ≃ − ⊗ R af (x 0 ) ⊗ R x 1Acan AA F : − ⊗ R A ⊗ A Σ ∗ ⊗ T Σ −→ − ⊗ R A ⊗ A A ⊗ R C ∼ = − ⊗ R A ⊗ R C− ⊗ R 1 A ⊗ A f ⊗ T x ↦→ − ⊗ R f (x 0 ) ⊗ R x 1 .W A can AA F Q : W LW A F Q = W LP Q → W ˜C A F QW A can AA F Q : − ⊗ T Σ R ⊗ R A ⊗ A Σ ∗ T ⊗ T Σ ⊗ A Σ ∗ → − ⊗ T Σ R ⊗ R A ⊗ A A ⊗ R C ⊗ A Σ ∗− ⊗ T x ⊗ R a ⊗ A f ⊗ T y ⊗ A g ↦→ − ⊗ T x ⊗ R a ⊗ A f (y 0 ) ⊗ R y 1 ⊗ A gW A can AA F Q : − ⊗ T Σ R ⊗ R A ⊗ A Σ ∗ T ⊗ T Σ ⊗ A Σ ∗ → − ⊗ T Σ R ⊗ R A ⊗ A A ⊗ R C ⊗ A Σ ∗179

180− ⊗ T x ⊗ R 1 A ⊗ A f ⊗ T y ⊗ A g ↦→ − ⊗ T x ⊗ R 1 A ⊗ A f (y 0 ) ⊗ R y 1 ⊗ A gW A can AA F Q : − ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T Σ ⊗ A Σ ∗ → − ⊗ T Σ R ⊗ R A ⊗ R C ⊗ A Σ ∗− ⊗ T x ⊗ R f ⊗ T y ⊗ A g ↦→ − ⊗ T x ⊗ R f (y 0 ) ⊗ R y 1 ⊗ A g.If we apply W A can AA F Q both to θ l and θ r we get the following[(W A can AA F Q) ◦ θ l] (− ⊗ T x ⊗ R f)= (W A can AA F Q) ( − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ⊗ A f )(( ) ) ( )= − ⊗ T x 0 ⊗ R x 1 1 x21 ⊗R0 x2 ⊗ 1 1 A f= − ⊗ T x 0 ⊗ R 1 A ⊗ R x 1 ⊗ A fand[(W A can AA F Q) ◦ θ r ] (− ⊗ T x ⊗ R f)= (W A can AA F Q) (− ⊗ T x ⊗ R f ⊗ T x i ⊗ A x ∗ i )= − ⊗ T x ⊗ R f ((x i ) 0) ⊗ R (x i ) 1⊗ A x ∗ i .Since A can A is an isomorphism, we get that(D, j) = Equ Fun((W A can AA F Q) ◦ θ l , (W A can AA F Q) ◦ θ r)Hence D ⊆ P Q = − ⊗ T Σ ⊗ R Σ ∗ . At this point we stop because it is not so clearwhat is the comonad D.We try to compute the functor Q which does not require the comonad D, but wecannot do it as well. In fact, we have the following:We calculate the equalizerQ q →P C(θ l P)◦(P can −11 )◦(P Cu A )⇒(θ r P )◦(P can −11 )◦(P Cu A )(Q, q)= EquFun((θ l P ) ◦ (P i) , (θ r P ) ◦ (P i) )= Equ Fun((θ l P ) ◦ ( P can −11BP QP)◦ (P CuA ) , (θ r P ) ◦ ( )P can −11 ◦ (P CuA ) ) .We have(θ l P ) ◦ ( )P can −11 ◦ (P CuA ) : − ⊗ R C ⊗ R Σ ∗ T −→ − ⊗ R Σ ∗ T ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T B− ⊗ R c ⊗ R f ↦→ − ⊗ R 1 A ⊗ R c ⊗ R f ↦→ − ⊗ R c 1 ⊗ T c 2 ⊗ R f(↦→ − ⊗ R c 1 ⊗ ) T c2 ⊗ ( )0 R c2 1⊗ ( )1 T c2 2· f 1(= − ⊗ R c 1 ⊗ ) T c2 ⊗ 0 R β 1 ⊗ T β 2 · fwhereso thatMoreover we havec 1 ( c 2 0)⊗R c 2 1 = 1 A ⊗ R c1 A ε C (c) = c 1 ( c 2 0ε C ( c 2 1))= c1 ( c 2) .β 1 ( β 2 0)⊗R β 2 1 = 1 A ⊗ R c 2 <strong>1.</strong>

181On the other side,(θ r P ) ◦ ( )P can −11 ◦ (P CuA ) : − ⊗ R C ⊗ R Σ ∗ T −→ − ⊗ R Σ ∗ T ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T B− ⊗ R c ⊗ R f ↦→ − ⊗ R 1 A ⊗ R c ⊗ R f ↦→ − ⊗ R c 1 ⊗ T c 2 ⊗ R f↦→ − ⊗ R c 1 ⊗ T c 2 ⊗ R f ⊗ T 1 B .MaybewhereQ = − ⊗ R XX = Ker (ϕ)ϕ : C ⊗ R Σ ∗ T → Σ R ⊗ R Σ ∗ T ⊗ T Bc ⊗ R f ↦→ c 1 ⊗ T(c2 ) 0 ⊗ R β 1 ⊗ T β 2 · f = c 1 ⊗ T c 2 ⊗ R f ⊗ T 1 BIn case all the computations above make sense, we could write a coherd associatedto the pretorsor. By Theorem 7.5, the coherd is defined byχ : = µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ ( QP [ can −11 ◦ (Cu A ) ] Q ) ◦ (QqQ)i.e.= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ ( QP can −11 Q ) ◦ (QP Cu A Q) ◦ (QqQ)χ : QQQ ⊆ QP CQ = − ⊗ T Σ R ⊗ R C ⊗ R Σ ∗ T ⊗ T Σ R → Q = − ⊗ T Σ R− ⊗ T x ⊗ R c ⊗ R f ⊗ T y ↦→ − ⊗ T x ⊗ R 1 A ⊗ R c ⊗ R f ⊗ T y↦→ − ⊗ T x ⊗ R c 1 ⊗ T c 2 ⊗ R f ⊗ T y ↦→ − ⊗ T x ⊗ R c 1 ⊗ T c 2 ⊗ R f (y)↦→ − ⊗ T x · c 1 ⊗ T c 2 ⊗ R f (y) ↦→ − ⊗ T x · c 1 ⊗ T c 2 f (y)↦→ − ⊗ T(x · c1 ) ( c 2 f (y) ) = − ⊗ T xc 1 ( c 2 f (y) ) = − ⊗ T xc 1 ( c 2) f (y)where for every x ∈ Σ A and h ∈ Σ ∗In particular, we have= − ⊗ T x ( 1 A ε C (c) ) f (y)x · h ∈ B = Hom A (Σ A , Σ A ) is defined by setting(x · h) (t) = xh (t) for every t ∈ Σ.x · c 1 : Σ A → Σ At ↦→ xc 1 (t) .9.<strong>1.</strong> H-Galois extension. In order to understand better the situation of the previousexample, we decide to consider a very particular case, the Schauenburg setting.Let H be a Hopf algebra and let A/k be a right H-Galois extension. Let us recallsome useful equalities related to the translation mapγ := can −1 (1 A ⊗ −) : H → A ⊗ A : h → can −1 (1 A ⊗ h) =: h 1 ⊗ h 2 .For every h, l ∈ H, a ∈ A, we have(206)(207)h 1 ( h 2) 0 ⊗ ( h 2) 1= 1 A ⊗ hh 1 ⊗ ( h 2) 0 ⊗ ( h 2) 1= (h 1 ) 1 ⊗ (h 1 ) 2 ⊗ h 2

182(208)(209)(210)(211)(h1 ) 0 ⊗ h2 ⊗ ( h 1) 1= (h 2 ) 1 ⊗ (h 2 ) 2 ⊗ S (h 1 )h 1 h 2 = ε H (h) 1 A(hl) 1 ⊗ (hl) 2 = l 1 h 1 ⊗ h 2 l 2a 0 (a 1 ) 1 ⊗ (a 1 ) 2 = 1 A ⊗ a.9.5. The Schauenburg situation is the particular case when T = A co(H) = k. Hencewe haveA = Mod-k,B = Mod-k where T = End eC (Σ) = A co(H) = k.A = − ⊗ k A : A = Mod-k −→ A = Mod-kC = − ⊗ k H : A = Mod-k −→ A = Mod-kB = − ⊗ k A = A : B = Mod-k −→ B = Mod-k where A = Hom A (A A , A A )Ψ = − ⊗ k ψ : AC = − ⊗ k H ⊗ k A −→ CA = − ⊗ k A ⊗ k H(ψ : H ⊗ k A → A ⊗ k H, h ⊗ k a ↦→ can C can−1C (1 A ⊗ k h) a )L = − ⊗ k A : Mod-k −→ Mod-A ∼ = A AW = Hom A (A, −) : Mod-A −→ Mod-k˜C = − ⊗ A A ⊗ k H ∼ = ⊗ k H : Mod-A −→ Mod-Al = eC ρ L : L = − ⊗ k A −→ ˜CL = − ⊗ k A ⊗ A A ⊗ k H ∼ = − ⊗ k A ⊗ k HLet us assume thatThen,k = commutative ringH = Hopf algebraA/k = faithfully flat H-Galois extension with ρ H A : A → A ⊗ HA co(H) = k1 Acan : A ⊗ A → A ⊗ H defined by setting can (a ⊗ b) = ab 0 ⊗ b 1γ : H → A ⊗ A defined by setting γ (h) := can −1 (1 A ⊗ h) = h 1 ⊗ h 2τ : A → A ⊗ A ⊗ Aa ↦→ a 0 ⊗ γ (a 1 ) = a 0 ⊗ a 1 1 ⊗ a 2 1is a pretorsor. We want to construct the comonads C and D as in Theorem 6.5where A = B = Mod-k and P = Q = − ⊗ A. First, let us consider ω l = − ⊗ k ̂ωland ω r = − ⊗ k ̂ω r : − ⊗ A ⊗ A → − ⊗ A ⊗ A ⊗ A wherêω l = ( σ A ⊗ A ⊗ A ) ◦ (A ⊗ τ) : A ⊗ A → A ⊗ A ⊗ Âω l (a ⊗ b) = ab 0 ⊗ b 1 1 ⊗ b 2 1̂ω r = (u A ⊗ A ⊗ A) : A ⊗ A → A ⊗ A ⊗ Âω r (a ⊗ b) = 1 A ⊗ a ⊗ b.

Let ω = ω l −ω r and ̂ω = ̂ω l −̂ω r . First of all we want to prove that for any k-moduleX we haveKer (ωX) = Ker (X ⊗ ̂ω) = X ⊗ Ker (̂ω) .Since A is faithfully flat over k we equivalently prove thatNote thatA ⊗ Ker (̂ω)Ker (X ⊗ ̂ω ⊗ A) = X ⊗ Ker (̂ω) ⊗ A. A ⊗ A ⊗ AA⊗ b ω lA⊗c ω r A ⊗ A ⊗ A ⊗ Awith respect to the map m A ⊗ A ⊗ A is a contractible equalizer so that alsoKer (̂ω) A ⊗ Abω lcω r A ⊗ A ⊗ Ais a contractible equalizer (see the dual case of [BW, Proposition 3.4 (c)]). Hence,by Proposition <strong>2.</strong>20, ( it is preserved by any functor. Since(C, i) = Equ Fun ω l = − ⊗ ̂ω) l , ω r = − ⊗ ̂ω r we have that C = − ⊗ Ĉ where{∑Ĉ = a i ⊗ b i | ∑ ( ) ( aib i) ⊗ ( b i) 1⊗ ( b i) 2= 1 0 1 1 A ⊗ ∑ }a i ⊗ b i .Note that γ is a fork for ̂ω l and ̂ω r , in fact( ̂ωl ◦ γ)(h) = ̂ω ( l h 1 ⊗ h 2) = h 1 h 2 0 ⊗ ( )h 2 1 ( )1 ⊗ h2 2 (206)1 = 1 A ⊗ h 1 ⊗ h 2and(̂ω r ◦ γ ) (h) = ̂ω ( r h 1 ⊗ h 2) = 1 A ⊗ h 1 ⊗ h 2 .) ( )Since(Ĉ, î = Equ ̂ωl , ̂ω r , by the universal property of the equalizer, there existsa unique functorial morphism ϕ : H → Ĉ such that î ◦ ϕ = γ. In our case thismeans that Imγ ⊆ C and henceϕ : H → Ĉh ↦→ h 1 ⊗ h 2 .We want to prove that ϕ is an isomorphism. We computeLet us setand let us computeand[(A ⊗ ϕ) ◦ can] (a ⊗ b) = (A ⊗ ϕ) (ab 0 ⊗ b 1 ) = ab 0 ⊗ b 1 1 ⊗ b 2 <strong>1.</strong>ψ : A ⊗ C → A ⊗ Aa ⊗ b ⊗ d ↦→ ab ⊗ d[ψ ◦ (A ⊗ ϕ) ◦ can] (a ⊗ b) = ψ ( ab 0 ⊗ b 1 1 ⊗ b 2 1)= ab0 b 1 1 ⊗ b 2 1[(A ⊗ ϕ) ◦ can ◦ ψ] (a ⊗ b ⊗ d) = [(A ⊗ ϕ) ◦ can] (ab ⊗ d)(211)= a ⊗ b183

184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d∈C= a ⊗ b ⊗ d.Therefore, A ⊗ ϕ is an isomorphism and since A/k is faithfully flat, also ϕ is anisomorphism, i.e. Ĉ ∼ = H.Now we want to compute the comonad D. We haveθ l = ( A ⊗ A ⊗ σ B) ◦ (τ ⊗ A) : A ⊗ A → A ⊗ A ⊗ Aθ l (a ⊗ b) = a 0 ⊗ a 1 1 ⊗ a 2 1bθ r = A ⊗ A ⊗ u B : A ⊗ A → A ⊗ A ⊗ Aθ r (a ⊗ b) = a ⊗ b ⊗ 1 A .((Since (D, j) = Equ ) )Fun A ⊗ A ⊗ σB◦ (τ ⊗ A) , A ⊗ A ⊗ u B , we have{∑D = a i ⊗ b i | ∑ ( ) ai ⊗ ( a i) 1⊗ ( a i) 20 1 1 bi = ∑ }a i ⊗ b i ⊗ 1 A .By applying A ⊗ can to θ l and θ r , for every ∑ a i ⊗ b i ∈ D, we get that[ ] ( ∑ )(∑ ((A ⊗ can) ◦ θl a i ⊗ b i = (A ⊗ can)) ai ⊗ ( a i) 1⊗ ( a i) )20 1 1 bi= ∑ ( ) ai ⊗ ( a i) (1 (a ) (i 2 (a0 1 1)0 bi i⊗) 21)1bi= ∑ ( ) ai ⊗ ( a i) (1 (a ) )i 2 ( ) (0 1 1 bi (a ⊗ ) )i 2 ( )00 1 bi11(206)= ∑ ( ) ai ⊗ ( b i) ⊗ ( a i) ( ) (∑ )0 0 1 bi = 1 ρH A⊗A a i ⊗ b iand(∑ )[(A ⊗ can) ◦ θ r ] a i ⊗ b i(∑ )= (A ⊗ can) a i ⊗ b i ⊗ 1 A = ∑ a i ⊗ b i ⊗ 1 H .Since (A ⊗ can) is an isomorphism, we get D = Equ ( ρ H A⊗A , A ⊗ A ⊗ u H)= (A ⊗ A) co(H) .By Theorem 6.5, ∆ D and ε D are uniquely determined by(P τ) ◦ j = (jj) ◦ ∆ D and σ B ◦ j = u B ◦ ε D .Let ∑ a i ⊗ b i ∈ D = (A ⊗ A) co(H) . Then we have(∑ ) (∑ )(jj) ◦ ∆ D a i ⊗ b i = [(τ ⊗ A) ◦ j] a i ⊗ b iand also(uA ◦ ε D) ( ∑a i ⊗ b i )= ∑ (ai ) 0 ⊗ ( a i) 11 ⊗ ( a i) 21 ⊗ bi(∑ )= (τ ⊗ A) a i ⊗ b i(∑ )= (m A ◦ j) a i ⊗ b i = ∑ a i · b iSince ∑ a i ⊗ b i ∈ (A ⊗ A) co(H) , we have∑ ( ) ai ⊗ ( b i) ⊗ ( a i) ( )0 0 1 bi = ∑ a i ⊗ b i ⊗ 11 Hso that, by applying m A ⊗ H, since A is an H-comodule algebra, we get∑ (a i b i) ⊗ ( a i b i) = ∑ ( ) ( )0 1 ai0 bi ⊗ ( a i) ( )0 1 bi = ∑ a i · b i ⊗ 11 H

so that (θ l P ) ◦ (P i) : H ⊗ A → A ⊗ A ⊗ A∑i.e. ai · b i ∈ A co(H) = k1 A∼ = k. Note that from mA ◦ (A ⊗ u A ) ◦ (r A ) −1 = Id Awe get that A ⊗ u A is a monomorphism and hence, since A is faithfully flat overk, also u A is a monomorphism. We denote by ν = u |k1 AA: k → k1 A the obviousisomorphism. Thus from(uA ◦ ε D) ( ∑ )a i ⊗ b i = ∑ a i · b i185we getLet us computei.e.ε D (∑a i ⊗ b i )= v −1 (∑ai · b i ).θ l P = A ⊗ θ l : a ⊗ b ⊗ c ↦→ a ⊗ b 0 ⊗ b 1 1 ⊗ b 2 1cθ r P = A ⊗ θ r : a ⊗ b ⊗ c ↦→ a ⊗ b ⊗ c ⊗ 1 AA ⊗ θ l : A ⊗ A ⊗ AP i : C ⊗ A → A ⊗ A ⊗ Ah ⊗ a ↦→ h 1 ⊗ h 2 ⊗ ah ⊗ a ↦→ h 1 ⊗ ( h 2) ⊗ ( h 2) 1h 2) 20 1 1 h1 1 ⊗ h 2 1 ⊗ h 1 2 ⊗ h 2 2a(θ r P ) ◦ (P i) : H ⊗ A → A ⊗ A ⊗ Ah ⊗ a ↦→ h 1 ⊗ h 2 ⊗ a ⊗ 1 A .Recall that H ⊆ C ⊆ A ⊗ A. Such Q = (C ⊗ A) ∩ [A ⊗ D] ∼[= ” (H ⊗ A) ∩A ⊗ (A ⊗ A) co(H)] ”. Note that, assuming that A preserves equalizers, forevery h ⊗ a ∈ H ⊗ A, i.e. h 1 ⊗ h 2 ⊗ a ∈ C ⊗ A, h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) ifand only if h 1 ⊗ h 2 ⊗ a ∈ Equ ( A ⊗ ρ H A⊗A , A ⊗ A ⊗ A ⊗ u H)whereEqu ( )A ⊗ ρ H A⊗A, A ⊗ A ⊗ A ⊗ u H= Equ ( (A ⊗ A ⊗ A ⊗ m H ) ◦ (A ⊗ A ⊗ f ⊗ H) ◦ ( ) )A ⊗ ρ H A ⊗ ρ H A , A ⊗ A ⊗ A ⊗ uH= {a ⊗ b ⊗ c | a ⊗ b 0 ⊗ c 0 ⊗ b 1 c 1 = a ⊗ b ⊗ c ⊗ 1 H }so that h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) = Equ ( A ⊗ ρ H A⊗A , A ⊗ A ⊗ A ⊗ u H)if andonly ifh 1 ⊗ ( h 2) 0 ⊗ a 0 ⊗ ( h 2) 1 a 1 = h 1 ⊗ h 2 ⊗ a ⊗ 1 H .Let us prove that Q = (H ⊗ A) co(H) . Now,(H ⊗ A) co(H) = {h ⊗ a | h 1 ⊗ a 0 ⊗ h 2 a 1 = h ⊗ a ⊗ 1 H }where h ∈ H, h ϕ↦→ h 1 ⊗ h 2 ∈ C.1) Let h ⊗ a ∈ (H ⊗ A) co(H) and let us prove that h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) .Since h ⊗ a ∈ (H ⊗ A) co(H) , we computeh 1 ⊗ ( h 2) 0 ⊗ a 0 ⊗ ( h 2) 1 a 1(207)= (h 1 ) 1 ⊗ (h 1 ) 2 ⊗ a 0 ⊗ h 2 a 1 = h 1 ⊗ h 2 ⊗ a ⊗ 1 H

186so that h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) .2) Now, let h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) , i.e. h 1 ⊗ (h 2 ) 0⊗ a 0 ⊗ (h 2 ) 1a 1 =h 1 ⊗ h 2 ⊗ a ⊗ 1 H or equivalently (h 1 ) 1 ⊗ (h 1 ) 2 ⊗ a 0 ⊗ h 2 a 1 = h 1 ⊗ h 2 ⊗ a ⊗ 1 H . Byapplying to this equality the map can ⊗ A we obtain1 A ⊗ h 1 ⊗ a 0 ⊗ h 2 a 1 = (can ⊗ A) ( (h 1 ) 1 ⊗ (h 1 ) 2 )⊗ a 0 ⊗ h 2 a 1= (can ⊗ A) ( h 1 ⊗ h 2 ⊗ a ⊗ 1 H)= 1A ⊗ h ⊗ a ⊗ 1 Hand henceh 1 ⊗ 1 A a 0 ⊗ h 2 a 1 = h ⊗ 1 A a ⊗ 1 Hso that h 1 ⊗ h 2 ⊗ a ∈ (H ⊗ A) co(H) . Therefore we proved that[(ϕ ⊗ A) (H ⊗ A) ∩ A ⊗ (A ⊗ A) co(H)] [= (ϕ ⊗ A) (H ⊗ A) co(H)] .We can take( ) ()Q, q = (H ⊗ A) co(H) , (ϕ ⊗ A) |(H⊗A)co(H) .By Theorem 7.5, using i ◦ ϕ = γ, we have thatχ := m A ◦ (A ⊗ m A ) ◦ (m A ⊗ A ⊗ A) ◦ (A ⊗ A ⊗ A ⊗ m A )()◦ (A ⊗ i ⊗ A ⊗ A) ◦ A ⊗ (ϕ ⊗ A) |(H⊗A)co(H) ⊗ Aχ : A ⊗ (H ⊗ A) co(H) ⊗ A → Aa ⊗ h ⊗ b ⊗ c ↦→ a ⊗ h 1 ⊗ h 2 ⊗ b ⊗ c ↦→ ( ah 1) ( h 2 (bc) ) = a ( h 1 h 2) (bc) (209)= abcε H (h)is a coherd in X = (H, (A ⊗ A) co(H) , (H ⊗ A) co(H) , A, δ H , δ D ) where δ H(H ⊗ A) co(H) ⊗ A is uniquely determined by[](ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ H = (H ⊗ i) ◦ ∆ H .: H →Since (C, i) = Equ Fun(ω l , ω r) , by the universal property of the equalizer, there existsa unique functorial morphism ϕ : H → C such that i◦ϕ = γ. In our case this meansthat Imγ ⊆ C and henceϕ : H → Ch ↦→ h 1 ⊗ h 2 .For every h ∈ H, we have([] )(ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ C = [ (C ⊗ i) ◦ ∆ C] = (C ⊗ i) ◦ (ϕ ⊗ ϕ) ◦ ∆ H ◦ ϕ −1= (ϕ ⊗ i ◦ ϕ) ◦ ∆ H ◦ ϕ −1 = (ϕ ⊗ γ) ◦ ∆ H ◦ ϕ −1and hence[](ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ = (ϕ ⊗ γ) ◦ ∆ H = (ϕ ⊗ A ⊗ A) ◦ (H ⊗ γ) ◦ ∆ Hso that(ϕ −1 ⊗ A ⊗ A ) []◦ (ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ= ( ϕ −1 ⊗ A ⊗ A ) ◦ (ϕ ⊗ A ⊗ A) ◦ (H ⊗ γ) ◦ ∆ H

187Now(ϕ ⊗ A) |(H⊗A)co(H) = (ϕ ⊗ A) ◦ i (H⊗A)co(H)where i (H⊗A)co(H) : (H ⊗ A) co(H) → H ⊗ A is the canonical inclusion and hence weget(ϕ −1 ⊗ A ⊗ A ) []◦ [(ϕ ⊗ A) ⊗ A] ◦ i (H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ= ( ϕ −1 ⊗ A ⊗ A ) ◦ (ϕ ⊗ A ⊗ A) ◦ (H ⊗ γ) ◦ ∆ Hi.e. []i (H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ = (H ⊗ γ) ◦ ∆ H .Now we have[() ] (hi (H⊗A)co(H) ⊗ A ◦ δ 1 C ⊗ h 2) [() ]= i (H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ (h)= ( (H ⊗ γ) ◦ ∆ H) (h) = h 1 ⊗ h 1 2 ⊗ h 2 2i.e. (i (H⊗A)co(H) ⊗ A) (δC (h 1 ⊗ h 2)) = h 1 ⊗ h 1 2 ⊗ h 2 2Let us compute δ D : D → QQ = ⊗A⊗(H ⊗ A) co(H) following Proposition 7.2 whichneeds to satisfy(κ ′ 0Q) ◦ δ D = (Dj) ◦ ∆ D .In our case this meanswhereand(jP Q) ◦ (κ ′ 0Q) ◦ δ D = (jP Q) ◦ (Dj) ◦ ∆ D = (jj) ◦ ∆ D = (τ ⊗ A)κ ′ 0 (h ⊗ a) = h 1 ⊗ h 2 ⊗ a(jj) ◦ ∆ D (a ⊗ b) = [(τ ⊗ A) ◦ j] (a ⊗ b) = (τ ⊗ A) (a ⊗ b) = a 0 ⊗ a 1 1 ⊗ a 2 1 ⊗ b.so thatδ D(∑a i ⊗ b i )= ∑ (ai ) 0 ⊗ ( a i) 1 ⊗ bi .In this more specific situation we could compute both the comonads C and D andthe functor Q so that we obtained a coherd. We now would like to compute themonads corresponding to the coherd following Theorem 6.29. But the computationsare not straightforward and it is not clear what these new monads are.9.<strong>2.</strong> H-Galois coextension. This is the most clear example of a coherd that wecould give. It gives also a description of the dual case of the Morita-Takeuchiequivalence studied by Schauenburg in [Scha4]. In fact we could understand theequivalence between the module categories over the two monads constructed fromthe coherd.Let H = ( H, m H , u H , ∆ H , ε H , S ) be a Hopf algebra and let L ⊆ H be a rightcoideal subalgebra i.e. ∆ H (L) ⊆ L ⊗ H. We can consider ε H : H → k as acharacter so that)J = J ε H =〈(hy) (1)ε((hy) H (2)− h (1) ε ( H h (2) y ) 〉| h ∈ H, y ∈ L

188= 〈( h (1) y (1))εH ( h (2) y (2))− h(1) ε H ( h (2))ε H (y) | h ∈ H, y ∈ L 〉= 〈 h (1) y (1) ε H ( h (2))εH ( y (2))− h(1) ε H ( h (2))ε H (y) | h ∈ H, y ∈ L 〉= 〈 hy − hε H (y) | h ∈ H, y ∈ L 〉 = 〈 hy ′ | h ∈ H, y ′ ∈ L +〉 = HL +where we denote L + = L ∩ Ker ( ε H) . Let us prove that such J is a coideal of H (seealso [BrHaj, Lemma 3.2]). In fact, since ∆ H (y) = y (1) ⊗ y (2) ∈ L ⊗ H we have∆ H ( hy − hε H (y) ) = h (1) y (1) ⊗ h (2) y (2) − h (1) ⊗ h (2) ε H (y)= h (1)(y(1) − ε H ( y (1)))⊗ h(2) y (2) + h (1) ε H ( y (1))⊗ h(2) y (2) − h (1) ⊗ h (2) ε H (y)= h (1)(y(1) − ε H ( y (1)))⊗ h(2) y (2) + h (1) ⊗ h (2) ε H ( y (1))y(2) − h (1) ⊗ h (2) ε H (y)= h (1)(y(1) − ε H ( y (1)))⊗ h(2) y (2) + h (1) ⊗ h (2)(y − ε H (y) ) ∈ HL + ⊗ H + H ⊗ HL +and obviouslyε H ( hy − hε H (y) ) = ε H (h) ε H (y) − ε H (h) ε H (y) = 0.Then we can construct the coalgebra C := H/J = H/HL + and we can considerthe canonical projection π : H → C = H/J which is a coalgebra map and a leftH-linear map. We can defineC ρ H := (π ⊗ H) ◦ ∆ H : H → C ⊗ H and ρ C H := (H ⊗ π) ◦ ∆H : H → H ⊗ Ch ↦→ π ( h (1))⊗ h(2) h ↦→ h (1) ⊗ π ( h (2))so that H is a C-bicomodule. Note that C ρ H : H → C□ C H in fact, using that π isa coalgebra map, the coassociativity and the naturality of ∆ H , we compute(∆ C ⊗ H ) ◦ C ρ H = ( ∆ C ⊗ H ) ◦ (π ⊗ H) ◦ ∆ H = (π ⊗ π ⊗ H) ◦ ( ∆ H ⊗ H ) ◦ ∆ H= (π ⊗ π ⊗ H) ◦ ( H ⊗ ∆ H) ◦ ∆ H = (C ⊗ π ⊗ H) ◦ (π ⊗ H ⊗ H) ◦ ( H ⊗ ∆ H) ◦ ∆ H= (C ⊗ π ⊗ H) ◦ ( C ⊗ ∆ H) ◦ (π ⊗ H) ◦ ∆ H = ( C ⊗ C ρ H)◦ C ρ H .Similarly, we also have that ρ C H : H → H□ CC in fact, using that π is a coalgebramap, the coassociativity and naturality of ∆ H , we have(H ⊗ ∆C ) ◦ ρ C H = ( H ⊗ ∆ C) ◦ (H ⊗ π) ◦ ∆ H = ( H ⊗ ∆ C ◦ π ) ◦ ∆ H= ( H ⊗ (π ⊗ π) ◦ ∆ H) ◦ ∆ H = (H ⊗ π ⊗ π) ◦ ( H ⊗ ∆ H) ◦ ∆ H= (H ⊗ π ⊗ π) ◦ ( ∆ H ⊗ H ) ◦ ∆ H = (H ⊗ π ⊗ C) ◦ (H ⊗ H ⊗ π) ◦ ( ∆ H ⊗ H ) ◦ ∆ HNow, the map= (H ⊗ π ⊗ C) ◦ ( ∆ H ⊗ C ) ◦ (H ⊗ π) ◦ ∆ H = ( ρ C H ⊗ C ) ◦ ρ C H.∆ H : H → H□ C Hh ↦→ h (1) ⊗ h (2)))is well defined. In fact h (1)(1) ⊗π(h (1)(2) ⊗h (2) = h (1) ⊗π(h (2)(1) ⊗h (2)(2) . Moreover,the map π : H → C is a counit for H, in fact[(π□C H) ◦ ∆ H] (h) = π ( )h (1) ⊗ h(2) = C ρ H (h) ≃ h[(H□C π) ◦ ∆ H] (h) = h (1) ⊗ π ( )h (2) = ρCH (h) ≃ h

189[(ε C ⊗ H ) ◦ (π ⊗ H) ◦ ∆ H] (h) = ( ε C π ) ( h (1))⊗ h(2) = ε H ( h (1))⊗ h(2) ≃ h[(H ⊗ εC ) ◦ (H ⊗ π) ◦ ∆ H] (h) = h (1) ⊗ ( ε C π ) ( h (2))= h(1) ⊗ ε H ( h (2))≃ hso that H is a C-coring. Moreover, H has a right L-module structurewhich is left C-colinear i.e.µ L H : H ⊗ L → Hh ⊗ b ↦→ m H (h ⊗ b) = hb(212) π ( h (1) b (1))⊗ h(2) b (2) = π (h 1 ) ⊗ h (2) b(see [BrHaj, Lemma 3.3]), so that [( H ⊗ µ L H)◦(∆ H ⊗ L )] (H ⊗ L) ⊆ H□ C H.Assume that H is a right L-Galois coextension over C, that iscocan = ( H ⊗ µ L H)◦(∆ H ⊗ L ) : H ⊗ L → H□ C Hh ⊗ b ↦→ h (1) ⊗ h (2) bis an isomorphism and assume also that H is flat over k. In particular, if H L isfaithfully flat, we know that co(C) H = L (see [Schn2, Lemma <strong>1.</strong>3 (2)] and [BrWi,34.2 p. 343]) where we denoteco(C) H = { h ∈ H | C ρ H (h) = π (1 H ) ⊗ h } .In this case, we can also define the inverse of the cocanonical map, i.e.cocan −1 : H□ C H → H ⊗ L∑h i ⊗ g i ↦→ ∑ h i (1) ⊗ S ( )h i (2) g i .For every ∑ h i ⊗g i ∈ H□ C H, we have ∑ )h i (1)(h ⊗π i (2)⊗g i = ∑ ( )h i ⊗π g(1)i ⊗g(2) i .By means of the left H-linearity of π and of this equality we have∑hi(1) ⊗ π ( S ( ) ) ( )h i (3) gi(1) ⊗ S hi(2) gi(2) = ∑ h i (1) ⊗ S ( ) ( ) ( )h i (3) π gi(1) ⊗ S hi(2) gi(2)= ∑ h i (1) ⊗ S ( ( ) ( )h(3)) i π hi(4) ⊗ S hi(2) g i = ∑ h i (1) ⊗ π ( S ( ) ) ( )h i (3) hi(4) ⊗ S hi(2) gi= ∑ h i (1) ⊗ π (1 H ) ⊗ S ( )h i (2) giso that∑hi(1) ⊗ S ( )h i (2) g i ∈ Ker ( H ⊗ [C ρ H − π (1 H ) ⊗ (−) ]) = H ⊗ co(C) H = H ⊗ Lwhere in the first equality we have used that H is flat over k. Therefore cocan −1 isa well-defined map. Note that, by applying ε H ⊗ L to this element, we also deducethat, for every ∑ h i ⊗ g i ∈ H□ C H, we have(213)∑S(hi ) g i ∈ L.Now, let k be a commutative ring, let H be a k-Hopf algebra and letL ⊆ H be a right coideal subalgebra. Assume that H is a right L-Galoiscoextension over the coalgebra C = H/HL + , assume that H L is faithfullyflat, so that co(C) H = L, assume that H k is faithfully flat and assume that

190H C is faithfully coflat. Assume also C H coflat. Then we can consider thefollowing formal codual structure X = (C, D, Q, P, δ C , δ D ) whereA = Mod-kB = Comod-CC = ( − ⊗ H, − ⊗ ∆ H , − ⊗ ε H) : A = Mod-k −→ A = Mod-k(D = −□ C H, −□ C ∆ H , −□ C ε C H C) : B = Comod-C −→ B = Comod-CQ = −□ C H : B = Comod-C → A = Mod-kP = − ⊗ H C : A = Mod-k → B = Comod-Cδ C : = − ⊗ ∆ H : C = − ⊗ H → QP = − ⊗ H□ C Hδ D : = −□ C ∆ H : D = −□ C H → P Q = −□ C H ⊗ H.Now, for every ∑ ∑i j ki,j ⊗ h i,j ⊗ g i ∈ (H ⊗ H) □ C H, we have that∑ ∑( )(214)i j ki,j ⊗ h i,j(1) ⊗ π h i,j(2)⊗ g i = ∑ ∑i j ki,j ⊗ h i,j ⊗ π ( )g(1)i ⊗ gi(2) .We want to prove that ∑ ∑i j ki,j ⊗ S (h i,j ) g i ∈ H ⊗ L. We compute, using the leftH-linearity of π and (214)∑ ∑ (i j ki,j ⊗ π S ( h i,j) )(1) gi (1) ⊗ S ( h i,j) (2) gi (2)= ∑ ∑ ( )i j ki,j ⊗ S h i,j(2)π ( ) )g(1)i ⊗ S(h i,j(1)g(2)i= ∑ ∑ ( ) ( ) ( )i j ki,j ⊗ S h i,j(2)π h i,j(3)⊗ S h i,j(1)g i= ∑ ∑ ( ( ) ) ( )i j ki,j ⊗ π S h i,j(2)h i,j(3)⊗ S h i,j(1)g i= ∑ ∑i j ki,j ⊗ π (1 H ) ⊗ S ( h i,j) g iso that we get∑ ∑ij ki,j ⊗πwhich means∑ ∑i(S ( h i,j) )(1) gi (1) ⊗S ( h i,j) (2) gi (2) = ∑ ∑i j ki,j ⊗π (1 H )⊗S ( h i,j) g ij ki,j ⊗ S ( h i,j) g i ∈ Ker ( H ⊗ [C ρ H − π (1 H ) ⊗ (−) ]) = H ⊗ co(C) H = H ⊗ Li.e.∑ ∑(215)i j ki,j ⊗ S ( h i,j) g i ∈ H ⊗ L.Similarly, for every ∑ ∑i j li,j ⊗ h i,j ⊗ g i ∈ (L ⊗ H) □ C H, we have that∑ ∑i j li,j ⊗ S ( h i,j) g i ∈ Ker ( L ⊗ [C ρ H − π (1 H ) ⊗ (−) ]) = L ⊗ co(C) H = L ⊗ Li.e.(216)∑i∑j li,j ⊗ S ( h i,j) g i ∈ L ⊗ L

so that, since L is a subalgebra of H we get that, for every ∑ ∑i j li,j ⊗ h i,j ⊗ g i ∈(L ⊗ H) □ C H,∑ ∑(217)i j li,j S ( h i,j) g i ∈ L.Let us consider the following mapi∑i∑̂χ : (H ⊗ H) □ C H → H∑j ki,j ⊗ h i,j ⊗ g i ↦→ ∑ ij ki,j S ( h i,j) g iwhich is left C-colinear. In fact, in view of (215), we have that ∑ iS (h i,j ) g i ∈ H ⊗ L and by (212), we get that∑ ( ∑j π [ (k ) i,j(1) S hi,jg i] [ ((1))⊗k ) i,j(2) S hi,jg i] = ∑ (k(2) i∑j π i,j(1)Therefore, we can define the coherd χ = −□ C ̂χ given byχ : QP Q = −□ C H ⊗ H□ C H → Q = −□ C H∑∑ ∑−□ Ci∑j ki,j ⊗ h i,j ⊗ g i ↦→ −□ Ci j ki,j S ( h i,j) g i .Let us prove the properties of χ. We have[χ ◦ (QP χ)](−□ C k ⊗ ∑ )h i ⊗ g i ⊗ l j ⊗ n j= [χ ◦ (χ ⊗ H□ C H)](−□ C k ⊗ ∑ )h i ⊗ g i ⊗ l j ⊗ n j∑ (= χ(−□ ) ) ∑ ( (C kS hig i ⊗ l j ⊗ n j = −□ ) C kS hig i) S ( l j) n jand[χ ◦ (χP Q)](−□ C k ⊗ ∑ )h i ⊗ g i ⊗ l j ⊗ n j= [χ ◦ (−□ C H ⊗ H□ C χ)](k ⊗ ∑ )h i ⊗ g i ⊗ l j ⊗ n j= χ(k ⊗ ∑ h i ⊗ g i S ( l j) ) ∑ (n j = −□ ) ( C kS hig i S ( l j) n j)191∑j ki,j ⊗)⊗k i,j(2) S ( h i,j) g i .so that χ is coassociative. Moreover, we have[χ ◦ (δ C Q)] (k ⊗ h) = [χ ◦ (−□ C H ⊗ δ C )] (−□ C k ⊗ h) = χ ( −□ C k ⊗ h (1) ⊗ h (2))and= −□ C kS ( h (1))h(2) = kε H (h) = ( −□ C H ⊗ ε H) (k ⊗ h) = ( ε C Q ) (k ⊗ h)∑ )[χ ◦ (Qδ D )](−□ C k i ⊗ h i∑ )= [χ ◦ (δ D □ C H)](−□ C k i ⊗ h i)hi∑ ) ∑= χ(−□ C ki(1) ⊗ k(2) i ⊗ h i = −□ C ki(1) S ( k(2)i= ∑ ε H ( k i) 1 H h i = ∑ ε H ( k i) h i = −□ C∑ (ε C ◦ π ) ( k i) h i∑= −□ ( C εCπ ( k i)) ∑ (h i ≃ −□ ) C π ki□ C h i∑ (= −□ ) ( ) ∑ )C ε C H C ki□ C h i = ε C H C □ C H(−□ C k i ⊗ h i

192=(Qε C H C) ( ∑ )−□ C k i ⊗ h iso that the counitality conditions are also satisfied, i.e. χ is really a coherd. SinceH k is faithfully flat, we have that ( k, ε H) (= Coequ Mod-k H ⊗ ε H , ε H ⊗ H ) and sinceH C is faithfully coflat, by [Schn1, Proposition <strong>1.</strong>1], we also have that(C, π) =(C, ε C H C) ()= Coequ Comod-C H□ C ε C H C , ε C H C □ C H= Coequ Comod-C (H□ C π, π□ C H)so that X is a regular formal codual structure and thus χ is a regular coherd. FollowingTheorem 6.29, we calculate the monad(A, x) = Coequ Fun(w l , w r)where w l = (χP ) ◦ (QP δ C ) and w r = QP ε C : QP C → QP . In our caseand− ⊗ ∑ i− ⊗ ∑ i∑∑w l : − ⊗ H ⊗ H□ C H → − ⊗ H□ C H∑j ki,j ⊗ ∑ h i,j ⊗ g i ↦→ − ⊗ ∑ iw r : − ⊗ H ⊗ H□ C H → − ⊗ H□ C H∑j ki,j ⊗ ∑ h i,j ⊗ g i ↦→ − ⊗ ∑ ij ki,j (1) ⊗ ki,j (2) S ( h i , j ) g ij εH ( k i,j) h i,j ⊗ g i .Assume now that k is a field, so that everything is flat over k. Hence, for everyX ∈ Mod-kand thusAX ==X ⊗ H□ C HIm (X ⊗ w l − X ⊗ w r ) = X ⊗ H□ C HIm (X ⊗ (w l − w r ))X ⊗ H□ C HX ⊗ Im (w l − w r ) = X ⊗ H□ CHIm (w l − w r )A = − ⊗ H□ CH〈 ∑where I w =i∑j ki,j (1) ⊗ ki,j (2) S (hi , j) g i − ∑ 〉i∑j εH (k i,j ) h i,j ⊗ g i . In the sequel,given elements ∑ i hi ⊗ g i ∈ H□ C H, we will use the notation][∑i hi ⊗ g i = ∑ i hi ⊗ g i + I w .AWe will prove that this new monad A on the category Mod-k is isomorphic to themonad coming from the algebra L. Consider the following mapI wϕ : H□ CH−→ LI] w[∑i hi ⊗ g i ↦→ ∑ S ( h i) g iiA

which is well-defined by (213), i.e. ∑ S (h i ) g i ∈ L. Note that, since L = co(C) H, forevery b ∈ L, we haveThe inverse of this map is given by1 H ⊗ π (1 H ) ⊗ b = 1 H ⊗ ∑ π (b 1 ) ⊗ b 2 .ϕ −1 : L −→ H□ CHI wb ↦→ [1 H ⊗ b] A.In fact we have (ϕ −1 ◦ ϕ) ( [ ∑ i hi ⊗ g i ] A)= ϕ −1 ( ∑ i S (hi ) g i ) = [1 H ⊗ ∑ i S (hi ) g i ] A=[ ∑ i hi ⊗ g i ] Aby definition of I w and (ϕ ◦ ϕ −1 ) (b) = ϕ ([1 H ⊗ b] A) = S (1 H ) b = band thus ϕ is bijective so thatA = − ⊗ H□ CH≃ − ⊗ L : Mod-k → Mod-k.I wThe functorial morphisms m A and u A of the monad A are uniquely determined byx ◦ (χP ) = m A ◦ (xx) and x ◦ δ C = u A ◦ ε Cwhere x : H□ C H → H□ CHI wdenotes the canonical projection. In our case we have( [ ]] )m H□ C HIw∑i hi ⊗ g i ⊗A[∑j kj ⊗ l j =A[∑i,j hi ⊗ g i S ( k j) ]l j A=[∑i,j 1 H ⊗ S ( h i) g i S ( k j) ] (∑l j = ϕ −1 S ( h i) g i S ( k j) )l jAi,j( (∑= ϕ −1 m L S ( h i) g i ⊗ ∑ S ( k j) ))l ji j= ( ϕ −1 ◦ m L ◦ (ϕ ⊗ ϕ) ) ( [∑]i hi ⊗ g]Ai ⊗[∑j kj ⊗ l jfrom which we deduce thatMoreoverso thatu H□ C HIwu H□ C HIwϕ ◦ m H□ C HIw= m L ◦ (ϕ ⊗ ϕ) .(ε H (h) ) = (x ◦ δ C ) (h) = [ h (1) ⊗ h (2)]((1 k ) = u H□ C H ε H (1 H ) ) = (x ◦ δ C ) (1 H ) = [ ]1 H(1) ⊗ 1 H(2) = [1 A H ⊗ 1 H ] AIw= [1 H ⊗ 1 L ] A= ϕ −1 (1 L ) = ( )ϕ −1 ◦ u L (1k )from which we deduce thatϕ ◦ u H□ C H = u L .IwThe two relations obtained say that ϕ : H□ CH−→ L is an algebra isomorphism sothatA = − ⊗ H□ CHI wI w≃ − ⊗ L as monads.AA)193

194Following Theorem 6.29, we now calculate the monad(E, y) = Coequ Fun(z l , z r)where z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q. In our case, let usconsider∑ii∑ẑ l : (H ⊗ H) □ C H −→ H ⊗ H∑j ki,j ⊗ h i,j ⊗ g i ↦→ ∑ ij ki,j S ( h i,j) g i (1) ⊗ g i (2)and let us prove that ẑl it is left C-colinear. By (215) we have that ∑ iS (h i,j ) g i ∈ H ⊗ L so that, in view of (212), we have∑ ( ∑j π [ (k ) ) [ ( i,j(1) S hi,jg(1)]i ⊗ k ) ]i,j(2) S hi,jg(1)iHenceand= ∑ i∑j π (k i,j(1))(1)⊗ k i,j(2) S ( h i,j) g i (1) ⊗ g i (2).z l : −□ C H ⊗ H□ C H −→ −□ C H ⊗ H∑∑ ∑−□ C∑j ki,j ⊗ h i,j ⊗ g i ↦→ −□ Cii(2) ⊗ gi (3)j ki,j S ( h i,j) g i (1) ⊗ g i (2)z r : −□ C H ⊗ H□ C H −→ −□ C H ⊗ H∑∑−□ C∑j ki,j ⊗ h i,j ⊗ g i ↦→ −□ C∑j ki,j ⊗ h i,j ε ( H g i)iare well-defined. For every ( X, ρ C X)∈ Comod-C we haveso thatwhereI X□C z =E ( X, ρ C X)=X□ C H ⊗ HIm (X□ C z l − X□ C z r ) =E ( X, ρ C X)=X□ C H ⊗ HI X□C ziX□ C H ⊗ HIm (X□ C (z l − z r ))〈 ∑i,j xj ⊗ k j S (h i ) g i (1) ⊗ gi (2) − ∑ i,j xj ⊗ k j ⊗ h i ε H (g i )| ∑ j xj ⊗ k j , ∑ i hi ⊗ g i ∈ H□ C H∑j ki,j ⊗Recall (see [BrHaj, Theorem 3.5]) that, associated to the cocanonical map, we havea unique canonical entwining structure given byψ = (̂τ ⊗ H) ◦ ( H ⊗ ∆ H) ◦ cocan : H ⊗ L −→ L ⊗ Hh ⊗ y ↦→ y (1) ⊗ hy (2)where ̂τ = ( ε H ⊗ L ) ◦cocan −1 : H□ C H −→ L is the cotranslation map. Since cocanis an isomorphism, in order to understand better the monad E we first composewith the isomorphism H ⊗ cocan and we compute for every h, g ∈ H, y ∈ L,(z l ◦ (H ⊗ cocan) ) (h ⊗ g ⊗ y) = hy (1) ⊗ gy (2) and (z r ◦ (H ⊗ cocan)) (h ⊗ g ⊗ y) =〉.h ⊗ gε H (y). Leti : (H ⊗ H) L + → (H ⊗ H)

denote the canonical inclusion. Then i is a left C-comodule map, in fact, for every∑i (h i ⊗ g i ) ( l i − ε H (l i ) ) = ∑ i h il i(1) ⊗ g i l i(2) − h i ⊗ g i ε H (l i ) ∈ (H ⊗ H) L + , since∆ H (l i ) = l i(1) ⊗ l i(2) ∈ L ⊗ H, we have∑i π ( h i(1) l i(1))⊗ hi(2) l i(2) ⊗ g i l i(3) − π ( h i(1))⊗ hi(2) ⊗ g i ε H (l i )= ∑ i π ( h i(1))⊗ hi(2) l i(1) ⊗ g i l i(2) − π ( h i(1))⊗ hi(2) ⊗ g i ε H (l i )= ∑ i π ( h i(1))⊗(hi(2) ⊗ g i) (li − ε H (l i ) ) ∈ C ⊗ (H ⊗ H) L + .Hence, for every ( X, ρ C X)∈ Comod-C, we can consider the mapX□ C i : X□ C (H ⊗ H) L + → X□ C (H ⊗ H) .so that, for every ( X, ρ C X)∈ Comod-C, we havewhereI X□C L =Let p : H ⊗ H →E ( X, ρ C X)=X□ C H ⊗ HI X□C z= X□ CH ⊗ HI X□C L〈 ∑i xi ⊗ h i y (1) ⊗ gy (2) − ∑ i xi ⊗ h i ⊗ gε H (y)| ∑ i xi ⊗ h i ⊗ g ⊗ y ∈ X□ C H ⊗ H ⊗ L= X□ C[(H ⊗ H) L+ ] .H⊗H(H⊗H)L +〉195be the canonical projection and let us assume that i isleft C-copure i.e. for every ( X, ρ C X)∈ Comod-C, the sequence0 → X□ C (H ⊗ H) L + X□ Ci−→ X□ C (H ⊗ H) X□ Cp−→ X□ Cis exact. In this case we get that, for every ( X, ρX) C ∈ Comod-C,E ( )X, ρ C ∼= H ⊗ HX X□C(H ⊗ H) L = X□ + C (H ⊗ H) LH ⊗ H(H ⊗ H) L → 0 +where (H ⊗ H) Ldenotes the invariants with respect to the algebra L. In the sequel,given h i , k i ∈ H we will use the notation][∑i hi ⊗ k i = ∑ i hi ⊗ k i + (H ⊗ H) L +ELet us denote E := −□ C (H ⊗ H) Land let us consider multiplication and unit ofE. Following Theorem 6.29, they are uniquely determined byi.e.wherêm E :m E ◦ (yy) = y ◦ (P χ) and y ◦ δ D = u E ◦ ε Dm E = −□ Ĉm E and u E = −□ C û EH ⊗ H(H ⊗ H) L □ H ⊗ H+ C(H ⊗ H) L −→ H ⊗ H+ (H ⊗ H) L and û + E : C −→ H ⊗ H(H ⊗ H) L +given by(∑ ∑ ∑ [̂m E k i,j ⊗ h i,j] □ [E C g i,s ⊗ l i,s] [∑E)= k i,j S ( h i,j) ]g i,s ⊗ l i,sE

196)û E(ε C H C (h) = û E (π (h)) = [ ]h (1) ⊗ h (2) . ELet us check that ̂m E is a well-defined map. Let us considerf : H ⊗ H → Hh ⊗ k ↦→ hS (k) .For every (h ⊗ k) · l ∈ (H ⊗ H) L + , we havef [( hl (1) ⊗ kl (2))− (h ⊗ k) ε (l)]= hl(1) S ( l (2))S (k) − hS (k) ε (l) = 0so that f induces a morphismH ⊗ Hf :(H ⊗ H) L → H +[h ⊗ k] ↦→ hS (k) .Now, let us consider the compositeH ⊗ H(H ⊗ H) L ⊗ H ⊗ H f⊗Id H⊗Id Hm−→ H ⊗ H ⊗ HH ⊗Id Hp−→ H ⊗ H −→H ⊗ H+ (H ⊗ H) L +where p denotes the canonical projection. Note that[p ◦ (m H ⊗ H)] ( H ⊗ (H ⊗ H) L +) = 0in fact, for every x ∈ H and (h ⊗ k) · l ∈ (H ⊗ H) L + we haveand thusx ⊗ [( hl (1) ⊗ kl (2))− (h ⊗ kε (l))]= x ⊗(hl(1) ⊗ kl (2))− x ⊗ (h ⊗ kε (l))xhl (1) ⊗ kl (2) − xh ⊗ kε (l) = (xh ⊗ k) (l − ε (l)) ∈ (H ⊗ H) L + .Therefore, the above composite map induces the mapH ⊗ H(H ⊗ H) L ⊗ H ⊗ H+ (H ⊗ H) L → H ⊗ H+ (H ⊗ H) L +[k ⊗ h] ⊗ [g ⊗ l] ↦→ [kS (h) g ⊗ l]which is well-defined and hence also the mapH ⊗ Ĥm E :(H ⊗ H) L □ H ⊗ H+ C(H ⊗ H) L → H ⊗ H+ (H ⊗ H) L +∑ [ ] k ⊗ hi ⊗ [ g i ⊗ l ] [∑ ↦→ ( )E E kS hig i ⊗ l].is well defined. Observe that, by using (213) and (212) we have∑ ( ( (π k ) (1) S hig i) ) ( (⊗ k ) (1) (2) S hig i) = ∑ π ( ) ( (k ) (2) (1) ⊗ k(2) S hig i)so that the maps∑ [ ] k ⊗ hi ⊗ [ g i ⊗ l ] [∑ ↦→ ( ) ]E E kS hig i ⊗ l∑k ⊗[h i ⊗ g i] A ↦→ ∑ kS ( h i) g iEE,and∑ [k ⊗ hi ] E ⊗ gi ↦→ ∑ kS ( h i) g i

are left C-colinear and hence ̂m E is also left colinear. Therefore the mapH ⊗ Hm E = −□ Ĉm E : −□ C(H ⊗ H) L □ H ⊗ H+ C(H ⊗ H) L → H ⊗ H+ (H ⊗ H) L +is well-defined. Moreover, Q = −□ C H can be equipped with the structure of an A-Bbimodulefunctor, i.e. in our setting, with a well-defined structure of L-E-bimodulefunctor given by A µ Q = −□ C µ Q and µ E Q = −□ C ̂µ E Q where∑iµ Q : H ⊗ L −→ Hk ⊗ l ↦→ kl̂µ E Q : H ⊗ H(H ⊗ H) L □ CH −→ H+∑ [k i,j ⊗ h i,j] ⊗ E gi ↦→ ∑ ∑ijj ki,j S ( h i,j) g i .Similarly one can prove that A µ Q and µ E Q are well-defined. Let us calculate the( )( (coequalizer ̂Q, l = Coequ Fun (P x) ◦ z l P ) , (P x) ◦ (z r P ) ) defined in Proposition7.6− ⊗ A ⊗ ĤQ =− ⊗ Im ((x ⊗ H) ◦ (H□ C z l ) − (x ⊗ H) ◦ (H□ C z r ))=− ⊗ H□ CHI w⊗ H− ⊗ Im ((x ⊗ H) ◦ (H□ C z l ) − (x ⊗ H) ◦ (H□ C z r )) .Since we are in the case when cocan is an isomorphism, we equivalently calculate,for every ∑ h i ⊗ g i ∈ H□ C H, k ∈ H and t ∈ L,((x ⊗ H) ◦(H□C z l) ◦ (H□ C H ⊗ cocan) ) ( ∑h i ⊗ g i ⊗ k ⊗ t)[∑ ]= h i ⊗ g i t (1) ⊗ kt (2)197and(∑ )((x ⊗ H) ◦ (H□ C z r ) ◦ (H□ C H ⊗ cocan)) h i ⊗ g i ⊗ k ⊗ t[∑ ]= h i ⊗ g i ⊗ kε H (t) .Having in mind that also ϕ is an isomorphism, we also compute(ϕ ⊗ H) ( (x ⊗ H) ◦ ( H□ C z l) ◦ (H□ C H ⊗ cocan) ) ( ∑ )h i ⊗ g i ⊗ k ⊗ t[∑ ]= (ϕ ⊗ H) h i ⊗ g i t (1) ⊗ kt (2) = ∑ S ( h i) g i t (1) ⊗ kt (2)and(∑ )(ϕ ⊗ H) ((x ⊗ H) ◦ (H□ C z r ) ◦ (H□ C H ⊗ cocan)) h i ⊗ g i ⊗ k ⊗ t[∑ ]= (ϕ ⊗ H) h i ⊗ g i ⊗ kε H (t) = ∑ S ( h i) g i ⊗ kε H (t) .

198Letandα l = (ϕ ⊗ H) ( (x ⊗ H) ◦ ( H□ C z l) ◦ (H□ C H ⊗ cocan) )α r = (ϕ ⊗ H) ((x ⊗ H) ◦ (H□ C z r ) ◦ (H□ C H ⊗ cocan)) .Then, for every ∑ h i ⊗ g i ∈ H□ C H, k ∈ H and t ∈ L,(∑ )(α l − α r ) h i ⊗ g i ⊗ k ⊗ t = ∑ S ( h i) g i t (1) ⊗ kt (2) − ∑ S ( h i) g i ⊗ kε H (t)[∑ (=) [∑ (S hig i ⊗ k]· t − ) ]S hig i ⊗ k · ε H (t) 1 L[∑ (=) S hig i ⊗ k]· (t )− ε H (t) 1 Lso that we getIm (α l − α r ) = (L ⊗ H) L +and hence the isomorphism ϕ : A = H□ CHI w−→ L induces an isomorphism( )( ( ̂Q, l = Coequ Fun (P x) ◦ z l P ) , (P x) ◦ (z r P ) )∼ = − ⊗ Coequ (αl , α r ) = − ⊗ L ⊗ H(L ⊗ H) L + .In the sequel, given elements l i ∈ L and h i ∈ H we will use the notation[∑ ]l i ⊗ h i = ∑ l i ⊗ h i + (L ⊗ H) L + .bQFollowing Proposition 7.6, the functor ̂Q can be equipped with the structure of aB-A-bimodule functor, i.e. in our setting, with a structure of E-L-bimodule functor.In particular E µb Q= − ⊗ ʵbQ: E ̂Q → ̂Q and µ L b Q= − ⊗ ̂µ L b Q: ̂QA → ̂Q where∑andi∑jE L ⊗ Hµb Q:(L ⊗ H) L □ H ⊗ H+ C(H ⊗ H) L −→ L ⊗ H+ (L ⊗ H) L +∑ [l i,j ⊗ k i,j] [Q b □ C h i,s ⊗ t i,s] ↦→ ∑ ∑ ∑ [E l i,j S ( k i,j) h i,s ⊗ t i,s] bi Qŝµ L b Q: L ⊗ L ⊗ H(L ⊗ H) L + −→ L ⊗ H(L ⊗ H) L +y ⊗ [y ′ ⊗ h]b Q↦→ [yy ′ ⊗ h]b Q.Such a bimodule functor ̂Q is the one giving rise, together with the functor Q, tothe equivalence of the categories of modules over the monads A ≃ L and B = Econstructed in the above Subsection 8.1 (in particular see Theorems 8.6 and 8.9).More explicitly,andLQ E = − ⊗ E H L : E B = E (Comod-C) → L A = L (Mod-k) = Mod-LE ̂Q L = − ⊗ LL ⊗ H(L ⊗ H) L + : LA = Mod-L → E B = E (Comod-C) .js

Now we will give details of the isomorphisms associated to the equivalence of categories.Given a right E-module functor F we will denote simply by − ⊗ E F thefunctor defined byLet us consider the functorE ̂Q LL Q E = − ⊗ E H L ⊗ LCoequ Fun(µEF EU, F E Uλ E).L ⊗ H(L ⊗ H) L + : EB = E (Comod-C) → E B = E (Comod-C) .We want to prove that E ̂QLL Q E is functorially isomorphic to Id E B. Now, for any(X, E µ X)∈ E B we have(X, E µ X)⊗E E = Coequ Fun(µEE EU ( X, E µ X), EE Uλ E(X, E µ X))= Coequ Fun(mE X, E E µ X) 3.14= ( X, E µ X).Thus to this aim it is enough to construct an isomorphism of left E-modules ̂β :L⊗HH L ⊗ L → H⊗H . This will imply that β = −□(L⊗H)L + (H⊗H)L + C ̂β : ̂QLL Q = −□ C H L ⊗ LL⊗HH⊗H→ E = −□(L⊗H)L + C gives rise to a functorial isomorphism(H⊗H)L +E ̂QLL Q E ≃Id E B. We want to show that ̂β is the following morphismL ⊗ H(L ⊗ H) L → H ⊗ H+ (H ⊗ H) L +h ⊗ L [x ⊗ h ′ ]b Q↦→ [hx ⊗ h ′ ] E.̂β : H ⊗ LFirst we have to prove that it is a well-defined map. Let us consider the mapβ : H ⊗ L ⊗ H →H ⊗ H(H ⊗ H) L +h ⊗ x ⊗ h ′ ↦→ [hx ⊗ h ′ ] E.For every (x ⊗ h ′ ) · (t− ε H (t) ) ∈ (L ⊗ H) L + we haveβ ( h ⊗ [ (x ⊗ h ′ ) · (t− ε H (t) )]) = β ( h ⊗ xt (1) ⊗ h ′ t (2) − h ⊗ x ⊗ h ′ ε H (t) )so that β factors through β : H ⊗we have= hxt (1) ⊗ h ′ t (2) − hx ⊗ h ′ ε H (t) ∈ (H ⊗ H) L +199L⊗H → H⊗H . Moreover, for every l ∈ L,(L⊗H)L + (H⊗H)L +β (hl ⊗ x ⊗ h ′ ) = [(hl) x ⊗ h ′ ] E= [h (lx) ⊗ h ′ ] E= β (h ⊗ lx ⊗ h ′ )so that β is also L-balanced and gives rise to the map ̂β : H ⊗ LThe inverse of ̂β is given bŷθ :H ⊗ H(H ⊗ H) L → H ⊗ L ⊗ H+ L(L ⊗ H) L +[x ⊗ y] E↦→ x ⊗ L [1 L ⊗ y]b Q.L⊗H→ H⊗H .(L⊗H)L + (H⊗H)L +

200This map is well-defined, in fact, let us consider the map θ : H ⊗ H → H ⊗ LL⊗H(L⊗H)L + defined by settingθ (x ⊗ y) = x ⊗ L [1 L ⊗ y]b Q.For every (h ⊗ g) · (t− ε H (t) ) ∈ (H ⊗ H) L + , we have (h ⊗ g) · (t− ε H (t) ) =ht (1) ⊗ gt (2) − h ⊗ gε H (t) and using that ∆ (L) ⊆ L ⊗ H, we computeht (1) ⊗ L(1L ⊗ gt (2))− h ⊗L(1L ⊗ gε H (t) )= h ⊗ L t (1) · (1L ⊗ gt (2))− h ⊗L(1L ⊗ gε H (t) )= h ⊗ L(t(1) ⊗ gt (2))− h ⊗L(1L ⊗ gε H (t) )= h ⊗ L(t(1) ⊗ gt (2) − 1 L ⊗ gε H (t) )= h ⊗ L((1L ⊗ g) · t − (1 L ⊗ g) ε H (t) )= h ⊗ L((1L ⊗ g) · (t− ε H (t) )) ∈ H ⊗ L (L ⊗ H) L +so that θ factors throughH⊗H(H⊗H)L + → H ⊗ LL⊗H(L⊗H)L +giving rise to the map ̂θ. Wecompute, using definition of ̂Q L = ⊗ L ̂Q and µ A b Q(̂θ ◦ ̂β) (h ⊗ L [x ⊗ h ′ ]b Q)= ̂θ ([hx ⊗ h ′ ] E) = hx ⊗ L [1 L ⊗ h ′ ]b Q= h ⊗ L x · [1 L ⊗ h ′ ]b Q= h ⊗ L [x1 L ⊗ h ′ ]b Q= h ⊗ L [x ⊗ h ′ ]b Qand (̂β ◦ ̂θ)([x ⊗ y] E) = ̂β)(x ⊗ L [1 L ⊗ y]b Q= [x ⊗ y] E.Let us show that ̂β is an isomorphism of left E-modules. Using definition of ̂µ E Q ,(215) i.e. ∑ ∑i j ki,j ⊗ S (h i,j ) g i ∈ H ⊗ L, definition of ̂m E we compute(∑ ∑ [̂βk i,j ⊗ h i,j] )·i jE gi ⊗ L [x ⊗ h ′ ]b Q= ̂β(∑i∑j ki,j S ( h i,j) )g i ⊗ L [x ⊗ h ′ ]b Q= ̂β(∑i∑j ki,j ⊗ L S ( h i,j) )g i · [x ⊗ h ′ ]b Q= ̂β(∑[ (i∑j ki,j ⊗ ) L S hi,jg i x ⊗ h ′] )Q b[∑=i∑j ki,j S ( h i,j) ]g i x ⊗ h ′ = ∑ ∑ [k i,j S ( h i,j) g i x ⊗ h ′]E i jE= ∑ ∑ [k i,j ⊗ h i,j] · [g i x ⊗ h ′] = ∑ ∑ [i jE E k i,j ⊗ h i,j] · ̂β)(g i ⊗i jE L [x ⊗ h ′ ]b Q.Similarly we want to understand the other isomorphism. Given a right L-modulefunctor G we will denote simply by − ⊗ L G the functor defined byCoequ Fun(µLG LU, G L Uλ L).Let us consider the functorL ⊗ HLQ EE ̂QL = − ⊗ L(L ⊗ H) L ⊗ + E H : L A = Mod-L → L A = Mod-L.We want to prove that L Q EE ̂QL is functorially isomorphic to Id L A. Now, for any(X, L µ X)∈ L A we have(X, L µ X)⊗L L = Coequ Fun(µLL LU ( X, L µ X), LL Uλ L(X, L µ X))

201= Coequ Fun(mL X, L L µ X) 3.14= ( X, L µ X).Thus to this aim it is enough to construct an isomorphism of left L-modules ̂ζ :L⊗H⊗(L⊗H)L + E H → L. This will imply that ζ = −⊗ ̂ζ : Q EE ̂Q = −⊗L⊗H⊗(L⊗H)L + E H →− ⊗ L gives rise to a functorial isomorphism L Q EE ̂QL ∼ = IdL A. We want to showthat ̂ζ is the following morphismLet us consider∑iL ⊗ Ĥζ :(L ⊗ H) L ⊗ + E H → L∑ [l i,j ⊗ h i,j] Q b ⊗ E g i ↦→ ∑ ∑i∑ij∑ζ : L ⊗ H□ C H → L∑j li,j ⊗ h i,j ⊗ g i ↦→ ∑ ij li,j S ( h i,j) g i .j li,j S ( h i,j) g iand let us prove that it is well-defined. By (217) we get that ∑ iNow we use that C H is coflat. Let us prove thatζ [ (L ⊗ H) L + □ C H ] = 0.∑j li,j S (h i,j ) g i ∈ L.Let ∑ i zi ⊗ h i ∈ (L ⊗ H) L + □ C H where, for each i, z i ∈ (L ⊗ H) L + . This meansthat there exist elements w i,j ∈ L ⊗ H and elements t i,j ∈ L + such thatz i = ∑ j wi,j · t i,jSince w i,j ∈ L ⊗ H there exist l i,j,k ∈ L and g i,j,k ∈ H such thatw i,j = ∑ k li,j,k ⊗ g i,j,k .Hence we have∑i zi ⊗ h i = ∑ ∑ ∑ (l i,j,k ⊗ g i,j,k) · t i,j ⊗ h ii j k= ∑ ∑ ∑ [l i,j,k t i,ji j k(1) ⊗ gi,j,k t i,j(2) − ( l i,j,k ⊗ g i,j,k) (ε )] H ti,j⊗ h i= ∑ ∑ ∑i j k li,j,k t i,j(1) ⊗ gi,j,k t i,j(2) ⊗ hi − ( l i,j,k ⊗ g i,j,k) (ε ) H ti,j⊗ h iso that)ζ(∑i zi ⊗ h i = ∑ ∑ ∑ ( )i j k li,j,k t i,j(1) S g i,j,k t i,j(2)h i − ( l i,j,k S ( g i,j,k)) (ε ) H ti,jh i= ∑ ∑ ∑ ( )i j k li,j,k t i,j(1) S t i,j(2)S ( g i,j,k) h i − ( l i,j,k S ( g i,j,k)) (ε ) H ti,jh i= ∑ ∑ ∑ (i j k li,j,k ε ) H ti,jS ( g i,j,k) h i − ( l i,j,k S ( g i,j,k)) (ε ) H ti,jh i = 0.L⊗Hhence we have a well defined map ζ : □(L⊗H)L + C H → L defined by setting(∑ ∑ [ζl i,j ⊗ h i,j] )Q b ⊗ g i = ∑ ∑i j li,j S ( h i,j) g i .ij

202We now have to prove that this map induces a map ζ :e ∈ E. We have to prove that(∑ ∑ [ζl i,j ⊗ h i,j] ) (∑ ∑Q b · e ⊗ g i = ζijijL⊗H(L⊗H)L +⊗ E H → L. Let[l i,j ⊗ h i,j] b Q⊗ e · g i ).Since e ∈ E, there exist x k , y k ∈ H such that e = [∑ k xk ⊗ y k] . Hence we have toEprove that(∑ ∑ [ζl i,j ⊗ h i,j] ] )bi jQ ·[∑k xk ⊗ y k ⊗ g iE(∑ ∑ [= ζl i,j ⊗ h i,j] ] )Q b ⊗[∑k xk ⊗ y k · g iand by using definition of E µb Qand µ E Qi.e.∑i∑j∑i(∑ ∑ ∑ζi j(∑ ∑ ∑= ζijk li,j S ( h i,j) x k S ( y k) g i = ∑ ijkkwe have to prove that[l i,j S ( h i,j) x k ⊗ y k] )Q b ⊗ g i[l i,j ⊗ h i,j] b Q⊗ x k S ( y k) g i )∑j∑which is true, so that we can conclude that the map ̂ζ :Ek li,j S ( h i,j) x k S ( y k) g iL⊗H(L⊗H)L +⊗ E H → L iswell-defined. Now, we want to prove that ̂ζ is bijective. The inverse of ̂ζ is given byΞ : L →L ⊗ H(L ⊗ H) L ⊗ + E Hl ↦→ [1 H ⊗ 1 H ]b Q⊗ E l.Now we compute(Ξ ◦ ̂ζ) (∑ ∑ [l i,j ⊗ h i,j] ) (∑bi jQ⊗ E g i = Ξi∑j li,j S ( h i,j) )g i∑ ∑= [1 H ⊗ 1 H ]b Q⊗ Ei j li,j S ( h i,j) g i= ∑ ∑[1 [H ⊗ 1 H ]bi j Q⊗ E l i,j ⊗ h i,j] · E gi = ∑ ∑i= ∑ ∑ [l i,j ⊗ h i,j] bi jQ⊗ E g iand ) ()(̂ζ ◦ Ξ (l) = ζ [1 H ⊗ 1 H ]b Q⊗ E l = 1 H S (1 H ) l = l.j [1 H ⊗ 1 H ]b Q[l i,j ⊗ h i,j] E ⊗ E g iLet us show that ̂ζ is an isomorphism of left L-modules. Let a ∈ L and let usconsider̂ζ(a · ∑ ∑ [l i,j ⊗ h i,j] )bi jQ⊗ E g i = ̂ζ(∑ ∑ [( ) a · li,j⊗ h i,j] )bi jQ⊗ E g i= ∑ ∑i j ali,j S ( h i,j) (∑g i = a ·i∑j li,j S ( h i,j) )g i= a · ̂ζ(∑ ∑ [l i,j ⊗ h i,j] )Q b ⊗ E g i .ij

As observed at the beginning of this section, this reproduces what happens in thedual case of the [Scha4] setting where, starting from a Hopf-Galois extension, onecan produce a new Hopf algebra such that the Hopf-Galois object turns into a Hopfbi-Galois object and Hopf algebras are Morita-Takeuchi equivalent. In our setting,coming from a coGalois coextension we get a coherd, which allows us to computethe monads and in particular a new monad together with the new bimodule functor.Following the theory developed in the previous sections, we could then calculate indetails also the equivalence between the module categories with respects to the twomonads.9.3. Galois comodules. Let B Σ A be a B-A-bimodule. Let L = − ⊗ B Σ A , R =Hom A ( B Σ A , −). Let C be an A-coring and let C = (− ⊗ A C, − ⊗ A ∆, r ◦ (− ⊗ A ε)).Assume that (Σ, ρ Σ ) is a B-C-comodule i.e. (Σ, ρ Σ ) is a C-comodule andρ Σ : Σ → Σ ⊗ A Cis a morphism of B-A-bimodules. In particular the mapλ : B → EndC (Mod-A) ((Σ, ρ Σ )) defined by setting λ (b) (x) = bxis well-defined and is a ring morphism. Moreover λ is a monomorphism. In thiscase β = − ⊗ B ρ Σ : − ⊗ B Σ A → − ⊗ B Σ A ⊗ A C is a left C-comodule functor. Theassociated functorial morphism can = ϕ = (Cɛ) ◦ (βR) : LR → C,can : Hom A ( B Σ A , −) ⊗ B Σ AβR→ HomA ( B Σ A , −) ⊗ B Σ A ⊗ A C Cɛ→ − ⊗ A Cf ⊗ B x ↦→ f ⊗ B x 0 ⊗ A x 1 ↦→ f (x 0 ) ⊗ A x 1can M : Hom A ( B Σ A , M) ⊗ B Σ A → M ⊗ A Cf ⊗ B x ↦→ f (x 0 ) ⊗ A x 1203We havecan = (Cɛ) ◦ (βR) = (ɛ ⊗ A C) ◦ Hom A ( B Σ A , −) ⊗ B ρ Σcan M = ϕ M (f ⊗ B t) = (ɛ ⊗ A C) (f ⊗ B t 0 ⊗ A t 1 ) = f (t 0 ) ⊗ A t 1 .K ϕ : Mod-B → C (Mod-A) = Comod-CM ↦→ (M ⊗ B Σ, M ⊗ B ρ Σ ) .Since Mod-B has all equalizers, K ϕ has a right adjointD ϕ (X, x) = Equ ((− ⊗ A C) ◦ ρ Σ , Hom A ( B Σ A , x))= {f ∈ Hom A ( B Σ A , X) | x ◦ f = (f ⊗ A C) ◦ ρ Σ }= HomC (Mod-A) ((Σ, ρ Σ ) , (X, x))Hence D ϕ = HomC (Mod-A) ((Σ, ρ Σ ) , −) : C (Mod-A) = Comod-C → Mod-B has a leftadjoint K ϕ = (− ⊗ B Σ, − ⊗ B ρ Σ ) .Theorem 9.6 ([GT, Theorem 3.1] ). HomC (Mod-A) ((Σ, ρ Σ ) , −) : C (Mod-A) →Mod-B is full and faithful if and only if

2041) − ⊗ B Σ A preserves the equalizerHomC (Mod-A) ((Σ, ρ Σ ) , (X, x))i Hom A (Σ, X)x◦−(−⊗ A C)◦ρ Σ Hom A (Σ, X ⊗ A C) .2) can : Hom A ( B Σ A , −) ⊗ B Σ A → − ⊗ A C is a comonad isomorphism.Proof. Apply Theorem 4.53 to the adjunction (− ⊗ B Σ A , Hom A ( B Σ A , −)) .Theorem 9.7 ([GT, Theorem 3.2]). K ϕ : Mod-B → C (Mod-A) = Comod-C is anequivalence of categories if and only if1) − ⊗ B Σ A preserves the equalizerHomC (Mod-A) ((Σ, ρ Σ ) , (X, x))i Hom A (Σ, X)x◦−(−⊗ A C)◦ρ Σ Hom A (Σ, X ⊗ A C) .2) − ⊗ B Σ A reflects isomorphisms and3) can : Hom A ( B Σ A , −) ⊗ B Σ A → − ⊗ A C is a comonad isomorphism.Proof. Apply Theorem 4.55 to the adjunction (− ⊗ B Σ A , Hom A ( B Σ A , −)) .Let us now consider a particular case of the previous situation.Let C be an A-coring and let Σ be a right C-comodule. Set T = EndC (Mod-A) ((Σ, ρ Σ )).Then it is easy to check that (Σ, ρ Σ ) is a T -C-comodule. Following [Wis], we saythat Σ is a Galois C-comodule whenever can : Hom A ( T Σ A , −) ⊗ T Σ → − ⊗ A C isan isomorphism. The adjunction (C U, C F ) for C = (− ⊗ A C, − ⊗ A ∆, r ◦ (− ⊗ A ε))gives us the followingProposition 9.8. Let C be an A-coring and let Σ be a right C-comodule.T = EndC (Mod-A) ((Σ, ρ Σ )). Then the mapψL : Hom A ( T Σ A , L) → HomC (Mod-A)((Σ, ρΣ ) , C F L ) defined by settingψL (f) = (f ⊗ A C) ◦ ρ Σis an isomorphism whose inverse is defined by setting (ψL) −1 (h) = r L ◦ (L ⊗ A ε) ◦ h, for every L ∈ Mod-A. In this way we get a functorial isomorphismψ : Hom A ( T Σ A , −) → HomC (Mod-A)((Σ, ρΣ ) , C F ) .9.9. Note that, in particular, we have(ψA : Hom A ( T Σ A , A) → HomC (Mod-A) (Σ, ρΣ ) , C F A )whereso thatHomC (Mod-A)((Σ, ρΣ ) , C F A ) = HomC (Mod-A) ((Σ, ρ Σ ) , A ⊗ A C)and is defined by setting≃ HomC (Mod-A) ((Σ, ρ Σ ) , C)ψA : Hom A ( T Σ A , A) → HomC (Mod-A) ((Σ, ρ Σ ) , C)[ψA (f)] (t) = [l C ◦ (f ⊗ A C) ◦ ρ Σ ] (t) = f (t 0 ) t 1 .□□Set

Theorem 9.10 ([GT]). Let C be an A-coring and let Σ be a right C-comodule.Assume that A C is flat. Set T = EndC (Mod-A) ((Σ, ρ Σ )). Then the following areequivalent:(a) The functor HomC (Mod-A) ((Σ, ρ Σ ) , −) : C (Mod-A) → Mod-T is full andfaithful where C = −⊗ A C.(b) ɛ : HomC (Mod-A) ((Σ, ρ Σ ) , − ⊗ T Σ) → C (Mod-A) is an isomorphism.(c) (Σ, ρ Σ ) is a generator of C (Mod-A).(d) can : HomC (Mod-A) ((Σ, ρ Σ ) , −) ⊗ T Σ → − ⊗ A C is an isomorphism and T Σ isflat.Proof. By Proposition A.12, A C is flat if and only if (Mod-A) C is a Grothendieckcategory and the forgetful functor U : (Mod-A) C → Mod-A is left exact. Also,by the foregoing, D ϕ = HomC (Mod-A) ((Σ, ρ Σ ) , −) : C (Mod-A) → Mod-T has a leftadjoint K ϕ = (− ⊗ T Σ, − ⊗ T ρ Σ ) .(a) ⇔ (b) It follows by Proposition <strong>2.</strong>3<strong>2.</strong>(a) ⇔ (c) It follows by Proposition A.3.(c) ⇒ (d) Since (Σ, ρ Σ ) is a generator of C (Mod-A) and since (− ⊗ T Σ, − ⊗ T ρ Σ ) :Mod-T → C (Mod-A) is a left adjoint of HomC (Mod-A) ((Σ, ρ Σ ) , −) : C (Mod-A) →Mod-T, by Gabriel-Popescu Theorem A.9, (− ⊗ T Σ, − ⊗ T ρ Σ ) is a left exact functor.Since the forgetful functor U : (Mod-A) C → Mod-A is also left exact, wededuce that − ⊗ T Σ : Mod-T → Mod-A is left exact i.e. T Σ is flat. SinceHomC (Mod-A) ((Σ, ρ Σ ) , −) is full and faithful, by Theorem 9.6, can is an isomorphism.(d) ⇒ (a) It follows by Theorem 9.6.□Theorem 9.11 ([GT]). Let C be an A-coring, let B be a ring and assume that A Cis flat. Let (Σ, ρ Σ ) be a B-C-comodule. Then the following are equivalent:(a) The functor − ⊗ B Σ A : Mod-B → C (Mod-A) is an equivalence of categorieswhere C = − ⊗ A C.(b) can : Hom A ( B Σ A , −) → − ⊗ A C is an isomorphism and B Σ is faithfully flat.(c) (Σ, ρ Σ ) is a generator of C (Mod-A) and the functor − ⊗ B Σ : Mod-B →C (Mod-A) is full and faithful.(d) (Σ, ρ Σ ) is a generator of C (Mod-A), the functor − ⊗ B Σ : Mod-B →C (Mod-A) is faithful and λ : B → T = EndC (Mod-A) ((Σ, ρ Σ )) is an isomorphism.Proof. (a) ⇒ (b) By Theorem 9.7, can is an isomorphism. Since A C is flat, byProposition A.12, the forgetful functor U : C (Mod-A) → Mod-A is exact. Since Uis also faithful, we get that the functor− ⊗ B Σ A : Mod-B → Mod-Ais faithful and exact.(b) ⇒ (a) It follows by Theorem 9.7.(a) ⇒ (c) Since B is a generator of Mod-B, B Σ ≃ B ⊗ B Σ is a generator ofC (Mod-A).(c) ⇒ (d) Since − ⊗ B Σ : Mod-B → C (Mod-A) is full and faithful and it isthe left adjoint of the adjunction ( − ⊗ B Σ, HomC (Mod-A) ((Σ, ρ Σ ) , −) ) , the unit is a205

206functorial isomorphism. In particular we have thatηB : B → HomC (Mod-A) ((Σ, ρ Σ ) , (B ⊗ B Σ, ρ Σ )) ≃ EndC (Mod-A) ((Σ, ρ Σ ))is an isomorphism. Note that ηB is exactly λ.(d) ⇒ (b) By Theorem 9.10, can : HomC (Mod-A) ((Σ, ρ Σ ) , −) ⊗ B Σ → − ⊗ A C is anisomorphism and B Σ is flat. Since − ⊗ B Σ : Mod-B → C (Mod-A) is faithful andU : C (Mod-A) → Mod-A is also faithful, the functor −⊗ B Σ A = U (− ⊗ B Σ) : Mod-B → (Mod-A) is faithful. Then B Σ is faithfully flat.□Remark 9.1<strong>2.</strong> By Theorem 9.11 we deduce that if −⊗ B Σ A : Mod-B → C (Mod-A)is an equivalence of categories then Σ is a Galois C-comodule.9.13. Let B Σ A be a B-A-bimodule. In the case that Σ A is finitely generated andprojective we have a natural isomorphismΛ : Hom A (Σ, −) → − ⊗ A Σ ∗f ↦→ f (x i ) ⊗ A x ∗ iwhere Σ ∗ = Hom A (Σ, A) and (x i , x ∗ i ) i=1,...,nis a dual basis for Σ A . We can considerthe adjunction (− ⊗ B Σ A , − ⊗ A Σ ∗ ) and the associated comonad − ⊗ A Σ ∗ ⊗ B Σ A :Mod-A → Mod-A, then the A-coring Σ ∗ ⊗ B Σ A is called the comatrix coring associatedto the bimodule B Σ A . Moreover, when (Σ, ρ Σ ) is a B-C-comodule then wehave the following commutative diagramΛ⊗ B Σ(218) Hom A (Σ, −) ⊗ B Σ− ⊗ A Σ ∗ ⊗ B Σcan−⊗A can− ⊗ A Cwherecan : Σ ∗ ⊗ B Σ → Cdefined by settingcan (φ ⊗ B s) = φ (s 0 ) s 1is a morphism of A-corings where Σ ∗ ⊗ B Σ is an A-coring via comultiplication∆ (ϕ ⊗ B t) = ∑ ϕ ⊗ B x i ⊗ A x ∗ i ⊗ B t and counit ε (ϕ ⊗ B t) = ϕ (t) .iRemark 9.14. Following [BrWi, pag 189] we say that Σ is a Galois C-comodulewhen Σ A is finitely generated and projective and ev : HomC (Mod-A) (Σ, C) ⊗ B Σ → Cis an isomorphism.Corollary 9.15 ([GT]). Let C be an A-coring, let B be a ring and let (Σ, ρ Σ ) be aB-C-bicomodule. Then the following are equivalent:(a) A C is flat and the functor − ⊗ B Σ : Mod-B → C (Mod-A) is an equivalenceof categories where C = − ⊗ A C(b) Σ A is finitely generated and projective, the canonical map can : Σ ∗ ⊗ B Σ → Cis an isomorphism and B Σ is faithfully flat(c) A C is flat, Σ is a finitely generated projective generator of C (Mod-A) andλ : B → T = EndC (Mod-A) ((Σ, ρ Σ )) is an isomorphism.

Proof. (a) ⇒ (c) Apply Proposition A.19 to the functor T = − ⊗ B Σ. Since B is afinitely generated and projective generator of Mod-B, Σ C ≃ B ⊗ B Σ C is a finitelygenerated and projective generator of C (Mod-A). By the equivalence (a) ⇔ (d) ofTheorem 9.11 we get that λ : B → T = EndC (Mod-A) ((Σ, ρ Σ )) is an isomorphism.(c) ⇒ (b) Let us consider U : C (Mod-A) → Mod-A which is the left adjointof the free functor − ⊗ A C : Mod-A → C (Mod-A). We have to prove that Σ A isfinitely generated and projective. Now, by Proposition A.18, we prove that Σ A isfinite, i.e. that Hom Mod-A (Σ A , −) preserves coproducts. Let us consider a family(A i ) i∈I∈ Mod-A. We have the followingΣfinite≃∐i∈IHomC (Mod-A)Hom Mod-A (U (Σ) , A i ) (U,−⊗ AC)adj≃(Σ, ∐ i∈I(A i ⊗ A C))∐i∈I−⊗ A Cright adj≃HomC (Mod-A) (Σ, A i ⊗ A C)HomC (Mod-A)((U,−⊗ A C)adj≃ Hom Mod-A U (Σ) , ∐ )A ii∈I(Σ,( ∐i∈IA i)207⊗ A C)Since Σ A = U (Σ) we deduce that Hom Mod-A (Σ A , −) preserves coproducts. Since byassumption A C is flat, by Theorem 9.10 (c) ⇒ (d) we get thatcan : HomC (Mod-A) ((Σ, ρ Σ ) , −)⊗ B Σ → −⊗ A C is an isomorphism and B Σ is flat. Bydiagram 218 we obtain that can is also an isomorphism. Since Σ is a finitely generatedprojective generator of C (Mod-A), by Corollary A.21 HomC (Mod-A) ((Σ, ρ Σ ) , −)is an equivalence of categories, hence so is − ⊗ B Σ : Mod-B → C (Mod-A) so thatBΣ is faithfully flat.(b) ⇒ (a) Since can is an isomorphism, we have that A C is flat if and only ifAΣ ∗ ⊗ B Σ is flat. By assumption we know that B Σ is flat. Since Σ A is finitelygenerated and projective, also A Σ ∗ is finitely generated and projective so A Σ ∗ is flat.Therefore the functor − ⊗ A Σ ∗ ⊗ B Σ is left exact and, since can is an isomorphism,−⊗ A C is also left exact. By diagram 218, since can is an isomorphism, can is also anisomorphism. Now, A C is flat and B Σ is faithfully flat, then we can apply Theorem9.11 (b) ⇒ (a) to deduce that − ⊗ B Σ : Mod-B → C (Mod-A) is an equivalence ofcategories.□Remark 9.16. By Corollary 9.15 we deduce that if A C is flat and − ⊗ B Σ A : Mod-B → C (Mod-A) is an equivalence of categories, then Σ is a Galois C-comodule.Corollary 9.17 ([GT, Theorem 3.10] Generalized Descent for Modules). Let B Σ Abe a B-A-bimodule such that Σ A is finitely generated and projective. Let Σ ∗ =Hom A (Σ, A) . Then the following are equivalent:(a) A (Σ ∗ ⊗ B Σ) is flat and the functor − ⊗ B Σ : Mod-B → C (Mod-A) is anequivalence of categories where C = − ⊗ A Σ ∗ ⊗ B Σ(b) B Σ is faithfully flat.Proof. By (9.13) we have that Σ ∗ ⊗ B Σ is an A-coring and thus C = − ⊗ A Σ ∗ ⊗ B Σis a comonad on Mod-A. Note that Σ is a B-C-bicomodule via a canonical right

208coaction ρ C Σ : Σ → Σ ⊗ A Σ ∗ ⊗ B Σ defined by setting ρ C Σ (s) = ∑ n x i ⊗ A x ∗ i ⊗ B s where(x i , x ∗ i ) i=1,...,nis a dual basis for Σ A . Then we can apply Corollary 9.15 (a) ⇔ (b) tothe case ”C” = Σ ∗ ⊗ B Σ so that can : Σ ∗ ⊗ B Σ → C is the identity map. □9.18. Let B → A a k-algebra extension. Let ” B Σ A ” = B A A in 9.13. Then thecomatrix coring becomes C = A ⊗ B A which is an A-coring with coproduct ∆ C : C =A ⊗ B A → C ⊗ A C = A ⊗ B A ⊗ A A ⊗ B A defined by setting∆ C (a ⊗ B a ′ ) = a ⊗ B 1 A ⊗ A 1 A ⊗ B a ′and counit ε C : C = A ⊗ B A → A defined by settingε C (a ⊗ B a ′ ) = aa ′for every a, a ′ ∈ A. Such A-coring C = A ⊗ B A is called canonical coring or Sweedlercoring associated to the algebra extension B → A.Definition 9.19. Let B be a k-algebra and let B → σ A be an algebra extension.A right descent datum from A to B is a right A-module M together with a rightA-module morphism δ : M → M ⊗ B A such that(219) (δ ⊗ B A) ◦ δ = (M ⊗ B σ ⊗ B A) ◦ ( M ⊗ B l −1Aand(220) µ M ◦ δ = Mi=1)◦ δwhere l −1A: B ⊗ B A → A is the canonical isomorphism and µ M : M ⊗ B A → M isinduced by the A-module structure of M, µ A M : M ⊗ A → A. Given (M, δ) , (M ′ , δ ′ )two right descent data from A to B, a morphism of right descent data from A to Bis a right A-module map f : M → M ′ such thatδ ′ ◦ f = (f ⊗ B A) ◦ δ.We will denote by D (A ↓ B) the category of right descent data. Similarly one candefine left descent data from A to B and their category (A ↓ B) D.Let A = (A, m A , u A ) be a monad on a category A. Then we can consider theadjunction ( A F , A U) , where A F : A → A A and A U : AA → A, with unit u Awhich is the unit of the monad and counit λ A determined by A U ( λ A(X, A µ X))=A µ X for every ( X, A µ X)∈ A A. Then A F A U is a comonad on the category A A byProposition 4.4. Hence we can consider the category of comodules for the comonadC = A F A U, A F A U ( A A) = C ( A A) which is the category of descent data with respect tothe monad A and it is denoted by Des A (A) .Example 9.20. Let B σ → A be a k-algebra extension. Then A is a B-ring. In factm A : A ⊗ A → A induces m : A ⊗ B A → A as follows. We have to prove thatm A (ab ⊗ a ′ ) = m A (a ⊗ ba ′ ) . We computem A (ab ⊗ a ′ ) = m A (aσ (b) ⊗ a ′ ) = (aσ (b)) a ′= a (σ (b) a ′ ) = m A (a ⊗ σ (b) a ′ ) = m A (a ⊗ ba ′ ) .

Moreover the unit is u = σ : B → A. Then A = ( )− ⊗ B A, − ⊗ B m, (− ⊗ R u) ◦ r−−1is a monad on the category of right B-modules, Mod-B, as in Example 3.3. Notethat we have an iso of categories K : Mod-A → A (Mod-B) given byMod-A −→ A (Mod-B)( )X, µAX ↦→( )X, µ A Xwhere µ A X : X ⊗ B A → X is well-defined starting from µ A X : X ⊗ A → X. In fact wehave(µ A X (xb ⊗ a) = µ A X µAX (x ⊗ σ (b)) ⊗ a )XisA-mod= µ A X (x ⊗ m A (σ (b) ⊗ a)) = µ A X (x ⊗ ba) .Now, since A = ( )− ⊗ B A, − ⊗ B m, (− ⊗ R u) ◦ r−−1 is a monad, we can considerAF = − ⊗ B A : Mod-B → A (Mod-B) ≃ Mod-A and A U = − ⊗ A A B : A (Mod-B) ≃Mod-A → Mod-B so that C = A F A U = −⊗ A A⊗ B A is a comonad on A (Mod-B) ≃Mod-A associated to the A-coring C = A ⊗ B A. The category of comodules for thecomonad C = A F A U = − ⊗ A A ⊗ B A is then the category of right comodules for theA-coring C = A ⊗ B AAF A U ( A (Mod-B)) = C ( A (Mod-B)) = ( A (Mod-B)) C ≃ (Mod-A) C= C (Mod-A) = A F A U (Mod-A)and it is the category of right descent data from A to B, usually denoted byD (A ↓ B) .Corollary 9.21 ([Scha4, Theorem 4.5.2] Faithfully flat descent). Let A be a k-algebra and let B ⊆ A be a k-algebra extension. Let C = A ⊗ B A be the canonicalA-coring. The following statements are equivalent:(a) B A is flat and the functor − ⊗ B A : Mod-B → D (A ↓ B) is an equivalenceof categories;(b) B A is faithfully flat.Proof. Apply Corollary 9.17 to the case ” B Σ A ” = B A A , noting that by Example9.20 C (Mod-A) = D (A ↓ B) where C = − ⊗ A C = − ⊗ A A ⊗ B A. □Remark 9.2<strong>2.</strong> The inverse equivalence of the induction functor − ⊗ B A : Mod-B → D (A ↓ B) = C (Mod-A) where C = − ⊗ A C = − ⊗ A A ⊗ B A, maps a descentdatum (M, δ) into M coδ = {m ∈ M | δ (m) = m ⊗ B 1 A } ≃ M coC . Moreover, sincewe have an equivalence, in particular the counit is an isomorphism, so that the mapis an isomorphism with inverse given byM coδ ⊗ B A ɛ M−→ Mm ⊗ B a ↦→ maM → M coδ ⊗ B Am ↦→ δ (m) .In fact we have [(δ ⊗ B A) ◦ δ] (m) = m 00 ⊗ B m 01 ⊗ B m 1 = m 0 ⊗ B 1 A ⊗ B m 1 so thatδ (m) ∈ M coδ ⊗ B A.209

210Now, we consider a particular case of the setting investigated above.Lemma 9.23. Let C be an A-coring. Then A can be endowed with a right C-comodulestructure ρ C A if and only if C has a grouplike element, namely [( l A C ◦ ρC A)(1A ) ] .Proof. Assume first that A has a right C-comodule structure given by ρ C A . We wantto prove that g = [( lC A ◦ A) ρC (1A ) ] is a grouplike element for C. First, fromg = [( )lC A ◦ ρ C A (1A ) ]we deduce that(221) ρ C A (1 A ) = ( l A C) −1(g) = 1A ⊗ A gLet us compute∆ (( C lC A ◦ ρA) C (1A ) ) = ( ∆ C ◦ lC A ◦ ρA) C (1A ) = [( lC A ⊗ A C ) ◦ ( A ⊗ A ∆ C) ◦ ρA] C (1A )[(lAC ⊗ A C ) ◦ ( ρ C A ⊗ A C ) ◦ ρA] C (1A ) = [( lC A ⊗ A C ) ◦ ( ρ C A ⊗ A C )] ( ρ C A (1 A ) )ArightC-com=Moreover(221)= [( l A C ⊗ A C ) ◦ ( ρ C A ⊗ A C )] (1 A ⊗ A g) = [( l A C ◦ ρ C A)⊗A C ] (1 A ⊗ A g)= ( l A C ◦ ρ C A)(1A ) ⊗ A g = g ⊗ A g.ε (( C lC A ◦ ρA) C (1A ) ) = ( ε C ◦ lC A ◦ ρA) C (1A )= [ l A ◦ ( A ⊗ A ε C) ]◦ ρ C A (1A ) ArightCcom= 1 A .Conversely, let us assume that g ∈ C is a grouplike element and let us define ρ C A :A → A ⊗ A C by settingρ C A (a) = 1 A ⊗ A g · a.We have to check that it defines a C-comodule structure on A. We compute, forevery a ∈ A,[(A ⊗A ∆ C) ◦ ρ C A](a) =(A ⊗A ∆ C) (1 A ⊗ A g · a) ∆C Alin= 1 A ⊗ A g ⊗ A g · aso that= ( ρ C A ⊗ A C ) (1 A ⊗ A g · a) = [( ρ C A ⊗ A C ) ◦ ρ C A](a)(A ⊗A ∆ C) ◦ ρ C A = ( ρ C A ⊗ A C ) ◦ ρ C A.We also have, for every a ∈ A,[rAA ◦ ( A ⊗ A ε C) ] [◦ ρ C A (a) = rAA ◦ ( A ⊗ A ε C)] (1 A ⊗ A g · a)so that= r A A(1A ⊗ A ε C (g · a) ) ε C Alin= 1 A ε C (g) a = ar A A ◦ ( A ⊗ A ε C) ◦ ρ C A = Id A .□

Let C be an A-coring and assume that g ∈ C is a grouplike element. Then we canconsider the map ρ A : A → A ⊗ A C defined by settingρ A (a) = 1 A ⊗ A (g · a) for every a ∈ A.We denote by C = − ⊗ A C. Then, by Lemma 9.23, (A, ρ A ) is a right C-comoduleandEndC (Mod-A) ((A, ρ A )) ≃ {b ∈ A | 1 A ⊗ A (g · b) = b ⊗ A g}= {b ∈ A | 1 A ⊗ A (g · b) = 1 A ⊗ A bg} = {b ∈ A | g · b = bg} = A coC .In this case the mapcan : Σ ∗ ⊗ B Σ = A ⊗ B A → Cis defined by settingcan (a ⊗ B a ′ ) = aga ′and C is called a Galois coring iff can is an isomorphism and B = A coC .Proposition 9.24. Let C be an A-coring and assume that g ∈ C is a grouplikeelement. Let B ⊆ A coC . Then the following statements are equivalent:(a) A C is flat and the functor − ⊗ B A : Mod-B → C (Mod-A) = (Mod-A) C is anequivalence of categories;(b) the canonical map can : A ⊗ B A → C is an isomorphism and B A is faithfullyflat;(c) A C is flat, A is a finitely generated projective generator of C (Mod-A) andλ : B → T = EndC (Mod-A) ((A, ρ A )) = A coC is an isomorphism.Theorem 9.25. [BRZ2002, Theorem 5.6]Let C be an A-coring and assume thatg ∈ C is a grouplike element.1) If C is a Galois coring and A coCA is faithfully flat, then the functor −⊗ A coC A :Mod-A coC → C (Mod-A) = (Mod-A) C is an equivalence of categories and A Cis flat.2) If the functor − ⊗ A coC A : Mod-A coC → C (Mod-A) is an equivalence ofcategories, then C is a Galois coring.3) If A C is flat and the functor − ⊗ A coC A : Mod-A coC → C (Mod-A) is anequivalence of categories, then A coCA is faithfully flat.Proof. 1) follows from Proposition 9.24 (b) ⇒ (a).2) follows from Theorem 9.7.3) follows from Proposition 9.24 (a) ⇒ (b). □Corollary 9.26. Let C be an A-coring and assume that g ∈ C is a grouplikeelement. Assume that A C is flat. Let B ⊆ A coC . Then the following statements areequivalent:(a) the functor − ⊗ B A : Mod-B → C (Mod-A) = (Mod-A) C is an equivalenceof categories;(b) the canonical map can : A ⊗ B A → C is an isomorphism and B A is faithfullyflat;(c) A is a finitely generated projective generator of C (Mod-A) and λ : B → T =EndC (Mod-A) ((A, ρ A )) = A coC is an isomorphism.211

212Definition 9.27. Let k be a commutative ring. An entwining structure (A, C, ψ)over k consists of• A = (A, m, u) a k-algebra• C = (C, ∆, ε) a k-coalgebra• ψ : C ⊗ A → A ⊗ C satisfying the following relations(222)(m ⊗ C)◦(A ⊗ ψ)◦(ψ ⊗ A) = ψ ◦(C ⊗ m) and ψ ◦(C ⊗ u)◦r −1Cand= (u ⊗ C)◦l−1C(223)(ψ ⊗ C) ◦ (C ⊗ ψ) ◦ (∆ ⊗ A) = (A ⊗ ∆) ◦ ψ and r A ◦ (A ⊗ ε) ◦ ψ = l A ◦ (ε ⊗ A) .Notation 9.28. Let (A, C, ψ) be an entwining structure over k. We will use sigmanotationψ (c ⊗ a) = ∑ a α ⊗ c αor with summation understoodψ (c ⊗ a) = a α ⊗ c α .Using this notation we can rewrite (222) and (223) as follows(224)(225)(ab) α⊗ c α = a α b β ⊗ c αβ , ψ (c ⊗ 1 A ) = (1 A ) α⊗ c α = 1 A ⊗ ca α ⊗ c α 1 ⊗ c α 2 = a αβ ⊗ c β 1 ⊗ c α 2 , a α ε C (c α ) = ε C (c) aMoreover we set, for every a, b, a ′ , b ′ ∈ A and c ∈ Ca (b ⊗ c) = ab ⊗ c and (b ⊗ c) b ′ = bψ (c ⊗ b ′ ) = bb ′ α ⊗ c α .We also define a map ∆ C : C = A ⊗ C → C ⊗ A C = A ⊗ C ⊗ A A ⊗ C, by settingand a map ε C : C → A, as follows∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2)ε C (a ⊗ c) = aε (c) .Definition 9.29. Let (A, C, ψ) be an entwining structure. An entwined (A, C, ψ)-module is a triple ( M, µ A M , ρC M)where(M, µAM)is a right A-module,(M, ρCM)is aright C-comodule such that the structures are compatible(µAM ⊗ C ) ◦ (M ⊗ ψ) ◦ ( ρ C M ⊗ A ) = ρ C M ◦ µ A Mi.e. for every m ∈ M and for every a ∈ A we have∑(226)(ma)0 ⊗ (ma) 1= ∑ m 0 a α ⊗ m α 1 .A morphism of entwined modules f : ( M, µ A M , (M) ρC → N, µAN , ρN) C is a morphism ofright A-modules and a morphism of right C-comodules. We denote by M C A (ψ) thecategory of entwined (A, C, ψ)-modules.Proposition 9.30 ([BrWi, 3<strong>2.</strong>6 pg. 325]). Let k be a commutative ring, let A =(A, m, u) be a k-algebra and let C = (C, ∆, ε) be a k-coalgebra.1) If (A, C, ψ) is an entwining structure, then, using the notations introducedin (9.28), ( C = A ⊗ C, ∆ C , ε C) is an A-coring that will be called the A-coringassociated to the entwining (A, C, ψ) .

2) If A ⊗ C is an A-coring then (A, C, ψ) is an entwining structure whereψ (c ⊗ a) = (1 A ⊗ c) · a.3) If C = A ⊗ C is the A-coring associated to the entwining (A, C, ψ), thenM C A = (Mod-A)C ≃ M C A (ψ).Proof. 1) Let us define the A-bimodule structures on C = A ⊗ C. Set, for everya, b, a ′ , b ′ ∈ A and c ∈ Ca (b ⊗ c) = ab ⊗ c and (b ⊗ c) b ′ = bψ (c ⊗ b ′ ) = bb ′ α ⊗ c αi.e.a ′ (b ⊗ c) b ′ = a ′ bψ (c ⊗ b ′ ) = a ′ bb ′ α ⊗ c α .We check the right module structure. Let us compute(a ⊗ c) (bb ′ ) = aψ (c ⊗ bb ′ ) = a [ψ (C ⊗ m) (c ⊗ b ⊗ b ′ )]ψentw= a [((m ⊗ C) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)) (c ⊗ b ⊗ b ′ )][= a [((m ⊗ C) ◦ (A ⊗ ψ)) (b α ⊗ c α ⊗ b ′ )] = a (m ⊗ C)(b α ⊗ b ′ β ⊗ (c α ) β)](= a b α b ′ β ⊗ (c α ) β) = ab α b ′ β ⊗ (c α ) β = (ab α ⊗ c α ) b ′ = ((a ⊗ c) b) b ′ .Let us calculate(a ⊗ c) 1 A = aψ (c ⊗ 1 A ) = a [(ψ ◦ (C ⊗ u)) (c ⊗ 1 k )]= a [( ) ]ψ ◦ (C ⊗ u) ◦ r −1 ψentwC (c) = a [( ) ](u ⊗ C) ◦ l −1C (c)= a [(u ⊗ C) (1 k ⊗ c)] = a (1 A ⊗ c) = a ⊗ c.Now, let us check that it is a bimodule(a ′ (a ⊗ c)) b ′ = a ′ aψ (c ⊗ b ′ ) = a ′ (aψ (c ⊗ b ′ )) = a ′ ((a ⊗ c) b ′ ) .We define the coproduct on C = A⊗C, ∆ C : C = A⊗C → C ⊗ A C = A⊗C ⊗ A A⊗C,by setting∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2)where we denote ∆ (c) = c (1) ⊗ c (2) . It is straightforward to check that it is leftA-linear. Let us check it is also right A-linear. Let us compute∆ C ((a ⊗ c) b ′ ) = ∆ C (aψ (c ⊗ b ′ )) = ∆ C (ab ′ α ⊗ c α )(225)= ab ′ α ⊗ c α (1) ⊗ A 1 A ⊗ c α (2) = a (b ′ α) β⊗ c β (1) ⊗ A 1 A ⊗ c α (2)= aψ ( c (1) ⊗ b α) ′ ⊗A 1 A ⊗ c α (224)(2) = aψ ( )c (1) ⊗ b ′ α ⊗A ψ ( )c α (2) ⊗ 1 A= ( )a ⊗ c (1) b′α ⊗ A ψ ( ) ( )c α (2) ⊗ 1 A = a ⊗ c(1) ⊗A b ′ αψ ( )c α (2) ⊗ 1 A= a ⊗ c (1) ⊗ A(b′α ⊗ c α (2))1A = a ⊗ c (1) ⊗ A b ′ α ⊗ c α (2)= a ⊗ c (1) ⊗ A ψ ( c (2) ⊗ b ′) = a ⊗ c (1) ⊗ A ψ ( c (2) ⊗ b ′)= a ⊗ c (1) ⊗ A(1A ⊗ c (2))b ′ = ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))b ′ = ( ∆ C (a ⊗ c) ) b ′Let us check the coassociativity(∆ C ⊗ C ) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= a ⊗ c(1)(1) ⊗ A 1 A ⊗ c (1)(2) ⊗ A 1 A ⊗ c (2)213

214∆coass= a ⊗ c (1) ⊗ A 1 A ⊗ c (2)(1) ⊗ A 1 A ⊗ c (2)(2)= ( C ⊗ ∆ C) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2)).We define the counit of C, ε C : C → A, as followsε C (a ⊗ c) = aε (c) .It is straightforward to check that ε C is left A-linear. Let us check it is also rightA-linear. Let us computeε C ((a ⊗ c) b ′ ) = ε C (aψ (c ⊗ b ′ )) = ε C (ab ′ α ⊗ c α ) = ab ′ αε (c α )(225)= aε (c) b ′ = ( ε C (a ⊗ c) ) b ′ .Let now check the counitality(rC ◦ ( C ⊗ ε C) ◦ ∆ C) (a ⊗ c) = ( r C ◦ ( C ⊗ ε C)) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= r C(a ⊗ c(1) ⊗ A ε ( c (2)))= a ⊗ cand similarly(lC ◦ ( ε C ⊗ C ) ◦ ∆ C) (a ⊗ c) = ( l C ◦ ( ε C ⊗ C )) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= l C(aε(c(1))⊗A 1 A ⊗ c (2))= a ⊗ cthe right counitality is proved.2) Assume that A ⊗ C is an A-coring with the coproduct and counit as above, i.e.∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2) and ε C (a ⊗ c) = aε (c) .Let us setψ (c ⊗ a) = (1 A ⊗ c) · a.We want to prove that ψ is an entwining for A and C. Since A ⊗ C is an A-coring,it is in particular a right A-module, so thati.e.((a ⊗ c) · a ′ ) · b ′ = (a ⊗ c) · (a ′ b ′ )(227) aa ′ αb ′ β ⊗ c αβ = a (a ′ b ′ ) α⊗ c αLet us compute, for every a, b ∈ A and c ∈ C[(m ⊗ C) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)] (c ⊗ a ⊗ b)= [(m ⊗ C) ◦ (A ⊗ ψ)] (ψ (c ⊗ a) ⊗ b)= [(m ⊗ C) ◦ (A ⊗ ψ)] (a α ⊗ c α ⊗ b) = (m ⊗ C)(a α ⊗ b β ⊗ (c α ) β)= a α b β ⊗ (c α ) β (227)= (ab) α⊗ c α= ψ (c ⊗ ab) = [ψ ◦ (C ⊗ m)] (c ⊗ a ⊗ b)and[ ]ψ ◦ (C ⊗ u) ◦ r−1C (c) = [ψ ◦ (C ⊗ u)] (c ⊗ 1k ) = ψ (c ⊗ 1 A )A⊗CrightAmod= (1 A ⊗ c) · 1 A = 1 A ⊗ c = (u ⊗ C) (1 k ⊗ c) = [ ](u ⊗ C) ◦ l −1 (c) .C

On the other hand, ∆ C and ε C are A-bilinear maps and in particular right A-modulemap so that we havei.e.∆ C ((a ⊗ c) b ′ ) = ( ∆ C (a ⊗ c) ) b ′ and ε C ((a ⊗ c) b ′ ) = ( ε C (a ⊗ c) ) b ′(228) ab ′ α ⊗ c α (1) ⊗ A 1 A ⊗ c α (2) = a ⊗ c (1) ⊗ A b ′ α ⊗ c α (2)and(229) ab ′ αε (c α ) = aε (c) b ′ .Then, for every c ∈ C and a ∈ A, we haveand[(A ⊗ ∆) ◦ ψ] (c ⊗ a) = (A ⊗ ∆) (a α ⊗ c α ) = a α ⊗ c α (1) ⊗ c α (2)(228)= 1 A ⊗ c (1) a α ⊗ c α (2) = ψ ( )c (1) ⊗ a α ⊗ cα(2)= (ψ ⊗ C) ( ) (c (1) ⊗ a α ⊗ c α (2) = [(ψ ⊗ C) ◦ (C ⊗ ψ)] c(1) ⊗ c (2) ⊗ a )= [(ψ ⊗ C) ◦ (C ⊗ ψ) ◦ (∆ ⊗ A)] (c ⊗ a)[r A ◦ (A ⊗ ε) ◦ ψ] (c ⊗ a) = [r A ◦ (A ⊗ ε)] (a α ⊗ c α ) = a α ε (c α )(229)= ε (c) a = [l A ◦ (ε ⊗ A)] (c ⊗ a) .3) Let M ∈ M C A (ψ) , that is ρC M is a right A-module map where A ⊗ C has a rightA-module structure given by(a ⊗ c) b ′ = aψ (c ⊗ b ′ ) = ab ′ α ⊗ c α .Since ρ C M is a right A-module map, then the comodule structure given by the compositeρ C M : M ρC M−→ M ⊗ C ≃ M ⊗ A A ⊗ C = M ⊗ A Cis a right A-module map and thus ( ( )M, ρM) C is a right C-comodule. Conversely, letM, ρCM be a right C-comodule, then we can consider215ρ C M : M ρC M−→ M ⊗ A C = M ⊗ A A ⊗ C ≃ M ⊗ Cas a right A-module map and thus we can see M as a (A, C, ψ)-entwined module.In fact, (226) just means that the map ρ C M is a right A-module map.□Theorem 9.31 ([SS, Lemma <strong>1.</strong>7]). Let C be a k-coalgebra and let A be a k-algebrasuch that (A, C, ψ) is an entwining structure. Then C = A ⊗ C is an A-coring.Assume that A C is flat (i.e. C is k-flat) and that ( A, m, ρA) C ∈ MCA (ψ). Let B =A coC . Then the following statements are equivalent:(a) the functor − ⊗ B A : Mod-B → M C A (ψ) is an equivalence of categories;(b) the canonical map can : A ⊗ B A → A ⊗ C is an isomorphism (i.e. B ⊆ A isa C-Galois extension) and B A is faithfully flat.Proof. By Proposition 9.30 we know that M C A = (Mod-A)C ≃ M C A (ψ) . By hypothesis( A, m, ρA) C ∈ MCA (ψ) and thus, by Lemma 9.23, C = A ⊗ C has a grouplikeelement, that is ρ C A (1 A) . Then we can apply Corollary 9.26 to conclude. □

216Definition 9.3<strong>2.</strong> Let H = ( )H, ∆ H , ε H , m H , u H be a k-bialgebra, letA = ( ) (((A, m A , u A ) , ρ H A be a right H-comodule algebra, let D = D, ∆ D , ε D) ), µ H Dbe a right H-module coalgebra and g ∈ D be a grouplike element. We definethe category of (D, A)-Hopf modules (or Doi-Koppinen Hopf modules) denoted byM D A (H) , as follows:• M ∈ Ob ( M D A (H)) is a right D-comodule via ρ D M , a right A-module via µA Msuch that for every m ∈ M we have( ) ∑(230) ρDM ◦ µ A M (m ⊗ a) = µAM (m 0 ⊗ a 0 ) ⊗ µ H D (m 1 ⊗ a 1 )where ρ D M (m) = ∑ m 0 ⊗ m 1 ∈ M ⊗ D and ρ H A (a) = ∑ a 0 ⊗ a 1 ∈ A ⊗ H, i.e.ρ D M is a morphism of right A-modules or equivalently, µA M is a morphism ofright D-comodules• f ∈ Hom M DA (H) (M, N) is both a morphism of right D-comodules and amorphism of right A-modules.Lemma 9.33. Let H = ( ) (H, ∆ H , ε H , m H , u H be a k-bialgebra, let A = (A, mA , u A ) , ρ H Abe a right H-comodule algebra, let D = (( D, ∆ D , ε D) , µ D) H be a right H-modulecoalgebra and g ∈ D be a grouplike element. Then A ∈ M D A (H) and A is a rightD-comodule algebra.Proof. We denote µ H D (d ⊗ h) = d · h. First of all we want to prove that A is a rightD-comodule. In fact we can considerdefined by settingρ D A : A → A ⊗ Dρ D A (a) = a 0 ⊗ g · a 1 .Let us compute, for every a ∈ A,[( ) ] (A ⊗ ∆D◦ ρ ) D A (a) = A ⊗ ∆D(a 0 ⊗ g · a 1 ) = a 0 ⊗ (g · a 1 ) (1)⊗ (g · a 1 ) (2)so thatDisHmodcoalg= a 0 ⊗ g (1) · a 1(1) ⊗ g (2) · a 1(2)A is Hcom= a 00 ⊗ g (1) · a 01 ⊗ g (2) · a 1ggrouplike= a 00 ⊗ g · a 01 ⊗ g · a 1 = ( ρ D A ⊗ D ) (a 0 ⊗ g · a 1 ) = [( ρ D A ⊗ D ) ◦ ρ D A](a)(A ⊗ ∆D ) ◦ ρ D A = ( ρ D A ⊗ D ) ◦ ρ D A.We compute, for every a ∈ A,[rA ◦ ( A ⊗ ε D) ] [◦ ρ D A (a) = rA ◦ ( A ⊗ ε D)] (a 0 ⊗ g · a 1 )(= r A a0 ⊗ ε D (ga 1 ) ) DisHmodcoalg (= r A a0 ⊗ ε D (g) ε H (a 1 ) )so that= a 0 ε D (g) ε H (a 1 ) ggrouplike= a1 k = ar A ◦ ( A ⊗ ε D) ◦ ρ D A = Id A .Note that A is a right A-module via m A . It remains to prove (230) . Recall thatρ D A (a) = ∑ a 0 ⊗ g · a 1 ∈ A ⊗ D)

217so that we have( ) ( )ρDA ◦ µ A A (a ⊗ b) = ρDA ◦ m A (a ⊗ b) = ρDA (ab)= ∑ ∑(ab) 0⊗ µ H D (g ⊗ (ab) 1) AisHcomalg= a0 b 0 ⊗ g · (a 1 b 1 )∑a0 b 0 ⊗ (g · a 1 ) · b 1 = ∑ ( )a 0 b 0 ⊗ µ H D µHD (g ⊗ a 1 ) ⊗ b 1= ∑ ( )m A (a 0 ⊗ b 0 ) ⊗ µ H D µHD (g ⊗ a 1 ) ⊗ b 1DisHmodcoalg== ∑ µ A A (a 0 ⊗ b 0 ) ⊗ µ H D (g · a 1 ⊗ b 1 ) .Then we deduce that A ∈ M D A (H). Note that this last computation says that m Ais a morphism of right D-comodules. It remains to prove that u A is also a morphismof right D-comodules. Let us compute(ρDA ◦ u A)(1k ) = ρ D A (1 A ) = (1 A ) 0⊗ g · (1 A ) 1AisHcomalgDisHmodcoalg= 1 A ⊗ g · 1 H = 1 A ⊗ g= (u A ⊗ D) (1 k ⊗ g) = [ ](u A ⊗ D) ◦ ρ D k (1k )so that we conclude that ( (A, m A , u A ) , ρA) D is a right D-comodule algebra. □Theorem 9.34 ([MeZu, Theorem 3.29 (a) ⇔ (f)]). Let H = ( )H, ∆ H , ε H , m H , u Hbe a k-bialgebra, let A = ( )(((A, m A , u A ) , ρ H A be a right H-comodule algebra, let D =D, ∆ D , ε D) ), µ H D be a right H-module coalgebra and let g ∈ D be a grouplikeelement. Then ( (A, m A , u A ) , ρA) D is a right D-comodule algebra and D = A ⊗ D isan A-coring. Assume that A D is flat (i.e. D is k-flat). Let B = A coD . Then thefollowing statements are equivalent:(a) the functor − ⊗ B A : Mod-B → M D A (H) is an equivalence of categories;(b) the canonical map can : A ⊗ B A → A ⊗ D is an isomorphism (i.e. B ⊆ A isa D-Galois extension) and B A is faithfully flat.Proof. We set µ H D (d ⊗ h) = d · h. By Lemma 9.33, we know that ( )(A, m A , u A ) , ρ D Ais a right D-comodule algebra. First of all we want to prove that D = A ⊗ D is anA-coring. Let us consider ψ : D ⊗ A → A ⊗ D defined by setting, for every a ∈ Aand d ∈ D,ψ (d ⊗ a) = a 0 ⊗ d · a 1where we denote ρ H A (a) = a 0 ⊗ a 1 . Let us prove that (A, D, ψ) is then an entwiningstructure over k. We have to prove (222). Let us compute, for every a, b ∈ A, d ∈ D,[(m A ⊗ D) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)] (d ⊗ a ⊗ b)= [(m A ⊗ D) ◦ (A ⊗ ψ)] (a 0 ⊗ d · a 1 ⊗ b)= (m A ⊗ D) (a 0 ⊗ b 0 ⊗ (d · a 1 ) · b 1 ) = a 0 b 0 ⊗ (d · a 1 ) · b 1so that(D,µ H D)Hmodcoalg= a 0 b 0 ⊗ d · (a 1 b 1 ) (A,ρH A)H-com alg= (ab) 0⊗ d · (ab) 1= ψ (d ⊗ ab) = [ψ ◦ (D ⊗ m A )] (d ⊗ a ⊗ b)(m A ⊗ D) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A) = ψ ◦ (D ⊗ m A ) .

218Let us compute, for every d ∈ D,[ ]ψ ◦ (D ⊗ uA ) ◦ r −1D (d) = [ψ ◦ (D ⊗ uA )] (d ⊗ 1 k ) = ψ (d ⊗ 1 A )so that we have(A,ρ H A)H-com algDisHmodcoal= (1 A ) 0⊗ d · (1 A ) 1= 1 A ⊗ d · 1 H = 1 A ⊗ d= (u A ⊗ D) (1 k ⊗ d) = [ ](u A ⊗ D) ◦ l −1D (d)ψ ◦ (D ⊗ u A ) ◦ r −1D= (u A ⊗ D) ◦ l −1D .Let us prove (223) . Let us compute, for every a ∈ A, d ∈ D[(ψ ⊗ D) ◦ (D ⊗ ψ) ◦(∆ D ⊗ A )] (d ⊗ a)= [(ψ ⊗ D) ◦ (D ⊗ ψ)] ( d (1) ⊗ d (2) ⊗ a )= (ψ ⊗ D) ( d (1) ⊗ a 0 ⊗ d (2) · a 1)= a00 ⊗ d (1) · a 01 ⊗ d (2) · a 1A isHcomod= a 0 ⊗ d (1) · a 1(1) ⊗ d (2) · a 1(2)DisHmodcoalg= a 0 ⊗ (d · a 1 ) (1)⊗ (d · a 1 ) (2)so that we getandso that we get= ( A ⊗ ∆ D) (a 0 ⊗ d · a 1 ) = [( A ⊗ ∆ D) ◦ ψ ] (d ⊗ a)(ψ ⊗ D) ◦ (D ⊗ ψ) ◦ ( ∆ D ⊗ A ) = ( A ⊗ ∆ D) ◦ ψ[rA ◦ ( A ⊗ ε D) ◦ ψ ] (d ⊗ a) = [ r A ◦ ( A ⊗ ε D)] (a 0 ⊗ d · a 1 )(= r A a0 ⊗ ε D (d · a 1 ) ) DisHmodcoal (= r A a0 ⊗ ε D (d) ε H (a 1 ) )= aε D (d) = ε D (d) a = l A(ε D (d) ⊗ a ) = [ l A ◦ ( ε D ⊗ A )] (d ⊗ a)r A ◦ ( A ⊗ ε D) ◦ ψ = l A ◦ ( ε D ⊗ A ) .Then by Proposition 9.30, ( D = A ⊗ D, ∆ D , ε D) is the A-coring associated to theentwining (A, D, ψ) and M D A = (Mod-A)D ≃ M D A (ψ). Note that M ∈ MD A (ψ) , issuch that ( ( )M, µ M) A is a right A-module, M, ρDM is a right D-comodule satisfying(µAM ⊗ D ) ◦ (M ⊗ ψ) ◦ ( ρ D M ⊗ A ) = ρ D M ◦ µ A Mi.e. for every m ∈ M and for every a ∈ A( )ρDM ◦ µ A M (m ⊗ a) = µAM (m 0 ⊗ a 0 ) ⊗ µ H D (m 1 ⊗ a 1 )which is exactly the condition (230) for M ∈ M D A (H) . Since morphisms in bothcategories M D A (ψ) and MD A (H) are right A-linear and right D-colinear morphismswe deduce thatM D A (ψ) ≃ M D A (H) .Since by Lemma 9.33 A ∈ M D A (H) ≃ MD A (ψ) , we can apply Theorem 9.31 to thecase ”C” = D and ”M C A (ψ) ” = MD A (ψ) ≃ MD A (H) .□

219Corollary 9.35 ([Schn1, Theorem I (2) ⇔ (4)]). Let H = ( )H, ∆ H , ε H , m H , u Hbe a k-bialgebra and let ( )(A, m A , u A ) , ρ H A be a right H-comodule algebra. ThenH = A⊗H is an A-coring. Assume that A H is flat (i.e. H is k-flat). Let B = A coH .Then the following statements are equivalent:(a) the functor − ⊗ B A : Mod-B → M H A is an equivalence of categories;(b) the canonical map can : A ⊗ B A → A ⊗ H is an isomorphism (i.e. B ⊆ A isan H-Galois extension) and B A is faithfully flat.Proof. We can apply Theorem 9.34 to the case ”D” = H so that M D A (H) = MH A .□10. BicategoriesIn this last part we will change some notations to be more clear and to give moreevidence to a new product we introduce here.Let C be a bicategory. For every 0-cell X in C, we denote by 1 X : X → X theidentity 1-cell over X. For every 1-cell A in C, we denote by 1 A : A → A the identity2-cell over A. We will use juxtaposition when we compose 2-cells vertically and wewill denote by · the horizontal composition of 1-cells and 2-cells.Let us assume that C is a bicategory with completeness requirement (all thecategories C ((X, A) , (Y, B)) have coequalizers which are preserved by compositionwith 1-cells).We keep denoting by (A, m A , u A ) a monad with its multiplication and unit.We now want to define the 2-category Mnd (C) following the definition given in[St]. For simplicity we will always assume to work with a 2-category C even if onecan prove similar <strong>results</strong> for an arbitrary bicategory.Definition 10.<strong>1.</strong> Let C be a 2-category. A monad in C is a pair (X, A) where Xis an object of C, A : X → X is a 1-cell in C together with 2-cells m A : 1 A · 1 A → 1 Aand u A : 1 X → 1 A satisfying associativity and unitality conditions, i.e.(231)(232)m A (1 A · m A ) = m A (m A · 1 A )m A (u A · 1 A ) = 1 A = m A (1 A · u A ) .Definition 10.<strong>2.</strong> Let C be a 2-category and let (X, A) , (Y, B) be monads in C. Amonad functor in C is a pair (Q, φ) : (X, A) → (Y, B) where Q : X → Y is a 1-celland φ : B · Q → Q · A satisfying the following conditions(233)(234)(1 Q · m A ) (φ · 1 A ) (1 B · φ) = φ (m B · 1 Q )φ (u B · 1 Q ) = 1 Q · u A .Definition 10.3. Let C be a 2-category, let (X, A) , (Y, B) be monads in C and let(Q, φ) , (Q ′ , φ ′ ) : (X, A) → (Y, B) be monad functors. A monad functor transformationσ : (Q, φ) → (Q ′ , φ ′ ) in C is a 2-cell σ : Q → Q ′ such that(235) φ ′ (1 B · σ) = (σ · 1 A ) φ.Definition 10.4. The 2-category Mnd (C) consists of• Objects: monads in C• 1-cells: monad functors in C

220• 2-cells: monad functor transformations in C.Remark 10.5. We denote by (X, 1 X ) in C the trivial monad on the object X withtrivial multiplication and unit m 1X : 1 1X · 1 1X → 1 1X and u 1X : 1 X → 1 1X .Definition 10.6. Let C, C ′ be two 2-categories and let G : C → C ′ be a pseudofunctor.Then the pseudofunctor Mnd (G) : Mnd (C) → Mnd (C ′ ) is defined asfollows:• Mnd (G) (X, A) = (G (X) , G (A))• Mnd (G) (Q, φ) = (G (Q) , G (φ))• Mnd (G) (σ) = G (σ) .Remark 10.7. Note that Cmd (C) = Mnd (C ∗ ) where C ∗ is the dual reversing2-cells of C.1<strong>1.</strong> Construction of BIM (C)The idea of defining this bicategory goes back to the strict monoidal category ofbalanced bimodule functors that we defined in Subsection 3.<strong>2.</strong> We observed that,considering bimodule functors with respect to the same monad on both sides, wehave a unit and a composition, so that they form a strict monoidal category. In thecase we consider a bimodule with respect to two different monads, the unit objectfails and the composition between them is no longer inside the class of objects of thecategory. The way to solve this problem is to look at balanced bimodule functorsas 0-cells of a bicategory, changing the definition of their product.Definition 1<strong>1.</strong><strong>1.</strong> Let X be a 0-cell and let (Y, B) be a monad in C. A left B-modulein C (or simply a left B-module) is a pair (Q, λ Q ) where Q : X → Y is a 1-cell andλ Q : B · Q → Q is a 2-cell in C, satisfying the associativity and unitality propertieswith respect to the monad B, i.e.λ Q (m B · 1 Q ) = λ Q (1 B · λ Q ) and λ Q (u B · 1 Q ) = 1 Q .Definition 1<strong>1.</strong><strong>2.</strong> Let (X, A) and (Y, 1 Y ) be monads in C. A right A-module in C(or simply a right A-module) is a monad functor in C ∗ , i.e. a 1-cell Q : X → Yand a 2-cell ρ Q : Q · A → 1 Y · Q = Q in C, satisfying the associativity and unitalityproperties with respect to the monad A, i.e.ρ Q (1 Q · m A ) = ρ Q (ρ Q · 1 A ) and ρ Q (1 Q · u A ) = 1 Q .Definition 1<strong>1.</strong>3. Let (X, A) and (Y, B) be monads in C. A B-A-bimodule in C (orsimply a B-A-bimodule) is a triple (Q, λ Q , ρ Q ) where• (Q, λ Q ) is a left B-module in C• (Q, ρ Q ) is a right A-module in C• the compatibility condition holdsλ Q (1 B · ρ Q ) = ρ Q (λ Q · 1 A ) .Lemma 1<strong>1.</strong>4. Let (X, A) , (Y, B) be monads in C and let (Q, λ Q ) be a left A-moduleand (Q, ρ Q ) be a right B-module. Then (Q, λ Q ) = Coequ C (m A · 1 Q , 1 A · λ Q ) and(Q, ρ Q ) = Coequ C (1 Q · m B , ρ Q · 1 B ).

Proof. We will only prove the statement for the left module. Similarly can be provedthe other one. Since λ Q is associative, we deduce thatNow, assume that (S, σ) is such thatThen we haveλ Q (m A · 1 Q ) = λ Q (1 A · λ Q ) .σ (m A · 1 Q ) = σ (1 A · λ Q ) .σ (u A · 1 Q ) λ Qu A= σ (1 A · λ Q ) (u A · 1 A · 1 Q )propσ= σ (m A · 1 Q ) (u A · 1 A · 1 Q ) Amonad= σ.Moreover, since λ Q is epi, we conclude that the 2-cell σ (u A · 1 Q ) is unique withrespect to the propertyσ (u A · 1 Q ) λ Q = σso that(Q, λ Q ) = Coequ C (m A · 1 Q , 1 A · λ Q ) .□Proposition 1<strong>1.</strong>5. Let (X, A) , (Y, B) and (W, C) be monads in C and let Q : Y →X and Q ′ : W → Y be respectively a A-B-bimodule with (Q, λ Q , ρ Q ) and a B-Cbimodulein C with (Q ′ , λ Q ′, ρ Q ′). Then (Q • B Q ′ , p Q,Q ′) = Coequ C (ρ Q · 1 Q ′, 1 Q · λ Q ′)is a A-C-bimodule in C via the actions λ Q•B Q ′ and ρ Q• B Q ′ uniquely determined by(236) λ Q•B Q ′ (1 A · p Q,Q ′) = p Q,Q ′ (λ Q · 1 Q ′)and(237) ρ Q•B Q ′ (p Q,Q ′ · 1 C) = p Q,Q ′ (1 Q · ρ Q ′) .Proof. Let us define the bimodule structures on Q• B Q ′ . Let us consider the followingdiagramQ · B · Q ′ · C1 Q·1 B·ρ Q ′ρ Q·1 Q ′·1 C1 Q·λ Q ′·1 CQ · B · Q ′ ρ Q·1 Q ′1 Q·λ Q ′Q · Q ′ · C1 Q·ρ Q ′p Q,Q ′·1 C Q · Q ′ p Q,Q ′Q • B Q ′ · Cρ Q•B Q ′ Q • B Q ′Note that the left square serially commutes. In fact we have(1 Q · ρ Q ′) (ρ Q · 1 Q ′ · 1 C ) ρ Q= (ρ Q · 1 Q ′) (1 Q · 1 B · ρ Q ′)and(1 Q · ρ Q ′) (1 Q · λ Q ′ · 1 C ) Qbim= (1 Q · λ Q ′) (1 Q · 1 B · ρ Q ′) .Therefore, we getp Q,Q ′ (1 Q · ρ Q ′) (ρ Q · 1 Q ′ · 1 C ) = p Q,Q ′ (1 Q · ρ Q ′) (1 Q · λ Q ′ · 1 C )and by the universal property of the coequalizer(Q • B Q ′ · C, p Q,Q ′ · 1 C ) = Coequ C (ρ Q · 1 Q ′ · 1 C , 1 Q · λ Q ′ · 1 C ), there exists a unique2-cell ρ Q•B Q ′ : Q • B Q ′ · C → Q • B Q ′ such thatρ Q•B Q ′ (p Q,Q ′ · 1 C) = p Q,Q ′ (1 Q · ρ Q ′) .221

222We now want to prove that ρ Q·BQ ′consider the following diagramQ · B · Q ′ · C · Cρ Q·1 Q ′·1 C·1 C1 Q·λ Q ′·1 C·1 Cdefines a structure of right C-module. Let usQ · Q ′ · C · C p Q,Q ′·1 C·1 C Q • B Q ′ · C · C1 Q·1 B·1 Q ′·m C1 Q·1 Q ′·m Cρ Q•B Q ′·1 C1 Q•B Q ′·m C1 Q·1 B·ρ Q ′·1 CQ · B · Q ′ · Cρ Q·1 Q ′·1 C1 Q·λ Q ′·1 C1 Q·ρ Q ′·1 CQ · Q ′ · Cp Q,Q ′·1 CQ • B Q ′ · C1 Q·1 B·ρ Q ′1 Q·ρ Q ′ρ Q•B Q ′Q · B · Q ′ ρ Q·1 Q ′1 Q·λ Q ′ Q · Q ′ p Q,Q ′ Q • B Q ′The diagram serially commutes and since the rows and the first two columns arecoequalizers, also the third column is a coequalizer. In particular,i.e. ρ Q•B Q ′ρ Q•B Q ′ (ρ Q• B Q ′ · 1 C) = ρ Q•B Q ′ (1 Q• B Q ′ · m C)is associative. Now, we also have that the following diagramQ · B · Q ′ · Cρ Q·1 Q ′·1 C1 Q·λ Q ′·1 CQ · Q ′ · C p Q,Q ′·1C Q • B Q ′ · C1 Q·1 B·ρ Q ′1 Q·1 B·1 Q ′·u C1 Q·ρ Q ′1 Q·1 Q ′·u C1 Q•B Q ′·u CQ · B · Q ′ ρ Q·1 Q ′1 Q·λ Q ′Q · Q ′ p Q,Q ′ρ Q•B Q ′ Q • B Q ′serially commutes. In particularso thatand since p Q,Q ′(1 Q•B Q ′ · u C) p Q,Q ′ = (p Q,Q ′ · 1 C ) (1 Q · 1 Q ′ · u C )ρ Q•B Q ′ (1 Q• B Q ′ · u C) p Q,Q ′ = ρ Q•B Q ′ (p Q,Q ′ · 1 C) (1 Q · 1 Q ′ · u C )= p Q,Q ′ (1 Q · ρ Q ′) (1 Q · 1 Q ′ · u C ) Q′ mod= p Q,Q ′is an epimorphism, we get thatρ Q•B Q ′ (1 Q• B Q ′ · u C) = 1 Q•B Q ′so that ρ Q•B Q ′ is also unital. Therefore (Q • B Q ′ , ρ Q•B Q ′) is a right C-module. Similarly,let us consider the following diagramA · Q · B · Q ′ 1 A·ρ Q·1 Q ′λ Q·1 B·1 Q ′1 A·1 Q·λ Q ′Q · B · Q ′ ρ Q·1 Q ′1 Q·λ Q ′A · Q · Q ′ 1 A·p Q,Q ′λ Q·1 Q ′ Q · Q ′ p Q,Q ′A · Q • B Q ′λ Q•B Q ′ Q • B Q ′

Since we are assuming that the coequalizers are preserved by the composition withany 1-cell, both the rows are coequalizers and the left square serially commutes. Infact(λ Q · 1 Q ′) (1 A · ρ Q · 1 Q ′) Qbim= (ρ Q · 1 Q ′) (λ Q · 1 B · 1 Q ′)(λ Q · 1 Q ′) (1 A · 1 Q · λ Q ′) λ Q= (1 Q · λ Q ′) (λ Q · 1 B · 1 Q ′) .By the universal property of the coequalizer(Q • B Q ′ , p Q,Q ′) = Coequ C (ρ Q · 1 Q ′, 1 Q · λ Q ′), there exists a unique 2-cell λ Q•B Q ′ :A · Q • B Q ′ → Q • B Q ′ such thatλ Q•B Q ′ (1 A · p Q,Q ′) = p Q,Q ′ (λ Q · 1 Q ′) .By similar computations, one can prove that (Q • B Q ′ , λ Q•B Q ′) is a left A-module.Finally, we prove that the structures are compatible. In factρ Q•B Q ′ (λ Q• B Q ′ · 1 C) (1 A · p Q,Q ′ · 1 C )(236)= ρ Q•B Q ′ (p Q,Q ′ · 1 C) (λ Q · 1 Q ′ · 1 C )(237)= p Q,Q ′ (1 Q · ρ Q ′) (λ Q · 1 Q ′ · 1 C ) λ Q= p Q,Q ′ (λ Q · 1 Q ′) (1 A · 1 Q · ρ Q ′)(236)= λ Q•B Q ′ (1 A · p Q,Q ′) (1 A · 1 Q · ρ Q ′)223(237)= λ Q•B Q ′ (1 A · ρ Q•B Q ′) (1 A · p Q,Q ′ · 1 C )and since 1 A · p Q,Q ′ · 1 C is epi, we get thatρ Q•B Q ′ (λ Q• B Q ′ · 1 C) = λ Q•B Q ′ (1 A · ρ Q•B Q ′)i.e. (Q • B Q ′ , λ Q•B Q ′, ρ Q• B Q ′) is an A-C-bimodule.□Proposition 1<strong>1.</strong>6. Let (X, A) and (Y, B) be monads in the 2-category C, let (Q, λ Q )be a left A-module and (Q, ρ Q ) be a right B-module. Then A• A Q ≃ Q and Q• B B ≃Q.Proof. Let us consider the trivial left A-module (A, m A ) and note that (A • A Q, p A,Q ) =Coequ C (m A · 1 Q , 1 A · λ Q ). We already observed, in Lemma 1<strong>1.</strong>4, that (Q, λ Q ) =Coequ C (m A · 1 Q , 1 A · λ Q ). Therefore, there exists an isomorphism l Q : A • A Q → Qsuch that(238) l Q p A,Q = λ Q .Similarly, if we consider the trivial right B-module (B, m B ), since (Q • B B, p Q,B ) =Coequ C (1 Q · m B , ρ Q · 1 B ) = (Q, ρ Q ) we deduce that there exists an isomorphismr Q : Q • B B → Q such that(239) r Q p Q,B = ρ Q .Proposition 1<strong>1.</strong>7. Let (X, A), (Y, B), (Z, C), (W, D) be monads in the 2-categoryC and let (Q, λ Q , ρ Q ) be an A-B-bimodule, (Q ′ , λ Q ′, ρ Q ′) be a B-C-bimodule and(Q ′′ , λ Q ′′, ρ Q ′′) be a C-D-bimodule. Then the coequalizers (Q • B Q ′ ) • C Q ′′ ≃ Q • B(Q ′ • C Q ′′ ) are isomorphic as A-D-bimodules.□

224Proof. Let us consider the following diagramQ · B · Q ′ · C · Q ′′ 1ρ Q·1 Q ′·1 C·1 Q ′′1 Q·λ Q ′·1 C·1 Q ′′Q · Q ′ · C · Q ′′p Q,Q ′·1 C·1 Q ′′Q • B Q ′ · C · Q ′′1 Q·1 B·1 Q ′·λ Q ′′1 Q·1 B·ρ Q ′·1 Q ′′1 Q·1 Q ′·λ Q ′′1 Q·ρ Q ′·1 Q ′′2ρ Q•B Q ′·1 Q ′′Q · B · Q ′ · Q ′′ ρ Q·1 Q ′·1 Q ′′1 Q·λ Q ′·1 Q ′′Q · Q ′ · Q ′′ p Q,Q ′·1 Q ′′1 Q•B Q ′·λ Q ′′ Q • B Q ′ · Q ′′1 Q·1 B·p Q ′ ,Q ′′31 Q·p Q ′ ,Q ′′4p Q•B Q ′ ,Q ′′Q · B · Q ′ • C Q ′′ ρ Q·1 Q ′ •C Q ′′ 1 Q·λ Q ′ •C Q ′′ Q · Q ′ • C Q ′′ p Q,Q ′ •C Q ′′ Q • B (Q ′ • C Q ′′ ) ≃ (Q • B Q ′ ) • C Q ′′Note that the left upper square serially commutes because of naturality of the 2-cells. The right upper square commutes because of naturality and of (237). The leftbottom square commutes because of naturality and of (236). The rows are coequalizersand, since the 1-cells preserves coequalizers, also the columns are coequalizers.By the commutativity of the diagram, we deduce thatp Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) (ρ Q · 1 Q ′ · 1 Q ′′)3= p Q,Q ′ • C Q ′′ (ρ Q · 1 Q ′ • C Q ′′) (1 Q · 1 B · p Q ′ ,Q ′′)coequ= p Q,Q ′ • C Q ′′ (1 Q · λ Q ′ • C Q ′′) (1 Q · 1 B · p Q ′ ,Q ′′)3= p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) (1 Q · λ Q ′ · 1 Q ′′)and since (Q • B Q ′ · Q ′′ , p Q,Q ′ · 1 Q ′′) = Coequ C (ρ Q · 1 Q ′ · 1 Q ′′, 1 Q · λ Q ′ · 1 Q ′′), thereexists a unique 2-cell ξ : Q • B Q ′ · Q ′′ → Q • B (Q ′ • C Q ′′ ) such that(240) ξ (p Q,Q ′ · 1 Q ′′) = p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) .Moreover, we haveξ (1 Q•B Q ′ · λ Q ′′) (p Q,Q ′ · 1 C · 1 Q ′′) 2 = ξ (p Q,Q ′ · 1 Q ′′) (1 Q · 1 Q ′ · λ Q ′′)(240)= p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) (1 Q · 1 Q ′ · λ Q ′′)p Q ′ ,Q ′′coequ= p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) (1 Q · ρ Q ′ · 1 Q ′′)(240)= ξ (p Q,Q ′ · 1 Q ′′) (1 Q · ρ Q ′ · 1 Q ′′) 2 = ξ (ρ Q•B Q ′ · 1 Q ′′) (p Q,Q ′ · 1 C · 1 Q ′′)and since p Q,Q ′ · 1 C · 1 Q ′′ is epi, we get that ξ is a fork for (1 Q•B Q ′ · λ Q ′′, ρ Q• B Q ′ · 1 Q ′′) .By the universal property of the coequalizer((Q • B Q ′ ) • C Q ′′ , p Q•B Q ′ ,Q ′′) = Coequ C (1 Q•B Q ′ · λ Q ′′, ρ Q• B Q ′ · 1 Q ′′), there exists a unique2-cell ζ : (Q • B Q ′ ) • C Q ′′ → Q • B (Q ′ • C Q ′′ ) such thatand thus we haveζp Q•B Q ′ ,Q ′′ = ξζp Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) = ξ (p Q,Q ′ · 1 (240)Q ′′) = p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′)

i.e.(241) ζp Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) = p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) .Similarly, we havep Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) (1 Q · ρ Q ′ · 1 Q ′′)2= p Q•B Q ′ ,Q ′′ (ρ Q• B Q ′ · 1 Q ′′) (p Q,Q ′ · 1 C · 1 Q ′′)p Q•B Q ′ ,Q ′′coequ= p Q•B Q ′ ,Q ′′ (1 Q• B Q ′ · λ Q ′′) (p Q,Q ′ · 1 C · 1 Q ′′)2= p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) (1 Q · 1 Q ′ · λ Q ′′)and since (Q · Q ′ • C Q ′′ , 1 Q · p Q ′ ,Q ′′) = Coequ C ((1 Q · ρ Q ′ · 1 Q ′′) , (1 Q · 1 Q ′ · λ Q ′′)) , thereexists a unique ξ ′ : Q · Q ′ • C Q ′′ → (Q • B Q ′ ) • C Q ′′ such that(242) ξ ′ (1 Q · p Q ′ ,Q ′′) = p Q• B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) .Moreover, we haveξ ′ (ρ Q · 1 Q ′ • C Q ′′) (1 Q · 1 B · p Q ′ ,Q ′′) ρ Q= ξ ′ (1 Q · p Q ′ ,Q ′′) (ρ Q · 1 Q ′ · 1 Q ′′)(242)= p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) (ρ Q · 1 Q ′ · 1 Q ′′)p Q,Q ′coequ= p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) (1 Q · λ Q ′ · 1 Q ′′)(242)= ξ ′ (1 Q · p Q ′ ,Q ′′) (1 Q · λ Q ′ · 1 Q ′′) 3 = ξ ′ (1 Q · λ Q ′ • C Q ′′) (1 Q · 1 B · p Q ′ ,Q ′′)and since 1 Q·1 B·p Q ′ ,Q ′′ is epi, we deduce that ξ′ is a fork for (ρ Q · 1 Q ′ • C Q ′′, 1 Q · λ Q ′ • C Q ′′) .Since (Q • B (Q ′ • C Q ′′ ) , p Q,Q ′ • C Q ′′) = Coequ C (ρ Q · 1 Q ′ • C Q ′′, 1 Q · λ Q ′ • C Q ′′) , there existsa unique 2-cell ζ ′ : Q • B (Q ′ • C Q ′′ ) → (Q • B Q ′ ) • C Q ′′ such thatand thusso thatζ ′ p Q,Q ′ • C Q ′′ = ξ′ζ ′ p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) = ξ′ (242)(1 Q · p Q ′ ,Q′′) = p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′)(243) ζ ′ p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) = p Q• B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) .We now want to prove that ζ and ζ ′ are two-sided inverse. We haveζζ ′ p Q,Q ′ • C Q ′′ (1 (243)Q · p Q ′ ,Q′′) = ζp Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′)(241)= p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′)and since p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q′′) is an epimorphism, we deduce thatSimilarly, we haveζζ ′ = 1 Q•B (Q ′ • C Q ′′ ).ζ ′ ζp Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 (241)Q ′′) = ζ ′ p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′)(243)= p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′)225

226and since p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) is an epimorphism, we get thatζ ′ ζ = 1 (Q•B Q ′ )• C Q ′′.Therefore, (Q • B Q ′ ) • C Q ′′ ≃ Q • B (Q ′ • C Q ′′ ) via ζ. Moreover, by Proposition 1<strong>1.</strong>5,we know that (Q • B Q ′ ) • C Q ′′ and Q • B (Q ′ • C Q ′′ ) are A-D-bimodules. We nowwant to prove that ζ is a morphism of left A-modules and right D-modules. Let uscomputeζλ (Q•B Q ′ )• C Q ′′ (1 A · p Q•B Q ′ ,Q ′′) (1 A · p Q,Q ′ · 1 Q ′′)defλ (Q•B Q ′ )• C Q ′′= ζp Q•B Q ′ ,Q ′′ (λ(Q•B Q ′ ) · 1 Q ′′)(1A · p Q,Q ′ · 1 Q ′′)defλ (Q•B Q ′ )= ζp Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) (λ Q · 1 Q ′ · 1 Q ′′)(241)= p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) (λ Q · 1 Q ′ · 1 Q ′′)λ Q=pQ,Q ′ • C Q ′′ (λ Q · 1 Q ′ • C Q ′′) (1 A · 1 Q · p Q ′ ,Q ′′)defλ (Q•B Q ′ )= λ Q•B (Q ′ • C Q ′′ ) (1 A · p Q,Q ′ • C Q ′′) (1 A · 1 Q · p Q ′ ,Q ′′)(241)= λ Q•B (Q ′ • C Q ′′ ) (1 A · ζ) (1 A · p Q•B Q ′ ,Q ′′) (1 A · p Q,Q ′ · 1 Q ′′)and since (1 A · p Q•B Q ′ ,Q ′′) (1 A · p Q,Q ′ · 1 Q ′′) is epi, we get thatζλ (Q•B Q ′ )• C Q ′′ = λ Q• B (Q ′ • C Q ′′ ) (1 A · ζ)i.e. ζ is a morphism of left A-modules. Similarly, one can prove that ζ is a morphismof right D-modules.□Notation 1<strong>1.</strong>8. In the setting of Proposition 1<strong>1.</strong>7, let us consider the isomorphismof bimodulesζ : (Q • B Q ′ ) • C Q ′′ → Q • B (Q ′ • C Q ′′ ) .In order to be more clear, in the following, we will denote it byζ Q,Q ′ ,Q ′′ : (Q • B Q ′ ) • C Q ′′ → Q • B (Q ′ • C Q ′′ )which is the unique satisfying the following(244) ζ Q,Q ′ ,Q ′′p Q• B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) = p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) .Proposition 1<strong>1.</strong>9. Let (X, A), (Y, B), (Z, C), (W, D), (U, E) be monads in the 2-category C and let (Q, λ Q , ρ Q ) be an A-B-bimodule, (Q ′ , λ Q ′, ρ Q ′) be a B-C-bimodule,(Q ′′ , λ Q ′′, ρ Q ′′) be a C-D-bimodule and (Q ′′′ , λ Q ′′′, ρ Q ′′′) be a D-E-bimodule. Then thePentagon Axiom holds, i.e. the following diagram is commutative((Q • B Q ′ ) • C Q ′′ ) • D Q ′′′ ζ Q,Q ′ ,Q ′′• D 1 Q ′′′(Q • B (Q ′ • C Q ′′ )) • D Q ′′′ζ Q•B Q ′ ,Q ′′ ,Q ′′′ζ Q,Q ′ •C Q ′′ ,Q ′′′(Q • B Q ′ ) • C (Q ′′ • D Q ′′′ )Q • B ((Q ′ • C Q ′′ ) • D Q ′′′ )ζ Q,Q ′ ,Q ′′ •D Q ′′′1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′Q • B (Q ′ • C (Q ′′ • D Q ′′′ ))

227Proof. We computeand(1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′ (ζ Q,Q ′ ,Q ′′ • D 1 Q ′′′) p (Q•B Q ′ )• C Q ′′ ,Q ′′′(p Q•B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)= (1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′p Q• B (Q ′ • C Q ′′ ),Q ′′′ (ζ Q,Q ′ ,Q ′′ · 1 Q ′′′)(p Q•B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)(244)= (1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′p Q• B (Q ′ • C Q ′′ ),Q ′′′ (p Q,Q ′ • C Q ′′ · 1 Q ′′′)(1 Q · p Q ′ ,Q ′′ · 1 Q ′′′)= (1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) p Q,(Q ′ • C Q ′′ )• D Q ′′′ (1 Q · p Q ′ • C Q ′′ ,Q ′′′) (1 Q · p Q ′ ,Q ′′ · 1 Q ′′′)= p Q,Q ′ • C (Q ′′ • D Q ′′′ ) (1 Q · ζ Q ′ ,Q ′′ ,Q ′′′) (1 Q · p Q ′ • C Q ′′ ,Q ′′′) (1 Q · p Q ′ ,Q ′′ · 1 Q ′′′)(241)= p Q,Q ′ • C (Q ′′ • D Q ′′′ ) (1 Q · p Q ′ ,Q ′′ • D Q ′′′) (1 Q · 1 Q ′ · p Q ′′ ,Q ′′′)ζ Q,Q ′ ,Q ′′ • D Q ′′′ζ Q• B Q ′ ,Q ′′ ,Q ′′′p (Q• B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)(241)= ζ Q,Q ′ ,Q ′′ • D Q ′′′p Q• B Q ′ ,Q ′′ • D Q ′′′ (1 Q• B Q ′ · p Q ′′ ,Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)p Q,Q ′= ζ Q,Q ′ ,Q ′′ • D Q ′′′p Q• B Q ′ ,Q ′′ • D Q ′′′ (p Q,Q ′ · 1 Q ′′ • D Q ′′′) (1 Q · 1 Q ′ · p Q ′′ ,Q ′′′)so that we get that(241)= p Q,Q ′ • C (Q ′′ • D Q ′′′ ) (1 Q · p Q ′ ,Q ′′ • D Q ′′′) (1 Q · 1 Q ′ · p Q ′′ ,Q ′′′)(1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′ (ζ Q,Q ′ ,Q ′′ • D 1 Q ′′′) p (Q•B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′)(p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)= ζ Q,Q ′ ,Q ′′ • D Q ′′′ζ Q• B Q ′ ,Q ′′ ,Q ′′′p (Q• B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)and since p (Q•B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′) is an epimorphism, wededuce that the Pentagon Axiom holds.□Proposition 1<strong>1.</strong>10. Let (X, A), (Y, B), (Z, C) be monads in the 2-category C andlet (Q, λ Q , ρ Q ) be an A-B-bimodule and (Q ′ , λ Q ′, ρ Q ′) be a B-C-bimodule. Then theTriangle Axiom holds, i.e. the following diagram is commutativeProof. We compute(Q • B B) • B Q ′ ζ Q,B,Q ′Q • B (B • B Q ′ )rQ • B 1 Q ′1 Q • B l Q ′Q • B Q ′(r Q • B 1 Q ′) (p Q•B B,Q ′) (p Q,B · 1 Q ) = p Q,Q ′ (r Q · 1 Q ′) (p Q,B · 1 Q )(239)= p Q,Q ′ (ρ Q · 1 Q ′) defp= p Q,Q ′ (1 Q · λ Q ′)(238)= p Q,Q ′ (1 Q · l Q ′) (1 Q · p B,Q ′)= (1 Q • B l Q ′) p Q,B•B Q ′ (1 Q · p B,Q ′)

228(241)= (1 Q • B l Q ′) ζ Q,B,Q ′p Q•B B,Q ′ (p Q,B · 1 Q ′)so that, since p Q•B B,Q ′ (p Q,B · 1 Q ′) is an epimorphism, we getr Q • B 1 Q ′ = (1 Q • B l Q ′) ζ Q,B,Q ′.Proposition 1<strong>1.</strong>1<strong>1.</strong> Let (X, A) , (Y, B) be monads in C, let (P, λ P , ρ P ) , (Q, λ Q , ρ Q )be A-B-bimodules in C, let (P ′ , λ P ′, ρ P ′) , (Q ′ , λ Q ′, ρ Q ′) be B-C-bimodules in C andlet f : P → Q, f ′ : P ′ → Q ′ be bimodule morphisms in C. Then there exists a uniqueA-C-bimodule morphism f • B f ′ : P • B P ′ → Q • B Q ′ .Proof. Since f is an A-B-bimodule morphism, we have that(245) λ Q (1 A · f) = fλ P and ρ Q (f · 1 B ) = fρ P .Since f ′ is a B-C-bimodule morphism, we have that(246) λ Q ′ (1 B · f ′ ) = f ′ λ P ′ and ρ Q ′ (f ′ · 1 C ) = f ′ ρ P ′.Let us consider the following diagramP · B · P ′ ρ P ·1 P ′ 1 P ·λ P ′f·1 B·f ′ P · P ′ p P,P ′f·f ′P • B P ′f• B f ′Q · B · Q ′ ρ Q·1 Q ′ Q · Q ′ p Q,Q ′ Q • B Q ′1 Q·λ Q ′Note that the left square serially commutes, in factandThus, we get that(f · f ′ ) (ρ P · 1 P ′) = (f · 1 Q ′) (1 P · f ′ ) (ρ P · 1 P ′)ρ P= (f · 1 Q ′) (ρ P · 1 Q ′) (1 P · 1 B · f ′ )(245)= (ρ Q · 1 Q ′) (f · 1 B · 1 Q ′) (1 P · 1 B · f ′ ) = (ρ Q · 1 Q ′) (f · 1 B · f ′ )(f · f ′ ) (1 P · λ P ′) = (f · 1 Q ′) (1 P · f ′ ) (1 P · λ P ′)(246)= (f · 1 Q ′) (1 P · λ Q ′) (1 P · 1 B · f ′ )f= (1 Q · λ Q ′) (f · 1 B · 1 Q ′) (1 P · 1 B · f ′ )= (1 Q · λ Q ′) (f · 1 B · f ′ ) .p Q,Q ′ (f · f ′ ) (ρ P · 1 P ′) = p Q,Q ′ (f · f ′ ) (1 P · λ P ′)and since (P • B P ′ , p P,P ′) = Coequ C (ρ P · 1 P ′, 1 P · λ P ′) we deduce that there exists aunique 2-cell f • B f ′ : P • B P ′ → Q • B Q ′ such that(247) (f • B f ′ ) p P,P ′ = p Q,Q ′ (f · f ′ ) .□

We now want to prove that f • B f ′ is a morphism of A-C-bimodules. Note that, byProposition 1<strong>1.</strong>5, P • B P ′ and Q • B Q ′ are A-C-bimodules. We computeλ Q•B Q ′ (1 A · f • B f ′ ) (1 A · p P,P ′) (247)= λ Q•B Q ′ (1 A · p Q,Q ′) (1 A · f · f ′ )(236)= p Q,Q ′ (λ Q · 1 Q ′) (1 A · f · f ′ )= p Q,Q ′ (λ Q · 1 Q ′) (1 A · f · 1 Q ′) (1 A · 1 P · f ′ )(245)= p Q,Q ′ (f · 1 Q ′) (λ P · 1 Q ′) (1 A · 1 P · f ′ )λ P= p Q,Q ′ (f · 1 Q ′) (1 P · f ′ ) (λ P · 1 P ′) = p Q,Q ′ (f · f ′ ) (λ P · 1 P ′)(247)= (f • B f ′ ) p P,P ′ (λ P · 1 P ′)(236)= (f • B f ′ ) λ P •B P ′ (1 A · p P,P ′)229and since 1 A · p P,P ′is an epimorphism, we get thatλ Q•B Q ′ (1 A · f • B f ′ ) = (f • B f ′ ) λ P •B P ′i.e. f • B f ′ is a morphism of left A-modules. Similarly, we also haveρ Q•B Q ′ (f • B f ′ · 1 C ) (p P,P ′ · 1 C ) (247)= ρ Q•B Q ′ (p Q,Q ′ · 1 C) (f · f ′ · 1 C )(237)= p Q,Q ′ (1 Q · ρ Q ′) (f · f ′ · 1 C )= p Q,Q ′ (1 Q · ρ Q ′) (1 Q · f ′ · 1 C ) (f · 1 P ′ · 1 C )(246)= p Q,Q ′ (1 Q · f ′ ) (1 Q · ρ P ′) (f · 1 P ′ · 1 C )f= p Q,Q ′ (1 Q · f ′ ) (f · 1 P ′) (1 P · ρ P ′) = p Q,Q ′ (f · f ′ ) (1 P · ρ P ′)and since p P,P ′ · 1 C is epi, we get that(247)= (f • B f ′ ) p P,P ′ (1 P · ρ P ′)(237)= (f • B f ′ ) ρ P •B P ′ (p P,P ′ · 1 C)ρ Q•B Q ′ (f • B f ′ · 1 C ) = (f • B f ′ ) ρ P •B P ′i.e. f • B f ′ is also a morphism of right C-modules.□Proposition 1<strong>1.</strong>1<strong>2.</strong> For any monad (Y, B) in C, the composition denoted by • B iscompatible with the vertical canonical composition.Proof. Let (X, A) , (Y, B) , (Z, C) be monads in C, let (P, λ P , ρ P ) , (Q, λ Q , ρ Q ) ,(W, λ W , ρ W ) be A-B-bimodules in C, let (P ′ , λ P ′, ρ P ′) , (Q ′ , λ Q ′, ρ Q ′) , (W ′ , λ W ′, ρ W ′)be B-C-bimodules in C and let f : P → Q, g : Q → W be A-B-bimodule morphisms,f ′ : P ′ → Q ′ , g ′ : Q ′ → W ′ be B-C-bimodule morphisms in C. By Proposition 1<strong>1.</strong>11we can consider the A-C-bimodule morphisms f • B f ′ : P • B P ′ → Q • B Q ′ andg • B g ′ : Q • B Q ′ → W • B W ′ and we can compose them in order to get(g • B g ′ ) (f • B f ′ ) : P • B P ′ → W • B W ′ .

230On the other hand, we can first consider the canonical vertical composites gf :P → W and g ′ f ′ : P ′ → W ′ , which are still bimodule morphisms, and then we cancompose them horizontally gettingWe have to prove thatLet us consider the following diagramsand(gf) • B (g ′ f ′ ) : P • B P ′ → W • B W ′ .(g • B g ′ ) (f • B f ′ ) = (gf) • B (g ′ f ′ ) .P · B · P ′f·1 B·f ′ ρ P ·1 P ′ 1 P ·λ P ′Q · B · Q ′ ρ Q·1 Q ′g·1 B·g ′ (gf)·1 B·(g ′ f ′ )1 Q·λ Q ′P · P ′ p P,P ′f·f ′Q · Q ′ p Q,Q ′g·g ′P • B P ′f• B f ′Q • B Q ′g• B g ′W · B · W ′ρ W ·1 W ′W · W ′ p W,W ′ W • B W ′P · B · P ′1 W ·λ W ′ρ P ·1 P ′ 1 P ·λ P ′P · P ′ p P,P ′(gf)·(g ′ f ′ )P • B P ′W · B · W ′ρ W ·1 W ′W · W ′ p W,WW •′ B W ′1 W ·λ W ′(gf)• B (g ′ f ′ )We have to prove that (gf) • B (g ′ f ′ ) makes the external square of the first diagramcommutative. Since Bim (C) is a bicategory, in particular we have that (gf)·(g ′ f ′ ) =(g · g ′ ) (f · f ′ ) so that, by the commutativity of the first diagram, we deduce thatalso the left square of the second one commutes, i.e.[(gf) · (g ′ f ′ )] (ρ P · 1 P ′) = (ρ W · 1 W ′) [(gf) · 1 B · (g ′ f ′ )][(gf) · (g ′ f ′ )] (1 P · λ P ′) = (1 W · λ W ′) [(gf) · 1 B · (g ′ f ′ )] .Therefore, the exists the unique 2-cell (gf) • B (g ′ f ′ ) : P • B P ′ → W • B W ′ such thatThen we have[(gf) • B (g ′ f ′ )] p P,P ′ = p W,W ′ [(gf) · (g ′ f ′ )] .[(gf) • B (g ′ f ′ )] p P,P ′ = p W,W ′ [(gf) · (g ′ f ′ )] = p W,W ′ [(g · g ′ ) (f · f ′ )]= p W,W ′ (g · g ′ ) (f · f ′ ) = (g • B g ′ ) (f • B f ′ ) p P,P ′and since p P,P ′ is an epimorphism, we get that(gf) • B (g ′ f ′ ) = (g • B g ′ ) (f • B f ′ ) .Definition 1<strong>1.</strong>13. The bicategory BIM (C) consists of• 0-cells are monads in C□

• 1-cells are bimodules in C together with their horizontal composition definedas follows. Let (X, A) , (Y, B) and (W, C) be monads in C and let Q : Y → Xand Q ′ : W → Y be respectively an A-B-bimodule with (Q, λ Q , ρ Q ) and aB-C-bimodule in C with (Q ′ , λ Q ′, ρ Q ′). Then the horizontal composition ofthe two bimodules is given by (Q • B Q ′ , p Q,Q ′) = Coequ C (ρ Q · 1 Q ′, 1 Q · λ Q ′)[Note that Q • B Q ′ is an A-C-bimodule in C by Proposition 1<strong>1.</strong>5. Moreover,such horizontal composition is weakly associative and unital by Propositions1<strong>1.</strong>7 and 1<strong>1.</strong>6.]• 2-cells are bimodule morphisms in C.Example 1<strong>1.</strong>14. Let us consider the bicategory SetMat as defined in [RW, <strong>2.</strong>1].The objects of this bicategory are sets, denoted by A, B, . . .. An arrow (1-cell)M : A → B is a set valued matrix which, to fix notation ,has entries M (a, b) forevery a ∈ A and b ∈ B. A 2-cell f : M → N : A → B is a matrix of functionsf (a, b) : M (a, b) → N (a, b). Moreover, for A −→ MB −→ LC we have L · M : A → Cdefined by(L · M) (a, c) = ∑ b∈BL (b, c) × M (a, b) .A monad in SetMat on an object A is thus a pair (A, M) where A is a set andM : A → A is a matrix whose entries are M (a, b) for every a, b ∈ A, i.e. itis a small category with set of objects A. Hence, a monad functor is a functorF : (A, M) → (B, N) where A and B are the sets of objects of the small categoriesM and N. Note that, since F is a functor between categories, F is just a mapF : A → B at the level of objects. This map induces a 1-cell Q F : A → B definedas follows{}∅ if b ≠ F (a)Q F (a, b) =.{(a, F (a))} if b = F (a)Moreover, we can consider the following 2-cell φ F : Q F A → BQ F defined, for every(a, b) ∈ A × B, by the mapnote thatQ F A (a, b) = ∑ a ′ ∈Aφ F (a, b) : Q F A (a, b) → BQ F (a, b) .Q F (a ′ , b) × A (a, a ′ ) =⋃a ′ ∈F ← (b)where F ← (b) = {a ∈ A | F (a) = b}. Similarly we have{(a ′ , F (a ′ ))} × A (a, a ′ )BQ F (a, b) = ∑ b ′ ∈BB (b ′ , b) × Q F (a, b ′ ) = B (F (a) , b) × {(a, F (a))} .We can identify the set⋃{(a ′ , F (a ′ ))} × A (a, a ′ ) = Q F A (a, b) =anda ′ ∈F ← (b)⋃a ′ ∈F ← (b)B (F (a) , b) × {(a, F (a))} = BQ F (a, b) = B (F (a) , b)A (a, a ′ )231

232so that we define the map φ F (a, b) : Q F A (a, b) =⋃a ′ ∈F ← (b)A (a, a ′ ) → BQ F (a, b) =B (F (a) , b) = B (F (a) , F (a ′ )). Clearly, such a map is induced by the matrix mapA (a, a ′ ) → B (F (a) , F (a ′ ))f ↦→ F (f) .Since F is a functor, F preserves composition, F (g ◦ f) = F (g) ◦ F (f) , i.e. F iscompatible with respect to the multiplications of the monads (A, M) and (B, N),and F preserves the identity, F (1 a ) = 1 F (a) , i. e. F is compatible with respect to theunits of the monads (A, M) and (B, N). Hence we get that F is a monad functor.Let now F, G : (A, M) → (B, N) be monad functors and let χ : (F, φ F ) → ( G, φ G)be a functor transformation. Then we have that χ : Q F → Q G is defined by setting{∅{(a, F (a))}Then, we have Q F (a, b) Q F (f)−→ Q F (a ′ , b ′ )χ (a, b) : Q F (a, b) → Q G (a, b)} {if b ≠ F (a)∅↦→if b = F (a) {(a, G (a))}}if b ≠ G (a).if b = G (a){∅{(a, F (a))}Q F (a, b) Q F (f)−→ Q F (a ′ , b ′ )} {}if b ≠ F (a)∅ if b↦→′ ≠ F (a ′ )if b = F (a) {(a ′ , F (a ′ ))} if b ′ = F (a ′ )and Q G (a, b) Q G(f)−→ Q G (a ′ , b ′ ) we have thati.e.χ is a monad functor transformation.Now, let us define the following mapχ (a ′ , b ′ ) (Q F (f)) = Q G (f) (χ (a, b))F (C) : Mnd (C) → BIM (C)(X, A) ↦→ (X, A)(X, A) (Q,φ)→ (Y, B) ↦→ (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A )(Q, φ) σ → (P, ψ) ↦→ σ · 1 A .Proposition 1<strong>1.</strong>15. The map F defined above is well-defined and it is a pseudofunctor.Proof. First, let us prove that (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A ) is a bimodule. Infact, we haveλ Q·A (1 B · λ Q·A ) = (1 Q · m A ) (φ · 1 A ) (1 B · 1 Q · m A ) (1 B · φ · 1 A )φ= (1 Q · m A ) (1 Q · 1 A · m A ) (φ · 1 A · 1 A ) (1 B · φ · 1 A )m A ass= (1 Q · m A ) (1 Q · m A · 1 A ) (φ · 1 A · 1 A ) (1 B · φ · 1 A )(233)= (1 Q · m A ) (φ · 1 A ) (m B · 1 Q · 1 A ) = λ Q·A (m B · 1 Q · 1 A )

233andλ Q·A (u B · 1 Q · 1 A ) = (1 Q · m A ) (φ · 1 A ) (u B · 1 Q · 1 A )(234)= (1 Q · m A ) (1 Q · u A · 1 A ) m Aunit= 1 Q · 1 A .For the right A-module structure, we haveandFinally, we computeρ Q·A (ρ Q·A · 1 A ) = (1 Q · m A ) (1 Q · m A · 1 A )m A ass= (1 Q · m A ) (1 Q · 1 A · m A ) = ρ Q·A (1 Q · 1 A · m A )ρ Q·A (1 Q · 1 A · u A ) = (1 Q · m A ) (1 Q · 1 A · u A ) m Aunit= 1 Q · 1 A .ρ Q·A (λ Q·A · 1 A ) = (1 Q · m A ) (1 Q · m A · 1 A ) (φ · 1 A · 1 A )m A ass= (1 Q · m A ) (1 Q · 1 A · m A ) (φ · 1 A · 1 A )φ= (1 Q · m A ) (φ · 1 A ) (1 B · 1 Q · m A ) = λ Q·A (1 B · ρ Q·A )so that (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A ) is a B-A-bimodule. Now, let us considerthe identity object (X, 1 X ) ∈ Mnd (C) . Then F ((X, 1 X )) = (X, 1 X ) which is anidentity object in BIM (C). Now, let us consider the composite of 1-cells in Mnd (C)We have to prove that(X, A) (Q,φ)−→ (Y, B) (P,ψ)−→ (Z, C) .F ((P, ψ) (Q, φ)) ≃ F ((P, ψ)) • B F ((Q, φ)) .We have that (P, ψ) (Q, φ) = (P · Q, (1 P · φ) (ψ · 1 Q )) where P · Q : X → Z and(1 P · φ) (ψ · 1 Q ) : C · P · Q → P · Q · A. Then we haveF ((P, ψ) (Q, φ)) = F ((P · Q, (1 P · φ) (ψ · 1 Q )))= (P · Q · A, (1 P · 1 Q · m A ) (1 P · φ · 1 A ) (ψ · 1 Q · 1 A ) , 1 P · 1 Q · m A ) .On the other hand, we haveand thusF ((P, ψ)) = (P · B, (1 P · m B ) (ψ · 1 B ) , 1 P · m B )F ((Q, φ)) = (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A )F ((P, ψ)) • B F ((Q, φ)) = (P · B) • B (Q · A) .By definition of ((P · B) • B (Q · A) , p P ·B,Q·A ) = Coequ C (ρ P ·B · 1 Q · 1 A , 1 P · 1 B · λ Q·A )we have the following diagramP · B · B · Q · A ρ P ·B·1 Q·1 A1 P ·1 B·λ Q·A P · B · Q · A p P ·B,Q·A (P · B) • B (Q · A)Note that, ρ P ·B = 1 P · m B so that we can rewrite it in the following way P · B · B · Q · A 1 P ·m B·1 Q·1 AP · B · Q · A p P ·B,Q·A (P · B) • B (Q · A)1 P ·1 B·λ Q·A

234Since we have(B • B (Q · A) , p B,Q·A ) = Coequ C (m B · 1 Q · 1 A , 1 B · λ Q·A )and we are assuming that the composition with 1-cells preserves coequalizer, we alsohave(P · (B • B (Q · A)) , 1 P · p B,Q·A ) = Coequ C (1 P · m B · 1 Q · 1 A , 1 P · 1 B · λ Q·A ) .Therefore, there exists a unique isomorphism h : (P · B)• B (Q · A) → P ·(B • B (Q · A))such that(248) h (p P ·B,Q·A ) = 1 P · p B,Q·A .Moreover, by Proposition 1<strong>1.</strong>6, P · (B • B (Q · A)) ≃ P · (Q · A) so that we get(P · B) • B (Q · A) ≃ P · (Q · A) = P · Q · A.Now, the left C-module structure λ (P ·B)•B (Q·A) of (P · B) • B (Q · A), by (236) isuniquely determined byλ (P ·B)•B (Q·A) (1 C · p P ·B,Q·A ) = p P ·B,Q·A (λ P ·B · 1 Q · 1 A ) .By (248) we getp P ·B,Q·A = h −1 (1 P · p B,Q·A )and thus we can rewrite the above relation(λ (P ·B)•B (Q·A) (1 C · p P ·B,Q·A ) = λ (P ·B)•B (Q·A) 1C · [h −1 (1 P · p B,Q·A ) ])andso that= λ (P ·B)•B (Q·A)(1C · h −1) (1 C · 1 P · p B,Q·A )p P ·B,Q·A (λ P ·B · 1 Q · 1 A ) = h −1 (1 P · p B,Q·A ) (λ P ·B · 1 Q · 1 A )defλ P ·B= h −1 (1 P · p B,Q·A ) (λ P · 1 B · 1 Q · 1 A )λ=Ph ( ) −1 λ P · 1 B•B (Q·A) (1C · 1 P · p B,Q·A )λ (P ·B)•B (Q·A)(1C · h −1) (1 C · 1 P · p B,Q·A ) = h −1 ( λ P · 1 B•B (Q·A))(1C · 1 P · p B,Q·A ) .Since 1 C · 1 P · p B,Q·A is epi, we get(λ (P ·B)•B (Q·A) 1C · h −1) = h ( )−1 λ P · 1 B•B (Q·A)and thusso that we get thatλ (P ·B)•B (Q·A) = h −1 ( λ P · 1 B•B (Q·A))(1C · h)λ (P ·B)•B (Q·A) ≃ λ P · 1 B•B (Q·A) ≃ λ P · 1 Q·A ≃ λ P ·Q·A .Similarly, the right A-module structure λ (P ·B)•B (Q·A) of (P · B) • B (Q · A), by (237)is uniquely determined byBy (248) we getρ (P ·B)•B (Q·A) (p P ·B,Q·A · 1 A ) = p P ·B,Q·A (1 P · 1 B · ρ Q·A ) .p P ·B,Q·A = h −1 (1 P · p B,Q·A )

and thus we can rewrite the above relation([ρ (P ·B)•B (Q·A) (p P ·B,Q·A · 1 A ) = ρ (P ·B)•B (Q·A) h −1 (1 P · p B,Q·A ) ] )· 1 Aandso that= ρ (P ·B)•B (Q·A)(h−1 · 1 A)(1P · p B,Q·A · 1 A )p P ·B,Q·A (1 P · 1 B · ρ Q·A ) = h −1 (1 P · p B,Q·A ) (1 P · 1 B · ρ Q·A )(237)= h −1 ( 1 P · ρ B•B (Q·A))(1P · p B,Q·A · 1 A )ρ (P ·B)•B (Q·A)(h−1 · 1 A)(1P · p B,Q·A · 1 A ) = h −1 ( 1 P · ρ B•B (Q·A))(1P · p B,Q·A · 1 A ) .Since 1 P · p B,Q·A · 1 A is epi, we get( )ρ (P ·B)•B (Q·A) h−1 · 1 ( )A = h−11 P · ρ B•B (Q·A)and thusso that we get thatρ (P ·B)•B (Q·A) = h −1 ( 1 P · ρ B•B (Q·A))(h · 1A )ρ (P ·B)•B (Q·A) ≃ 1 P · ρ B•B (Q·A) ≃ 1 P · ρ Q·A = ρ P ·Q·A .1<strong>2.</strong> Entwined modules and comodulesLet (X, 1 X ) , (Y, B) be monads in C and let us compute the categoryMnd (C) ((X, 1 X ) , (Y, B)) . Note that C (X, B) : C (X, Y ) → C (X, Y ) is a monadover the category C (X, Y ) . In fact, we set multiplication and unit of the monadto be C (X, m B ) = m B (−) : C (X, B · B) → C (X, B) and C (X, u B ) = u B (−) :C (X, 1 Y ) → C (X, B). In fact we haveandC (X, m B ) C (X, m B · 1 B ) = m B (m B · 1 B ) = m B (1 B · m B )= C (X, m B ) C (X, 1 B · m B )C (X, m B ) C (X, u B · 1 B ) = m B (u B · 1 B ) = 1 B = m B (1 B · u B )= C (X, m B ) C (X, 1 B · u B )The objects of such category are the monad functors (Q, φ) from (X, 1 X ) to (Y, B) ,i.e. the 1-cells Q : X → Y together with the 2-cells φ : B · Q = C (X, B) Q → Qsatisfying the following conditionsφ (1 B · φ) = φ (m B · 1 Q )φ (u B · 1 Q ) = 1 Qwhich says that φ gives a structure of left C (X, B)-module to the 1-cell Q : X → Y .Therefore, we can conclude thatMnd (C) ((X, 1 X ) , (Y, B)) = C(X,B) C (X, Y ).Now, following [St, pg. 158], we define the bicategory of comonads as follows:Cmd (C) = Mnd (C ∗ ) ∗where (−) ∗denotes the bicategory obtained by reversing235□

2362-cells. This means that a comonad (X, C) in C is a 1-cell C : X → X togetherwith 2-cells ∆ C : C → C · C and ε C : C → 1 X called comultiplication and counitsatisfying the reversed diagrams, i.e.(1C · ∆ C) ∆ C = ( )∆ C · 1 C ∆C(249)(1C · ε C) ∆ C = 1 C = ( )(250)ε C · 1 C ∆ C .A comonad functor is a pair (P, ψ) : (X, C) → (Y, D) where P : X → Y is a 1-cellin C and ψ : P · C → D · P is a 2-cell in C satisfying(εD · 1 P)ψ = 1P · ε C and (1 D · ψ) (ψ · 1 C ) ( 1 P · ∆ C) = ( ∆ D · 1 P)ψ.Finally, a comonad functorial morphism ω : (P ′ , ψ ′ ) → (P, ψ) is ω : P ′ → P is a2-cell in C satisfyingψ (ω · 1 C ) = (1 D · ω) ψ ′ .Now, we consider the category Cmd (C) ((X, 1 X ) , (Y, C)) where (X, 1 X ) and (Y, C)are 0-cells in Cmd (C) respectively with trivial comultiplication and counit theformer and ∆ C , ε C the latter. Note that C (X, C) : C (X, Y ) → C (X, Y ) is acomonad over the category C (X, Y ) with comultiplication and counit given byC ( X, ∆ C) = ∆ C () : C (X, C) → C (X, C · C) and C ( X, ε C) = ε C () : C (X, C) →C (X, 1 Y ). The objects of such category are the comonad functors (Q, ψ) : (X, 1 X ) →(Y, C) where Q : X → Y is a 1-cell and ψ : Q · 1 X → C · Q = C (X, C)Q is a 2-cellsatisfying ( ε C · 1 Q)ψ = 1Q and ( ∆ C · 1 Q)ψ = (1C · ψ) ψ so thatCmd (C) ((X, 1 X ) , (Y, C)) = C(X,C) C (X, Y ).Following the definition of the 2-category Mnd (C) for any bicategory C, we canconsider the 2-categories Mnd (Mnd (C)) and Mnd (BIM (C)) and the functorbetween themMnd (F (C)) : Mnd (Mnd (C)) → Mnd (BIM (C)) .Let us first consider Mnd (Mnd (C)):• 0-cells: pairs ((X, A) , (Q, φ)) where (X, A) is an object in Mnd (C) and(Q, φ) is a 1-cell in Mnd (C) together with a pair of 2-cells in Mnd (C) m (Q,φ)and u (Q,φ) satisfying associativity and unitality conditions. Therefore we havethat A : X → X is a 1-cell in C together with 2-cells m A = m (X,A) : A·A → Aand u A = u (X,A) : 1 X → A satisfying associativity and unitality conditionsand we have that Q : X → X is a 1-cell in C together with the 2-cell of Cφ : A · Q → Q · A satisfying the following conditions(251)(252)(253)(254)φ (m A · 1 Q ) = (1 Q · m A ) (φ · 1 A ) (1 A · φ)φ (u A · 1 Q ) = 1 Q · u AFinally, the 2-cells of Mnd (C)m (Q,φ) : (Q, φ) · (Q, φ) = (Q · Q, (1 Q · φ) (φ · 1 Q )) → (Q, φ) and u (Q,φ) :(1 X , 1 1X ) → (Q, φ) satisfying the associativity and unitality conditions, needsto satisfy also the followingφ ( 1 A · m (Q,φ))=(m(Q,φ) · 1 A)(1Q · φ) (φ · 1 Q )φ ( 1 A · u (Q,φ))= u(Q,φ) · 1 A .

An object, or 0-cell, of Mnd (Mnd (C)) is called distributive law and it givesrise to a monad structure on Q·A. In fact, the monad functor transformationm (Q,φ) induces a multiplication on Q · Adefined by settingm Q·A : Q · A · Q · A → Q · Am Q·A = ( m (Q,φ) · m A)(1Q · φ · 1 A ) = ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) (1 Q · φ · 1 A ) .Using naturality and associativity of m (Q,φ) , naturality of φ, associativity ofm A , we havem Q·A (m Q·A · 1 Q · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) (1 Q · φ · 1 A ) ( m (Q,φ) · 1 A · 1 Q · 1 A)(1 Q · 1 Q · m A · 1 Q · 1 A ) (1 Q · φ · 1 A · 1 Q · 1 A )= ( m (Q,φ) · 1 A) (m(Q,φ) · 1 Q · 1 A)(1Q · 1 Q · 1 Q · m A ) (1 Q · 1 Q · φ · 1 A )(1 Q · 1 Q · m A · 1 Q · 1 A ) (1 Q · φ · 1 A · 1 Q · 1 A )= ( m (Q,φ) · 1 A) (1Q · m (Q,φ) · 1 A)(1Q · 1 Q · 1 Q · m A ) (1 Q · 1 Q · φ · 1 A )(1 Q · 1 Q · m A · 1 Q · 1 A ) (1 Q · φ · 1 A · 1 Q · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) ( 1 Q · m (Q,φ) · 1 A · 1 A)(1Q · 1 Q · φ · 1 A )(1 Q · 1 Q · m A · 1 Q · 1 A ) (1 Q · φ · 1 A · 1 Q · 1 A )(251)= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) ( 1 Q · m (Q,φ) · 1 A · 1 A)(1Q · 1 Q · 1 Q · m A · 1 A )(1 Q · 1 Q · φ · 1 A · 1 A ) (1 Q · 1 Q · 1 A · φ · 1 A ) (1 Q · φ · 1 A · 1 Q · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) (1 Q · 1 Q · m A · 1 A ) ( 1 Q · m (Q,φ) · 1 A · 1 A · 1 A)(1 Q · 1 Q · φ · 1 A · 1 A ) (1 Q · φ · 1 Q · 1 A · 1 A ) (1 Q · 1 A · 1 Q · φ · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) (1 Q · 1 Q · 1 A · m A ) ( 1 Q · m (Q,φ) · 1 A · 1 A · 1 A)(1 Q · 1 Q · φ · 1 A · 1 A ) (1 Q · φ · 1 Q · 1 A · 1 A ) (1 Q · 1 A · 1 Q · φ · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) ( 1 Q · m (Q,φ) · 1 A · 1 A)(1Q · 1 Q · 1 Q · 1 A · m A )(1 Q · 1 Q · φ · 1 A · 1 A ) (1 Q · φ · 1 Q · 1 A · 1 A ) (1 Q · 1 A · 1 Q · φ · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) ( 1 Q · m (Q,φ) · 1 A · 1 A)(1Q · 1 Q · φ · 1 A )(1 Q · 1 Q · 1 A · 1 Q · m A ) (1 Q · φ · 1 Q · 1 A · 1 A ) (1 Q · 1 A · 1 Q · φ · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) ( 1 Q · m (Q,φ) · 1 A · 1 A)(1Q · 1 Q · φ · 1 A ) (1 Q · φ · 1 Q · 1 A )(1 Q · 1 A · 1 Q · 1 Q · m A ) (1 Q · 1 A · 1 Q · φ · 1 A )(253)= ( )m (Q,φ) · 1 A (1Q · 1 Q · m A ) (1 Q · φ · 1 A ) ( )1 Q · 1 A · m (Q,φ) · 1 A(1 Q · 1 A · 1 Q · 1 Q · m A ) (1 Q · 1 A · 1 Q · φ · 1 A )so that m Q·A is associative.u (Q,φ) induces a unit of Q · A= m Q·A (1 Q · 1 A · m Q·A )237Similarly, the monad functor transformationu Q·A : 1 X → Q · A

238defined by settingu Q·A = ( u (Q,φ) · 1 A)uA .Using naturality of u (Q,φ) , unitality of m (Q,φ) and m A we havem Q·A (u Q·A · 1 Q · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) (1 Q · φ · 1 A ) ( u (Q,φ) · 1 A · 1 Q · 1 A)(uA · 1 Q · 1 A )= ( m (Q,φ) · 1 A) (u(Q,φ) · 1 Q · 1 A)(1Q · m A ) (φ · 1 A ) (u A · 1 Q · 1 A )(252)= (1 Q · m A ) (1 Q · u A · 1 A ) = 1 Q · 1 A .so that we have a monad(Q · A, m Q·A , u Q·A ) = ( Q · A, ( )m (Q,φ) · m A (1Q · φ · 1 A ) , ( ) )u (Q,φ) · 1 A uA . Wewill see that such a monad is taken to an A-ring in the bimodule category.• 1-cells: pairs ((U, ϕ) , τ) : ((X, A) , (Q, φ)) → ((Y, B) , (P, ψ)) where (U, ϕ) :(X, A) → (Y, B) is a 1-cell in Mnd (C), i.e. a monad functor where ϕ :B · U → U · A satisfies ϕ (u B · 1 U ) = 1 U · u A and (1 U · m A ) (ϕ · 1 A ) (1 B · ϕ) =ϕ (m B · 1 U ), and τ is 2-cell in Mnd (C), i.e. a monad functor transformationτ : (P, ψ) (U, ϕ) → (U, ϕ) (Q, φ), i.e. τ : (P · U, (1 P · ϕ) (ψ · 1 Q )) →(U · Q, (1 U · φ) (ϕ · 1 Q )) satisfying(1 U · φ) (ϕ · 1 Q ) (1 B · τ) = (τ · 1 A ) (1 P · ϕ) (ψ · 1 Q ) .• 2-cells: σ : ((U, ϕ) , τ) → ((U ′ , ϕ ′ ) , τ ′ ) where σ : (U, ϕ) → (U ′ , ϕ ′ ) is a 2-cellin Mnd (C) i.e.ϕ ′ (1 B · σ) = (σ · 1 A ) ϕ,satisfyingLet us now consider Mnd (BIM (C)):τ ′ (1 P · σ) = (σ · 1 Q ) τ• 0-cells: pairs ((X, A) , (Q, λ Q , ρ Q )) where (X, A) is an object in BIM (C),i.e. a monad in C and (Q, λ Q , ρ Q ) : (X, A) → (X, A) is a 1-cell in BIM (C)together with 2-cells in BIM (C), i.e. (Q, λ Q , ρ Q ) is an A-bimodule in Ctogether with bimodule morphisms m (Q,λQ ,ρ Q) : Q• A Q → Q and u (Q,λQ ,ρ Q) :1 X → Q satisfying associativity and unitality conditions)()m (Q,λQ ,ρ Q)(m (Q,λQ ,ρ Q) • A 1 Q = m (Q,λQ ,ρ Q)1 Q • A m (Q,λQ ,ρ Q))()m (Q,λQ ,ρ Q)(u (Q,λQ ,ρ Q) • A 1 Q = 1 Q = m (Q,λQ ,ρ Q)1 Q • A u (Q,λQ ,ρ Q)• 1-cells: pairs ((U, λ U , ρ U ) , δ) : ((X, A) , (Q, λ Q , ρ Q )) → ((Y, B) , (P, λ P , ρ P ))where (U, λ U , ρ U ) : (X, A) → (Y, B) is a 1-cell in BIM (C) andδ : (P, λ P , ρ P ) (U, λ U , ρ U ) = P • B U → (U, λ U , ρ U ) (Q, λ Q , ρ Q ) = U • A Qsatisfies the following conditionsδ ( )u (P,λP ,ρ P ) • B 1 U = 1U • A u (Q,λQ ,ρ Q) and()1 U • A m (Q,λQ ,ρ Q)(δ • A 1 Q ) (1 P • B δ) = δ ( )1 U • B m (P,λP ,ρ P )

• 2-cells: ω : ((U, λ U , ρ U ) , δ) → ((U ′ , λ U ′, ρ U ′) , δ ′ ) where ω : (U, λ U , ρ U ) →(U ′ , λ U ′, ρ U ′) is a 2-cell in BIM (C), i.e. it is a B-A-bimodule morphism,satisfyingδ ′ (1 P • B ω) = (ω • A 1 Q ) δ.Now, let us apply the functorMnd (F (C)) : Mnd (Mnd (C)) → Mnd (BIM (C))to the distributive law ((X, A) , (Q, φ)). We getMnd (F (C)) (((X, A) , (Q, φ))) = (F (C) (X, A) , F (C) (Q, φ))where= ((X, A) , (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A ))Mnd (F (C)) ( m (Q,φ))= F (C)(m(Q,φ))= m(Q,φ) · 1 AMnd (F (C)) ( u (Q,φ))= F (C)(u(Q,φ))= u(Q,φ) · 1 AMnd (F (C)) ( m (Q,φ)):Mnd (F (C)) (Q · Q)= Q · Q · A−→Mnd (F (C)) (Q)= Q · AMnd (F (C)) ( u (Q,φ)): Mnd (F (C)) (1X ) = A −→ Mnd (F (C)) (Q) = Q · A.In particular, (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A ) comes together with bimodule morphismsMnd (F (C)) ( m (Q,φ))= m(Q,φ) · 1 A and Mnd (F (C)) ( u (Q,φ))= u(Q,φ) · 1 A .Note thatMnd (F (C)) (Q · Q) = Q · Q · A ≃ (Q · A) • A (Q · A)= Mnd (F (C)) (Q) • A Mnd (F (C)) (Q) .where the isomorphism is given by the following: by definition,((Q · A) • A (Q · A) , p Q·A,Q·A ) = Coequ C (1 Q · m A · 1 Q · 1 A , 1 Q · 1 A · λ Q·A )and since we are assuming that the coequalizers are preserved, by Lemma 1<strong>1.</strong>4 wealso have(Q · Q · A, 1 Q · λ Q·A ) = Coequ C (1 Q · m A · 1 Q · 1 A , 1 Q · 1 A · λ Q·A )so that there exists a unique isomorphismsuch thatα : (Q · A) • A (Q · A) → Q · Q · A(255) αp Q·A,Q·A = 1 Q · λ Q·A = (1 Q · 1 Q · m A ) (1 Q · φ · 1 A ) .Recall that we can consider the monad(Q · A, m Q·A , u Q·A ) = ( Q · A, ( m (Q,φ) · m A)(1Q · φ · 1 A ) , ( u (Q,φ) · 1 A)uA)and thusm Q·A = ( m (Q,φ) · m A)(1Q · φ · 1 A )= ( m (Q,φ) · 1 A)(1Q · 1 Q · m A ) (1 Q · φ · 1 A )= ( m (Q,φ) · 1 A)αpQ·A,Q·Athat is, m Q·A factorizes through (Q · A) • A (Q · A) and we denote by(256) m (Q·A,(1 Q·m A)(φ·1 A ),1 Q·m A) = ( m (Q,φ) · 1 A)α239

240the unique A-bimodule morphism such thatm Q·A = m (Q·A,(1 Q·m A)(φ·1 A ),1 Q·m A) p Q·A,Q·A.Note that also u (Q,φ) · 1 A is an A-bimodule morphism, in factλ Q·A(1A · u (Q,φ) · 1 A)= (1Q · m A ) (φ · 1 A ) ( 1 A · u (Q,φ) · 1 A)andTherefore,(252)= (1 Q · m A ) ( u (Q,φ) · 1 A · 1 A) u (Q,φ)= ( u (Q,φ) · 1 A)mA( )ρ Q·A u(Q,φ) · 1 A · 1 A = (1Q · m A ) ( ) u (Q,φ)u (Q,φ) · 1 A · 1 A = ( )u (Q,φ) · 1 A mA .()Q · A, m (Q·A,(1Q·m A)(φ·1 A ),1 Q·m A) , u (Q·A,(1 Q·m A)(φ·1 A ),1 Q·m A)= ( Q · A, ( ) )m (Q,φ) · 1 A α, u(Q,φ) · 1 Ais an A-ring, so that the functor Mnd (F (C)) : Mnd (Mnd (C)) → Mnd (BIM (C))associates distributive laws to A-rings.Let us now consider Cmd (Mnd (C)) :• 0-cells: ((X, A) , (C, γ)) where (X, A) is a monad, C : X → X, γ : A · C →C·A together with ∆ C : C → C·C and ε C : C → 1 X satisfying coassociativityand counitality and satisfying(1 C · γ) (γ · 1 C ) ( 1 A · ∆ C) = ( ∆ C · 1 A)γ(257)1 A · ε C = ( ε C · 1 A)γ.(258)Note that, if we consider ( C · A, ∆ C·A , ε C·A) where∆ C·A = (γ · 1 C · 1 A ) ( 1 A · ∆ C · 1 A)(uA · 1 C · 1 A ) and ε C·A : C · A → A coassociativityand counitality properties are not satisfied. But, by applying thefunctor Cmd (F (C)) to the comonad ( C, ∆ C , ε C) we getCmd (F (C)) (( C, ∆ C , ε C)) = ( C · A, ∆ C · 1 A , ε C · 1 A)where∆ C · 1 A ≃ ∆ C·A : C · A → C · A • A C · Aand ∆ C · 1 A and ε C · 1 A are A-bimodule morphisms, clearly satisfying coassociativityand counitality conditions. Hence, ( )C · A, ∆ C · 1 A , ε C · 1 A is anA-coring.SinceMnd (C) ((X, 1 X ) , (Y, B)) = C(X,B) C (X, Y )dually we getCmd (C) ((X, 1 X ) , (Y, C)) = C(X,C) C (X, Y ) .Consider the objects ( (X, 1 X ) , ( ))1 (X,1X ), 1 X , ((X, A) , (C, γ)) ∈ Cmd (Mnd (C))[(C, γ) : (X, A) → (X, A) , γ : A · C → C · A] and let((Q, φ) , σ) ∈ Cmd ( Mnd (C) ( (X, 1 X ) , ( )) )1 (X,1X ), 1 X , ((X, A) , (C, γ)) be a comonadfunctor, where (Q, φ) : (X, 1 X ) → (X, A) is a 1-cell in Mnd (C), i.e. φ : A · Q → Qsatisfies φ (u A · 1 Q ) = 1 Q and φ (1 A · φ) = φ (m A · 1 Q ) and σ : (Q, φ) ( )1 (X,1X ), 1 X =(Q, φ) → (C, γ) (Q, φ) = (C · Q, (1 C · φ) (γ · 1 Q )) is a 2-cell in Mnd (C), i.e.

(1 C · φ) (γ · 1 Q ) (1 A · σ) = σφ, i.e. σ is an A-linear map. Since ((Q, φ) , σ) is acomonad functor, the 2-cell σ : Q → C · Q satisfies ( )ε C · 1 Q σ = 1Q and (1 C · φ) φ =( )C(X,C)∆C · 1 Q φ. This means that ((Q, φ) , σ) ∈C(X,A)C (X, X) (γ) is an entwined module.By applying the functor Cmd (F (C)) : Cmd (Mnd (C)) −→ Cmd (BIM (C))to the element((Q, φ) , σ) ∈ Cmd ( Mnd (C) ( (X, 1 X ) , ( )) )1 (X,1X ), 1 X , ((X, A) , (C, γ)) we getCmd (F (C)) (((Q, φ) , σ)) ∈ Cmd ( BIM (C) ( (X, 1 X ) , ( )) )1 (X,1X ), 1 X , ((X, A) , (C, γ))which is an element of C(X,C·A) BIM (X, X), i.e. it is a left C · A-comodule with respectto • A .241Appendix A. Gabriel Popescu TheoremNotation A.<strong>1.</strong> Let A be a Grothendieck category, let U be an object of A andlet B = Hom A (U, U). Assume that U is a generator of A i.e. that the functorHom A (U, −) : A → Mod-B is faithful.Lemma A.<strong>2.</strong> In the assumptions and notations of A.1, let X ∈ A and let λ :U (Hom A(U,X)) → X be the codiagonal morphism of the family (f) f∈(HomA (U,X)) . ThenIm (λ) = X.(HomProof. Let J : Ker (λ) → U A(U,X))be the canonical monomorphism and letλ : U (HomA(U,X)) → X be the codiagonal morphism of the family (f) f∈(HomA (U,X))and, for every f ∈ Hom A (U, X) let i f : U → U (HomA(U,X)) the f-th canonicalinjection. Then we have λ ◦ i f = f. Let χ : X → Coker (λ) be the canonicalprojection and let us assume that χ ≠ 0. Then there exists h : U → X such thatχ ◦ h ≠ 0. Then we have0 ≠ χ ◦ h = χ ◦ λ ◦ i h = 0 ◦ i h = 0.Contradiction. Thus Coker (λ) = 0 and hence X = KerCoker (λ) = Im (λ).□Proposition A.3. In the assumptions and notations of A.1, the functor Hom A (U, −) :A → Mod-B is full.Proof. Let ϕ ∈ Hom B (Hom A (U, X) , Hom A (U, Z)). We have to prove that thereexists a morphism g : X → Z such that ϕ = Hom A (U, g). For any subset F ofHom A (U, X) we denote byi F : U (F ) → U (Hom A(U,X))the canonical injection. If F = {f} we will write i f instead of i {f} . Let λ :U (HomA(U,X)) → X be the codiagonal morphism of the family (f) f∈(HomA (U,X))and letµ : U (HomA(U,X)) → Z be the codiagonal morphism of the family (ϕ (f)) f∈(HomA (U,X)) .Then, for every f ∈ Hom A (U, X) we haveλ ◦ i f = f and µ ◦ i f = ϕ (f) .

242Let F be a finite subset of Hom A (U, X) and let us consider the commutative diagram0 Ker (λ F )j FU (F ) λ F Xh F0 Ker (λ)i Fj U (Hom A(U,X)) λId X Xwhere λ F : U (F ) → X is the codiagonal morphism of the family (f) f∈F, j and j Fare the canonical inclusions and h F is the morphism that factorizes i F ◦ j F throughKer (λ). We haveµ ◦ i F ◦ j F = µ ◦ j ◦ h F .For every f ∈ F let α f : U → U (F ) and π f : U (F ) → U be respectively the canonicalinjections and projections. ThenId U (F ) = ∑ f∈F α f ◦ π f .Let σ : U → Ker (λ F ) be any morphism. We compute0 = λ F ◦ j F ◦ σ = λ F ◦ Id U (F ) ◦ j F ◦ σ = ∑ f∈F λ F ◦ α f ◦ π f ◦ j F ◦ σ= ∑ f∈F f ◦ π f ◦ j F ◦ σ.Since π f ◦ j F ◦ σ ∈ B = Hom A (U, U) and ϕ ∈ Hom B (Hom A (U, X) , Hom A (U, Z)) ,we get that(∑))0 = ϕ f ◦ π f ◦ j F ◦ σ =f∈F(∑f∈F ϕ (f) ◦ π f ◦ j F ◦ σand hence, since U is a generator of A, we get that∑f∈F ϕ (f) ◦ π f ◦ j F = 0.On the other hand we have thatµ ◦ i F ◦ α f = µ ◦ i f = ϕ (f)and hence we obtain0 = ∑ )µ ◦ i F ◦ α f ◦ π f ◦ j F = µ ◦ i F ◦f∈F(∑f∈F α f ◦ π f ◦ j F= µ ◦ i F ◦ j F = µ ◦ j ◦ h F .Let u : Ker (µ) → U (HomA(U,X)) be the canonical injection.morphism β F : Ker (λ F ) → Ker (µ) such thatj ◦ h F = u ◦ β FThen there exists aand since both j and h F are mono, also β F is mono. We want to check that thefamily (β F ) F ⊆HomA (U,X) is compatible. For every F, G ⊆ Hom A (U, X) finite subsets,let us denote i G F : U (F ) → U (G) . Thus we have λ G ◦ i G F = λ F and0 = λ F ◦ j F = λ G ◦ i G F ◦ j F .

Since j G : Ker (λ G ) → U (G) there exists a unique morphism i ̂GF : Ker (λ F ) → Ker (λ G )such thati G F ◦ j F = j G ◦ i ̂GF .We want to prove that β G ◦ ̂ i G F = β F for every F, G finite subsets of Hom A (U, X) .Let us computeu ◦ β G ◦ i ̂GF= j ◦ h G ◦ i ̂GF = i G ◦ j G ◦ i ̂GF = i G ◦ i G F ◦ j F = i F ◦ j F= j ◦ h F = u ◦ β F .Since u is mono we conclude. Let us consider the exact sequence0 → Ker (λ F ) j F−→ U (F ) λ−→FXSince A is a Grothendieck category, we have that lim → are exact and hence we getthe exact sequence0 → lim Ker (λ F ) lim→ j F−→ lim U (F ) = U (Hom A(U,X)) lim→ λ F =λ−→ X.→ →It follows that Ker (λ) = lim → Ker (λ F ) and hence there exists a unique monomorphismβ = lim → β F : lim → Ker (λ F ) = Ker (λ) → Ker (µ) such thatβ ◦ h F = β F for every finite subset F of Hom A (U, X) .Since for every finite subset F of Hom A (U, X)243we get thatand henceu ◦ β ◦ h F = u ◦ β F = j ◦ h Fu ◦ β = jµ ◦ j = µ ◦ u ◦ β = 00 Ker (λ)j U (Hom A(U,X))λXtp∼ k∼Coker (j) Ker (χ)eµZλχ=0 Coker (λ)Therefore there exists a unique morphism ˜µ : Im (λ) ≃ Coker (j) → Z such that˜µ◦p = µ where p : U (Hom A(U,X)) → Coker (j) is the canonical projection. By LemmaA.2, we have that Im (λ) = X and then X = Im (λ) = Coker (j). Then there existsan isomorphism t : X → Coker (j) such that t ◦ λ = p. Set g = ˜µ ◦ t and for everyf ∈ Hom A (U, X), we computeg ◦ f = ˜µ ◦ t ◦ f = ˜µ ◦ t ◦ λ ◦ i f = ˜µ ◦ p ◦ i f = µ ◦ i f = ϕ (f) .This means that ϕ = Hom A (U, g).□

244Lemma A.4. Let A be an abelian category and let U ∈ A. Then, for every exactsequence in Athe sequence0 → Hom A (U, K) Hom A(U,k)−→0 → K k−→ Xis an exact sequence of abelian groups.f−→ Y,Hom A (U, X) Hom A(U,f)−→ Hom A (U, Y )Proof. Let h ∈ Hom A (U, K). Then Hom A (U, k) (h) = kh. Since k is a monomorphismit follows that kh = 0 if and only if h = 0. Hence Hom A (U, k) is also amonomorphism. Also (Hom A (U, f) ◦ Hom A (U, k)) (h) = fkh = 0. This impliesthat Im (Hom A (U, k)) ⊆ Ker (Hom A (U, f)). Let now g ∈ Hom A (U, X) and assumethat Hom A (U, f) (g) = 0 i.e. fg = 0. Since (K, k) = Ker (f) there exists a morphismg ′ : U → K such that g = kg ′ = Hom A (U, k) (g ′ ) ∈ Im (Hom A (U, k)). Thereforewe get that Ker (Hom A (U, f)) ⊆ Im (Hom A (U, k)) and hence Ker (Hom A (U, f)) =Im (Hom A (U, k)).□Lemma A.5. In the assumptions and notations of A.1, let (T, H) be an adjunctionwhere T : B → A and H : A → B and let f : X → Y be a morphism in A. Then fis a monomorphism (resp. epimorphism) if and only if T H (f) is a monomorphism(resp. epimorphism).Proof. First of all, for every X ∈ A let ɛX : T H (X) → X be the counit of theadjunction (T, H). Then, in view of Proposition A.3 and Proposition <strong>2.</strong>32, ɛX is anisomorphism and for every morphism f : X → Y in A we havef ◦ ɛX = ɛY ◦ T H (f) .Thus f is mono (resp. epi) if and only if T H (f) is mono (resp. epi).Lemma A.6. In the assumptions and notations of A.1, let m, n ∈ N, m, n ≥ 1,let f : B m → B n be a morphism of right B-modules, let X = Coim ( f ) and letj : X → B n be the canonical injection. Let T : Mod-B → A be a left adjoint of thefunctor Hom A (U, −) : A → Mod-B. Then T (j) is a monomorphism.Proof. Let m, n ∈ N, m, n ≥ 1, let f : U m → U n be a morphism in A. Let usconsider the diagram□0 Ker (f)k U m fU n iCoim (f)00where p is the canonical epimorphism and i is the canonical monomorphism. Byapplying to it the functor H = Hom A (U, −), in view of Lemma A.4, we obtain thep

245diagramH(k)H(f)0 Ker (H (f)) = H (Ker (f)) H (U m ) H (U n )H(p)H(i)H (Coim (f))0Since Coim (H (f)) = Coker (H (k)) and H (p) ◦ H (k) = 0 there exists a uniquemorphism ζ : Coim (H (f)) → H (Coim (f)) such that(259) H (p) = ζ ◦ qwhere q : H (U m ) → Coim (H (f)) is the canonical epimorphism. Let j : Coim (H (f)) →H (U n ) be the canonical monomorphism such that j ◦ q = H (f). Then fromj ◦ q ◦ H (k) = H (f ◦ k) = 0 we get that(260) q ◦ H (k) = 0.H(k)0 Ker (H (f)) = H (Ker (f)) H (U m ) H (UqH(p)n )H(i) jH (Coim (f)) Coim (H (f)) = Coker (H (k))ζ000From H (i) ◦ ζ ◦ q (259)= H (i) ◦ H (p) = H (f) = j ◦ q, since q is an epimorphism, weget that(261) H (i) ◦ ζ = j.Let us apply T to it having in mind that T is right exactH(f)0 T H (Ker (f))T H(k) T H (U m T H(f))T H (U n )T (q)T H(p)T H(i) T (j)ξT H (Coim (f)) T (Coim (H (f)))T (ζ)Let us prove that T (j) is mono. From formula (260) we obtain that T (q) ◦T H (k) = T (q ◦ H (k)) = 0. SinceT H (Coim (f)) = Coim (T H (f)) = Coker (Ker (T H (f)))= Coker (T H (Ker (f))) = Coker (T H (k))there exists a unique ξ : T H (Coim (f)) → T (Coim (H (f))) such that(262) ξ ◦ T H (p) = T (q) .

246We haveT (ζ) ◦ ξ ◦ T H (p) (262)= T (ζ) ◦ T (q) (259)= T H (p)and since T H (p) is epi by Lemma A.5, we getWe computeand since T (q) is epi, we obtain thatT (ζ) ◦ ξ = Id T H(Coim(f)) .ξ ◦ T (ζ) ◦ T (q) (259)= ξ ◦ T H (p) (262)= T (q)ξ ◦ T (ζ) = Id T (Coim(H(f))) .Therefore T (ζ) is an isomorphism. From formula (261) we get that T H (i) ◦ T (ζ) =T (j) and by Lemma A.5 we conclude that T (j) is a monomorphism.Let m, n ∈ N, m, n ≥ 1, let f : B m → B n be a morphism of right B-modulesand let X = Coim ( f ) . Since U is a generator, by Proposition A.3, there existsa unique morphism f : U m → U n such that H (f) = f. Then, by the foregoing,X = Coim (H (f)) and T (j) is a monomorphism where j : X → B n is the canonicalmonomorphism.□Lemma A.7. Let B be a ring and let X be a submodule of a free module A (Z) .Let P 0 (X) be the set of finite subsets of X and let j F : X F → X be the canonicalinclusion of the submodule of X spanned by F ∈ P 0 (X). Then for every F ∈ P 0 (X)there exists a finite subset Z F of Z and a monomorphism i F : X F → A (Z F ) such thatthe diagramj FX F X i Fj h FA (Z F )A (Z)where j : X → A (Z) is the canonical inclusion and h F : A (Z F ) → A (Z) is the canonicalsection of the canonical projection π F : A (Z) → A (Z F ) , is commutative. Moreover(i F ) F ∈P0 (X) is a family of morphisms between the direct systems (X F )( ) F ∈P0 (X) andA(Z F ) and (h F ∈P 0 (X) F ◦ i F ) F ∈P0 (X)is a compatible family of morphisms such thatlim (h F ◦ i F ) = j.−→Proof. Let (e z ) z∈Zbe the canonical basis of A (Z) . Then, for every x ∈ X there existsa finite subset F x of Z such thatFor every F ∈ P 0 (X) let us setx = ∑ z∈F xe z a z where a z ∈ A for every z ∈ F x .Z F = ⋃ x∈FF xand let ( e F z)z∈Z Fbe the canonical basis of A (Z F ) . Then the assignmente F z↦→ e z where z ∈ Z F

yields the canonical section h F : A (Z F ) → A (Z) of the canonical projection π F :A (Z) → A (Z F ) since π F (e z ) = e F z for every z ∈ F . Set i F = π F ◦ j ◦ j F . Then wehaveIm (j ◦ j F ) ⊆ ∑ z∈F e zA = ∑ z∈F (h F ◦ π F ) (e z ) Aand sincewe obtain that(h F ◦ π F ) (e z ) = e z for every z ∈ F(263) h F ◦ i F = h F ◦ π F ◦ j ◦ j F = j ◦ j F .Assume now that F, G ∈ P 0 (X) and that F ⊆ G. Then Z F ⊆ Z G so that wecan consider the canonical section h G F : A(Z F ) → A (ZG) of the canonical projectionπF G : A(ZG) → A (Z F ) . We have( ) ( )πFG eGz = eFz and h G F eFz = eGz for every z ∈ Z F .Moreoverh G ◦ h G F = h F .Let j G F : X F → X G be the canonical inclusion. Thenso that we getj G ◦ j G F = j F and π G ◦ h F = h G Fi G ◦ j G F = π G ◦ j ◦ j G ◦ j G F = π G ◦ j ◦ j F =(263)= π G ◦ h F ◦ i F = h G F ◦ i Fand henceh G ◦ i G ◦ jF G = h G ◦ h G F ◦ i F = h F ◦ i Fwhich proves that (h F ◦ i F ) F ∈P0 (X)is a compatible family of morphisms. Sincelim −→ (X F ) F ∈P0 (X)= X, to prove thatlim (h F ◦ i F ) = j−→it is enough to prove thatj ◦ j F = h F ◦ i Ffor every F ∈ P 0 (X) . This holds in view of (263).Lemma A.8. Let f : X → Y and p : W → X be morphisms in an abelian categoryA. Assume that p is an epimorphism and thatThen f is a monomorphism.Ker (f ◦ p) = Ker (p) .Proof. Since p is an epimorphism, we have that Coker (Ker (p)) = (X, p). It followsthat Coker (Ker (f ◦ p)) = (X, p). Let f ◦ p : Coker (Ker (f ◦ p)) → Ker (Coker (f ◦ p))be the isomorphism such that(264) f ◦ p = k ◦ f ◦ p ◦ p247□

248where k : Ker (Coker (f ◦ p)) → Y is the canonical monomorphism. Since p is anepimorphism, from formula (264) we obtain that f = k ◦ f ◦ p and hence f is amonomorphism.□Theorem A.9 ([Po, page 112]). Let A be a Grothendieck category, let U be an objectof A and let B = Hom A (U, U). Assume that U is a generator of A and that thereexists a left adjoint T : Mod-B → A of the functor Hom A (U, −) : A → Mod-B.Then T is an exact functor.Proof. By Proposition A.3, H is full and faithful. Since T is a left adjoint so that itpreserves epimorphisms, we have only to prove that it is left exact.Now let X be a submodule of a free right B-module B (Z) . Let P 0 (X) be the set offinite subset of X and let j F : X F → X be the canonical inclusion of the submoduleof X spanned by F ∈ P 0 (X). By Lemma A.7, for every F ∈ P 0 (X) there exists afinite subset Z F of Z and a monomorphism i F : X F → A (Z F ) such that the diagramj FX F X i Fj h FA (Z F )A (Z)where j : X → A (Z) is the canonical inclusion and h F : A (Z F ) → A (Z) is the canonicalsection of the canonical projection π F : A (Z) → A (Z F ) , is commutative. Moreover(h F ◦ i F ) F ∈P0 (X)is a compatible family of morphisms such thatlim (h F ◦ i F ) = j.−→Since T is a left adjoint functor, we have()T (j) = T lim (h F ◦ i F )−→= lim−→T (h F ◦ i F ) .By Lemma A.6 we know that T (i F ) is a monomorphism. On the other hand π F ◦h F = Id A (Z F ) and hence also T (h F ) is a monomorphism. Since A is a Grothendieckcategory, direct limits are exact in A and hence T (j) is a monomorphism.Finally let0 → L f−→ Mbe a monomorphism in Mod-B. Then we can construct the following commutativediagram with exact rows and columns000 Ker (p)i ′Pp ′L0Id Ker(p)0 Ker (p)f ′ i B (M) pf M 0

where p : B (M) → M is the usual epimorphism of right B-modules and (P, f ′ , p ′ ) isthe pullback of (p, f). Recall thatP = { (x, y) ∈ B (M) × L | p (x) = f (y) }and f ′ : P → B (M) is defined by setting f ′ ((x, y)) = x while p ′ : P → L is defined bysetting p ′ ((x, y)) = y. Moreover i ′ : Ker (p) → P is defined by setting i ′ (x) = (x, 0).Since f is a monomorphism we have that also f ′ is a monomorphism and since p isan epimorphism, also p ′ is an epimorphism. Then, by the foregoing, both T (f ′ ) andT (i) are monomorphism. Since T (i) is a monomorphism we get that T (i ′ ) is alsoa monomorphism so that (T (Ker (p)) , T (i ′ )) is a kernel of T (p ′ ). Since T (f ′ ) isa monomorphism and T (f ′ ) T (i ′ ) = T (i) we get that (T (Ker (p)) , T (i ′ )) is also akernel of T (p) T (f ′ ). In fact T (p) T (f ′ ) T (i ′ ) = T (p) T (i) = 0 and if ζ : Z → T (P )is a morphism such that T (p) T (f ′ ) ζ = 0 there exists a unique morphism ζ ′ : Z →T (Ker (p)) such that T (f ′ ) ζ = T (i) ζ ′ so that T (f ′ ) ζ = T (i) ζ ′ = T (f ′ ) T (i ′ ) ζ ′and since T (f ′ ) is mono we get that ζ = T (i ′ ) ζ ′ . Since T (p) T (f ′ ) = T (f) T (p ′ )we deduce that (T (Ker (p)) , T (i ′ )) is a kernel of T (f) T (p ′ ). Therefore we obtainthatKer (T (f) T (p ′ )) = Ker (T (p ′ )) .Since T is right exact we know that T (p ′ ) is an epimorphism and hence, in view ofLemma A.8, we deduce that T (f) is a monomorphism.02490 T (Ker (p)) T (i′ )T (P )T (p ′ )T (L)0Id T (Ker(p))T (f ′ )0 T (Ker (p)) T (i) T ( B (M)) T (p)T (f) T (M) 0Lemma A.10. Let C be an A-coring. Then the category (Mod-A) C has coproductsand cokernels so that it is cocomplete. Moreover if U : (Mod-A) C → (Mod-A) isthe forgetful functor we have((∐ ))U (Mi , ρ i ) i∈I, ρ M = ⊕ M i and U (( Coker (f) , ρ Coker(f))) = Coker (f) .i∈I□Proof. Let (M ι , ρ i ) i∈Ibe a family of right C-comodules and let ε i : M i → M = ⊕ M ii∈Ibe the canonical injection and π i : M → M i the canonical projection. Since π i ε i =Id Mi the mapε i ⊗ A C : M i ⊗ A C → M ⊗ A Cis injective and hence also the mapρ i = (ε i ⊗ A C) ◦ ρ i : M i → M ⊗ A Cis injective. Letρ M : M → M ⊗ A C

250be the codiagonal map of the ρ i . Then ρ M is uniquely defined byρ M ◦ ε i = ρ i .Then ( M, ρ M) ∈ (Mod-A) C . In fact, for every i ∈ I we have(ρ M ⊗ A C ) ◦ ρ M ◦ ε i = ( ρ M ⊗ A C ) ◦ ρ i = ( ρ M ⊗ A C ) ◦ (ε i ⊗ A C) ◦ ρ i= (ρ i ⊗ A C) ◦ ρ i = (ε i ⊗ A C) ◦ (ρ i ⊗ A C) ◦ ρ i = (ε i ⊗ A C) ◦ ( M i ⊗ A ∆ C) ◦ ρ iand= ( M ⊗ A ∆ C) ◦ (ε i ⊗ A C) ◦ ρ i = ( M ⊗ A ∆ C) ◦ ρ i = ( M ⊗ A ∆ C) ◦ ρ M ◦ ε ir M ◦ ( M ⊗ A ε C) ◦ ρ M ◦ ε i = r M ◦ ( M ⊗ A ε C) ◦ ρ i = r M ◦ ( M ⊗ A ε C) ◦ (ε i ⊗ A C) ◦ ρ i= r M ◦ (ε i ⊗ A A) ◦ ( M i ⊗ A ε C) ◦ ρ i = ε i ◦ r Mi ◦ ( M i ⊗ A ε C) ◦ ρ i = ε i .Let f : ( M, ρ M) → ( N, ρ N) be a morphism in (Mod-A) C so that f : M → N is amorphism in Mod-A and let us considerMf−→ Np−→ Coker (f) → 0the cokernel of f in Mod-A. Then we have the following diagram in Mod-AMfNpCoker (f)0We computeρ Mρ Nρ Coker(f)M ⊗ A C f⊗ AC N ⊗ A C p⊗ AC Coker (f) ⊗A C 0(p ⊗ A C) ◦ ρ N ◦ f = (p ⊗ A C) ◦ (f ⊗ A C) ◦ ρ M = (pf ⊗ A C) ◦ ρ M = 0by the universal property of the cokernel there exists a unique morphism ρ Coker(f) :Coker (f) → Coker (f) ⊗ A C such thatρ Coker(f) ◦ p = (p ⊗ A C) ◦ ρ N .Let us check that ( Coker (f) , ρ Coker(f)) ∈ (Mod-A) C . Let us compute(ρ Coker(f) ⊗ A C ) ◦ ρ Coker(f) ◦ p = ( ρ Coker(f) ⊗ A C ) ◦ (p ⊗ A C) ◦ ρ Nand= (p ⊗ A C ⊗ A C) ◦ ( ρ N ⊗ A C ) ◦ ρ N = (p ⊗ A C ⊗ A C) ◦ ( N ⊗ A ∆ C) ◦ ρ N= ( Coker (f) ⊗ A ∆ C) ◦ (p ⊗ A C) ◦ ρ N = ( Coker (f) ⊗ A ∆ C) ◦ ρ Coker(f) ◦ pr Coker(f) ◦ ( Coker (f) ⊗ A ε C) ◦ ρ Coker(f) ◦ p= r Coker(f) ◦ ( Coker (f) ⊗ A ε C) ◦ (p ⊗ A C) ◦ ρ N= r Coker(f) ◦ (p ⊗ A A) ◦ ( N ⊗ A ε C) ◦ ρ N = p ◦ r N ◦ ( N ⊗ A ε C) ◦ ρ N = pand since p is epi we conclude. Now, let ζ : ( N, ρ N) → ( Z, ρ Z) be a morphism in(Mod-A) C such that ζ◦f = 0. Then, there exists a unique morphism ζ ′ : Coker (f) →Z in Mod-A such that ζ ′ ◦p = ζ. We want to prove that ζ ′ ∈ (Mod-A) C . We computeρ Z ◦ ζ ′ ◦ p = ρ Z ◦ ζ = (ζ ⊗ A C) ◦ ρ N

251= (ζ ′ ⊗ A C) ◦ (p ⊗ A C) ◦ ρ N = (ζ ′ ⊗ A C) ◦ ρ Coker(f) ◦ pand since p is an epimorphism we get that ζ ′ ∈ (Mod-A) C .Lemma A.1<strong>1.</strong> Let C be an A-coring and assume that A C is flat. Then the category(Mod-A) C has kernels. Moreover if U : (Mod-A) C → (Mod-A) is the forgetfulfunctor we haveU (( Ker (f) , ρ Ker(f))) = Ker (f) .Proof. Since by Lemma A.10 the preadditive category (Mod-A) C has coproducts,it also has products. Now, let f : ( M, ρ M) → ( N, ρ N) a morphism in (Mod-A) C .Then in Mod-A we can consider the exact sequence0 → Ker (f)k−→ Mand, since A C is flat, we get the exact sequence0 Ker (f)ρ Ker(f)kMf−→ N0 Ker (f) ⊗ A C k⊗ AC M ⊗ A C f⊗ AC N ⊗ A CWe have0 = ρ N ◦ f ◦ k = (f ⊗ A C) ◦ ρ M ◦ k.By the properties of the kernel of f there exists a unique morphism ρ Ker(f) : Ker (f) →Ker (f) ⊗ A C such thatρ M ◦ k = (k ⊗ A C) ◦ ρ Ker(f) .We have to prove that ( Ker (f) , ρ Ker(f)) ∈ (Mod-A) C . Let us compute(k ⊗ A C ⊗ A C) ◦ ( ρ Ker(f) ⊗ A C ) ◦ ρ Ker(f) = ( ρ M ⊗ A C ) ◦ (k ⊗ A C) ◦ ρ Ker(f)= ( ρ M ⊗ A C ) ◦ ρ M ◦ k = ( M ⊗ A ∆ C) ◦ ρ M ◦ k = ( M ⊗ A ∆ C) ◦ (k ⊗ A C) ◦ ρ Ker(f)andρ M= (k ⊗ A C ⊗ A C) ◦ ( Ker (f) ⊗ A ∆ C) ◦ ρ Ker(f)k ◦ r Ker(f) ◦ ( Ker (f) ⊗ A ε C) ◦ ρ Ker(f) = r M ◦ (k ⊗ A C) ◦ ( Ker (f) ⊗ A ε C) ◦ ρ Ker(f)= r M ◦ ( M ⊗ A ε C) ◦ (k ⊗ A C) ◦ ρ Ker(f) = r M ◦ ( M ⊗ A ε C) ◦ ρ M ◦ k = k.Since k is mono we conclude. Let now ζ : ( Z, ρ Z) → ( M, ρ M) be a morphism in(Mod-A) C such that f ◦ ζ = 0. Then by the universal property of the kernel of f inMod-A there exists a unique morphism ζ ′ : Z → Ker (f) such thatk ◦ ζ ′ = ζ.We want to prove that ζ ′ ∈ (Mod-A) C . Let us compute(k ⊗ A C) ◦ ρ Ker(f) ◦ ζ ′ = ρ M ◦ k ◦ ζ ′ = ρ M ◦ ζ = (ζ ⊗ A C) ◦ ρ Z= (k ⊗ A C) ◦ (ζ ′ ⊗ A C) ◦ ρ Zand since k⊗ A C is mono we conclude that ζ ′ ∈ (Mod-A) C and U (( Ker (f) , ρ Ker(f))) =Ker (f) .□fNρ N□

252Proposition A.12 ([ELGO2, Proposition <strong>1.</strong>2]). Let C be an A-coring. Then thefollowing are equivalent(a) A C is flat.(b) (Mod-A) C is an abelian category and the forgetful functor U : (Mod-A) C →Mod-A is left exact (and hence exact).(c) (Mod-A) C is a Grothendieck category and the forgetful functor U : (Mod-A) C →Mod-A is left exact (and hence exact).Proof. By Lemma A.10, the category (Mod-A) C has coproducts and cokernels.(a) ⇒ (c) By Lemma A.11 has kernels. Consider the following diagram( ) Ker (f) , ρKer(f) k() M, ρMf ( ) N, ρNχ ( ) Coker (f) , ρCoker(f)χ ′ ρ k′( ) ( )Coker (k) , ρCoker(k) Ker (χ) , ρKer(χ).¯fin (Mod-A) C . Then we get the diagramKer (f)k M χ ′ Coker (k)fρ¯f Nk ′ Ker (χ) .χ Coker (f)in Mod-A. Since Mod-A is preabelian, f is an isomorphism in Mod-A and hence alsoin (Mod-A) C . Thus also the category (Mod-A) C is preabelian and moreover abelian(there exist products of every finite family of objects in the category). Moreover,by Lemma A.10 and Lemma A.11 U is left exact. Further, the direct limits areexacts for module categories and thus also for (Mod-A) C . We now have to find agenerator for (Mod-A) C . Let ( M, ρ M) ∈ (Mod-A) C and let p : A (M) → M is theusual epimorphism of right A-modules. Let us consider the epimorphism l given bythe following compositel : C (M) −→ A (M) ⊗ A C p⊗ AC−→ M ⊗ A C −→ 0where the first arrow is the canonical isomorphism and the second one is the usualepimorphism so that l ( (c m ) m∈M)=∑m∈M m ⊗ A c m where c m are almost all zero.Then we have the following diagramρ0Pgρ M0MC (M) l M ⊗ A C 00

253where (P, ρ, g) is the pullback of ( l, ρ M) . Recall that P is the submodule ofdefined by settingC (M) × MP = { (x, m) ∈ C (M) × M | l (x) = ρ M (m) }and ρ : P → C (M) is defined by setting ρ ((x, m)) = x while g : P → M bysetting g ((x, m)) = m. Since ρ M is mono, ρ is also mono (thus ρ (P ) = H is asubcomodule of C (M) ) and since l is epi, g is epi. Denote by h M F : C(F ) → C (M) thecanonical inclusion for any F ⊆ M. Let m ∈ M, then there exist n ∈ N and F m ={y 1 , y 2 , . . . , y n } ⊆ M such that ρ M (m) = ∑ ))y∈F my ⊗ A c y = l(h M F m((c y ) y∈Fm.Then, for every m ∈ M, there exists z ∈ P such that m = g (z) = g ((ρ (z) , m))where ρ (z) = h M F m((c y ) y∈Fm)∈ h M F m(C(F m) ) ⊆ C (M) . Thus, for every m ∈ M, thereexists x m = (c y ) y∈Fm∈ such that m = g (( h M F m(x m ) , m )) . Then we have definedthe following homomorphismsuch thatν xm : x m A −→ Mx m ↦→ g (( h M F m(x m ) , m )) .m = ν xm (x m ) .Since x m A ⊆ C (Fm) , we deduce that the subcomodules of C (N) form a set of generatorsfor (Mod-A) C ⊕i.e. H is a generator for (Mod-A) C .H⊆C (N)(c) ⇒ (b) Obvious.(b) ⇒ (a) By Example 4.3 and Definition 4.10 F : Mod-A → (Mod-A) C is a rightadjoint of U and then F is left exact. Then using the hypothesis that U is left exact,we deduce that U ◦ F : Mod-A → Mod-A is also left exact, i.e. A C is flat. □Definition A.13. Let A be a Grothendieck category. An object A ∈ A is calledfinitely generated if, for every direct family of subobjects {A i } i∈Iof A such thatA = ∑ A i , there exists an index i 0 ∈ I such that A = A i0 .i∈IProposition A.14. Let A be a Grothendieck category. An object A ∈ A is finitelygenerated if and only if, for every family of subobjects {A i } i∈Iof A such that ∑ i∈IA i =∑A, there exists a finite number of subobjects A 1 , . . . , A n such that A = n A i .Proof. (⇐) Let {A i } i∈Ibe a direct family of subobjects of A closed under sumsand such that A = ∑ A i . By hypothesis there exists a finite number of subobjectsi∈I∑A 1 , . . . , A n such that A = n A i . Since the family is direct and closed under sums,i∈I∑there exists an index i 0 ∈ I such that A = n A i ⊆ A i0 . Then A is finitely generated.i∈Ii∈I

254(⇒) Let {A i } i∈Ibe a family of subobjects of A closed under sums and whichcontains A itself. Assume that A = ∑ A i . Since A is finitely generated, there existsi∈Ian index i 0 ∈ I such that A i0 = A.Lemma A.15. Let A be a Grothendieck category and let 0 → A ′ −→ A −→ pA ′′ →0 be an exact sequence in A. Then if A is finitely generated A ′′ is also finitelygenerated.Proof. Let (A ′′i ) i∈Ibe a direct family of subobjects of A ′′ such that A ′′ = ∑ A ′′i . Theni∈Iwe have, for every i ∈ I, A ′′i = A i A ′ for A i subobject of A such that A ′ ⊆ A i .Hence (A i ) i∈Iis a direct family of subobjects of A such that A = ∑ A i and since Ai∈Iis finitely generated there exists an index i 0 ∈ I such that A = A i0 . Then we haveA ′′ = A i0 A ′ = A ′′i 0, i.e. A ′′ is also finitely generated.□Lemma A.16. Let A be a Grothendieck category and let A ∈ A be a finitely generatedobject. Let f : A → ∐ i∈IB i be a morphism in A. Then there exist a finite subsetF ⊆ I such that Im (f) ⊆ ∑ ε i (B i ).i∈F)B ii∈Ii∈IProof. Let ε i : B i → ∐ B j the canonical inclusions and consider ∇ (ε i ) i∈I: ∐ B i →j∈Ii∈I( )∐∐B j defined by setting ∇ (ε i ) i∈IB j = ∑ ε i (B i ). We prove ∇ (ε i ) i∈I=j∈I j∈Ii∈IId`B i. In fact we have that ∇ (ε j ) j∈I◦ ε i = ε i = Id ` B j◦ ε i . Thus, ∐ B i =i∈Ij∈I(i∈IIm Id` = Im ( ) ∑∇ (ε i ) i∈I = ε i (B i ) where (ε i (B i )) i∈Idefine a family of subobjectsof ∐ B i . Let f : A → ∐ B j . By Lemma A.15, since A is finitely generatedi∈Ij∈Ialso Coim (f) is finitely generated and, since Coim (f) ≃ Im (f) ⊆ ∑ ε i (B i ), therei∈I□exists a finite subset F ⊆ I such that Im (f) ⊆ ∑ ε i (B i ).i∈F□Definition A.17. Let A be an abelian category. A projective object P ∈ A iscalled finite if the functor Hom A (P, −) preserves coproducts.Proposition A.18. Let A be a Grothendieck category and let P be a projectiveobject. Then P is finite if and only if P is finitely generated.Proof. Assume first that P ∈ A is finite. Let {P i } i∈Ibe a family of subobjects of Psuch that ∑ i∈IP i = P. Let p i : P i → P be the canonical inclusion for every i ∈ I andconsider p = ∇ (p i ) i∈I: ∐ i∈IP i → P. Then we have( ) ( )∐∇ (pi ) i∈IP i = ∑i∈Ii∈Ip i (P i ) = ∑ i∈IP i = P

and thus p is an epimorphism. Since P is projective there exists i : P → ∐ P i such(i∈Ithat p ◦ i = Id P . Note that i ∈ Hom A P, ∐ )P i and since P is finite we havei∈Ithat ∇ (Hom (P, ε i )) i∈I: ∐ (Hom A (P, B i ) → Hom A P, ∐ )B i is an isomorphism.i∈Ii∈IThus there exist n ∈ N and i 1 ∈ Hom A (P, P 1 ) , . . . , i n ∈ Hom A (P, P n ) such that∐i = ε 1 i 1 + · · · + ε n i n . Hence Id P = p ◦ i = p ◦ (ε 1 i 1 + · · · + ε n i n ) : P → n P i → P∐and then P = n P i so that P is finitely generated.ε F ii∈1Assume now that P is finitely generated. Let us denote by ε j : B j → ∐ B i ,i∈I: B i → ∐ B i the canonical injections for every F ⊆ I finite subset, and let usprove thati∈F∇ (Hom (P, ε i )) i∈I: ∐ i∈IHom A (P, B i ) −→ Hom A(P, ∐ i∈IB i)i∈1255(f i ) i∈I↦→ ∑ i∈Iε i ◦ f iis an isomorphism. First of all we prove that it is epi. Since P is finitely generated, iff : P → ∐ i∈IB i is a morphism in A, by Lemma A.16, there exists a F ⊆ I finite subsetsuch that Im (f) ⊆ ∑ ε i (B i ) . Let us denote by f : P → Im (f) and s : Im (f) ↩→∑i∈Fε i (B i ) the canonical inclusion. Let us consider h = ∇ (ε i ) i∈F: ∐ B i → ∐ B ii∈Fi∈F i∈Isatisfying(265) h ◦ ε F i = ∇ (ε i ) i∈F◦ ε F i = ε i for every i ∈ F .Then, by definition of the codiagonal morphism, we have that Im (h) = ∑ ε i (B i )i∈Fand thus, if we call ĥ : ∐ B i → Im (h) = ∑ ε i (B i ) the canonical projection, wei∈F i∈Fcan write(266) h = t ◦ hwhere t : ∑ ε i (B i ) → ∐ B i is the inclusion. Thus also f can be factorized throughi∈Fi∈If by(267) f = t ◦ s ◦ f.With these notations we can rewrite (265) as follows(268) ε i = h ◦ ε F i = t ◦ h ◦ ε F i .

256We will prove that h : ∐ B i → ∑ ε i (B i ) is in fact an isomorphism. We define thei∈F i∈Ffamily (η j ) j∈Iby settingη j = 0 for every j ∈ I\F and η j = ε F j for every j ∈ F .Then we can take ∇ (η i ) i∈I: ∐ B i → ∐ B i . Let us compute for every j ∈ Fi∈I i∈F∇ (η i ) i∈I◦ h ◦ ε F jdefh= ∇ (η i ) i∈I◦ ∇ (ε i ) i∈F◦ ε F j= ε F j = Id ` B i◦ ε F ji∈F(265)= ∇ (η i ) i∈I◦ ε jand thus ∇ (η i ) i∈I◦ h = Id ` B i. Therefore we deduce that h is mono and then h isi∈Fan isomorphism. Let us consider (δ ij : B i → B j ) j∈Ithe family defined by settingδ ii = Id Bi and δ ij = 0 for every j ≠ i.Since ∐ B i is a finite coproduct, we can view it as a product and call π j : ∐ B i =i∈Fi∈F∏B i = × i∈F→ B j the projections for every j ∈ F satisfyingi∈F(269) π j ◦ ε F i = δ ij and∑ε F i π i = Id ×.i∈FNote that, by the universal property of the coproduct, there exist q j : ∐ i∈IB i → B jsuch thati∈F(270) q j ◦ ε i = δ ij .Let us compute, for every i ∈ F and for every j ∈ I,and thusq j ◦ h ◦ ε F i = q j ◦ ε i(270)= δ ij(269)= π j ◦ ε F i(271) q j ◦ h = π j .We define the family (f i ) i∈I∈ ∐ i∈IHom A (P, B i ) by settingNote thati.e.f i = π i ◦ h −1 ◦ s ◦ f for every i ∈ F and f i = 0 for every i ∈ I\F .f i = π i ◦ h −1 ◦ s ◦ f (271)= q i ◦ h ◦ h −1 ◦ s ◦ f(266)= q i ◦ t ◦ h ◦ h −1 ◦ s ◦ f = q i ◦ t ◦ s ◦ f (267)= q i ◦ f(272) f i = q i ◦ f.For such a family (f i ) i∈I, we have to prove thatf = ∑ i∈Iε i ◦ f i .

257Since f i = 0 for every i ∈ I\F we have∑ε i ◦ f i = ∑i∈Ii∈Fso that it is sufficient to prove thatε i ◦ f if = ∑ i∈Fε i ◦ f iand by (267) and (268)t ◦ s ◦ f = ∑ i∈Ft ◦ h ◦ ε F i ◦ f i .Since t is mono we only need to prove thats ◦ f = ∑ i∈Fh ◦ ε F i◦ f iand thus, for every j ∈ F, thatπ j ◦ h −1 ◦ s ◦ f = π j ◦ h −1 ◦ ∑ i∈Fh ◦ ε F i ◦ f i .Let us computeOn the other handπ j ◦ h −1 ◦ s ◦ f (271)= q j ◦ h ◦ h −1 ◦ s ◦ f(266)= q j ◦ t ◦ h ◦ h −1 ◦ s ◦ f= q j ◦ t ◦ s ◦ f (267)= q j ◦ f (272)= f j .π j ◦ h −1 ◦ ∑ i∈Fh ◦ ε F iπ j ◦ h −1 ◦ h ◦ ε F iπ j ◦ ε F i◦ f i◦ f i = ∑ i∈F◦ f i = ∑ i∈F(269)= ∑ δ ij ◦ f i = f ji∈Fso that we conclude that ∐ (Hom A (P, B i ) ∇(Hom(P,ε i)) i∈I−→ HomA P, ∐ )B i is an epimorphism.Let now (f i ) i∈Ii∈Ii∈I∈ ∐ Hom A (P, B i ) where f i ’s are almost all zero, be( i∈I ) ∑such that ∇ (Hom (P, ε i )) i∈I (fi ) i∈I = ε i ◦ f i = 0. Since f i ’s are almost all zero,i∈Ilet F ⊆ I be a finite subset such that f i ≠ 0 for every i ∈ F and f i = 0 for everyi ∈ I\F . Then 0 = ∑ (265)ε i ◦ f i = ∑ h ◦ ε F i ◦ f i . Since h is mono, we also havei∈Ii∈F0 = ∑ (ε F i ◦ f i ∈ Hom A P, ∐ )B i so that, for every j ∈ F,i∈Fi∈F0 = π j ◦ ∑ i∈Iε F i◦ f i = ∑ i∈Iπ j ◦ ε F i ◦ f i(269)= f jand thus f i = 0 for every i ∈ I.□

258Proposition A.19. Let (T, H) be an adjunction where T : A → B, H : B → Aand A, B are Grothendieck categories. If T is an equivalence of categories, then1) if P is a generator of A then T P is a generator of B2) if P is projective in A then T P is projective B3) if P is finite in A then T P is finite B4) if P is finitely generated in A then T P is finitely generated in B5) if P is finitely generated and projective in A then T P is finitely generatedand projective B6) if P is finite projective in A then T P is finite projective in B.Proof. 1) Let f : Y → Y ′ be a non zero morphism in B. Since T is an equivalence,there exists a non zero morphism g : X → X ′ in A such that f = T (g). Since Pis a generator of A there exists a morphism p : P → X such that g ◦ p ≠ 0. Then0 ≠ T (g ◦ p) = T (g) ◦ T (p) = f ◦ T (p) and T (p) : T P → T X = Y so that T P is agenerator of B.2) Let f : Y → Y ′ be a morphism in B. Since T is an equivalence, there existsa morphism g : X → X ′ in A such that f = T (g), Y = T X and Y ′ = T X ′ . Letl : T P → Y a morphism in B, then l : T P → T X then there exists h : P → X suchthat l = T (h). Since P is projective A, there exists k : P → X ′ such that g ◦ h = k.By applying the functor T we get T (g ◦ h) = T (g) ◦ T (h) = f ◦ l = T (k) then T Pis projective in B.3) Let (N i ) i∈Ibe a family of objects in B. Since T is an equivalence, there existsa family (M i ) i∈Iof objects in A such that (N i ) i∈I= (T M i ) i∈I. Denote by ε i :M i → ∐ (M i and by Hom A (P, ε i ) : Hom A (P, M i ) → Hom A P, ∐ M i). Then wei∈Ii∈I (can consider the codiagonal morphism ∇ (Hom A (P, ε i )) i∈I: Hom A P, ∐ )M i →∐i∈IHom A (P, M i ). Since P is finite in A we have that P preserves coproducts, i.e.i∈I∇ (Hom A (P, ε i )) i∈Iis an isomorphism. Letφ X,X ′ : Hom A (X, X ′ ) −→ Hom B (T X, T X ′ )f ↦→ T (f)Since T is an equivalence φ X,X ′ is bijective for every X, X ′ ∈ A and ∐ (T M i ) =( )i∈I∐T M i . Then we can consideri∈I ( )) (∐Hom B (T P, T (ε i )) : Hom B (T P, T (M i )) → Hom B(T P, T M i = Hom B T P, ∐i∈I (i∈Iand their codiagonal morphism ∇ (Hom B (T P, T (ε i ))) i∈I: Hom B T P, ∐ )(T M i ) →i∈I)(T M i )

∐Hom B (T P, T (M i )). Then we have the following commutative diagrami∈I(Hom A P, ∐ )M ii∈Iφ ` P, M ii∈I∇(Hom A (P,ε i )) i∈I(Hom B T P, ∐ ∇(HomB (T P,T (ε i ))) i∈I(T M i ))i∈I∐Hom A (P, M i )i∈I`φ P,Mii∈I∐Hom B (T P, T (M i ))i∈I259where we observed that the first row is an isomorphism and also the φ’s are isomorphism.Then we deduce that ∇ (Hom B (T P, T (ε i ))) i∈Iis an isomorphism, so thatT P preserves coproducts, i.e. T P is finite.4) Let {Q i } i∈Ibe a direct family of subobjects of T P such that T P = ∑ i∈IQ i .Then, since T is an equivalence, there exists a direct family {P i } i∈Iof subobjects ofP such that T P i = Q i for every i ∈ I and P = ∑ i∈IP i . Since P is finitely generated,there exists an index i 0 ∈ I such that P = P i0 and then T P = T P i0 = Q i0 , i.e. T Pis finitely generated.5) By Proposition A.18, P is finite and thus by 3) we deduce that T P is alsofinite. Since T P is moreover projective, by Proposition A.18 we conclude that T Pis finitely generated.6) By Proposition A.18, since P is finite projective, P is finitely generated andprojective. Then we conclude by 5).□Definition A.20. Let A be an abelian category. A finite projective generator P inA is called progenerator.Corollary A.2<strong>1.</strong> Let A be a Grothendieck category. There exists an equivalenceF : A → Mod-B, where B is a ring, if and only if A contains a progenerator P .Moreover• If P is a progenerator of A, then Hom A (P, −) : A → Mod-T where T =Hom A (P, P ) .• If F is an equivalence, there exists a progenerator P in A such thatHom A (P, P ) ≃ B and F ≃ Hom A (P, −).Proof. Assume first that A contains a progenerator P . Let B = Hom A (P, P ) andconsider the functor Hom A (P, −) : A → Mod-B. Since P is a generator and A is aGrothendieck category, by Proposition A.3 we deduce that Hom A (P, −) is full andfaithful. Hence we only have to prove that Hom A (P, −) is surjective on the objects.Let M ∈ Mod-B. Then we have the following exact sequence in Mod-BB (X) −→ B (M) −→ M → 0.Since B = Hom A (P, P ) we can rewrite is asHom A (P, P ) (X) −→ Hom A (P, P ) (M) −→ M → 0.

260Since P is finite Hom A (P, P ) (X) ≃ Hom A(P, P(X) ) andHom A (P, P ) (M) ≃ Hom A(P, P(M) ) and then we have an exact sequence in Mod-B(273) Hom A(P, P(X) ) f−→ Hom A(P, P(M) ) −→ Q → 0where Q = Coker (f). Then Q ≃ M. Since Hom A (P, −) is full (and faithful) wehaveHom A (A, A ′ ) ≃ Hom Mod-B (Hom A (P, A) , Hom A (P, A ′ )) ,hence there exists a unique morphism g : P (X) → P (M) in A such that f =Hom A (P, −) (g). Let us consider in A(274) P (X) g−→ P (M) −→ X → 0where X = Coker (g). Since P is projective, Hom A (P, −) is exact, and applying itto (274) we get the exact sequenceHom A(P, P(X) ) f=Hom A (P,g)−→ Hom A(P, P(M) ) → Hom A (P, X) → 0.From this sequence and (273) we deduce that Q ≃ Hom A (P, X) where X =Coker (g) ∈ A.Conversely, let us assume that F : A → Mod-B is an equivalence of categories. LetG : Mod-B → A be its inverse equivalence. Since B is a progenerator and G is anequivalence of categories, by Proposition A.19 1) and 6), we deduce that G (B) is aprogenerator in A. Moreover we haveB ≃ Hom Mod-B (B, B) ≃ Hom A (G (B) , G (B)) .Observe that G is a left adjoint to F and thus we haveHom A (G (B) , −) ≃ Hom Mod-B (B, F−) .Since Hom Mod-B (B, F−) ≃ F as functors, we deduce thatF ≃ Hom A (G (B) , −)where G (B) is a progenerator in A.□Theorem A.2<strong>2.</strong> Let A be an abelian category. There exists an equivalence F : A →Mod-B, where B is a ring, if and only if A contains a progenerator P and arbitrarycoproducts of copies of P . If F is an equivalence, there exists a progenerator P inA such that Hom A (P, P ) ≃ B and F ≃ Hom A (P, −).Proof. Assume first that A contains a progenerator P and arbitrary coproducts ofcopies of P . Let B = Hom A (P, P ) and consider the functor Hom A (P, −) : A → Ab.Let us endow any abelian group Hom A (P, A) , for every A ∈ A, with a right B-module structure given by the composition with morphisms of Hom A (P, P ) = B.This means we have the following mapHom A (P, A) × Hom A (P, P ) → Hom A (P, A)(h, ξ) ↦→ h ◦ ξ.For every morphism f : A → B in A we define a morphism in Mod-B as followsHom A (P, f) : Hom A (P, A) → Hom A (P, B)

ξ ↦→ f ◦ ξ.Then it is well-defined a functor Hom A (P, −) : A → Mod-B. We want to prove thatHom A (P, −) is an equivalence of category. To be full and faithful for Hom A (P, −)means that the mapφ A,A ′ : Hom A (A, A ′ ) −→ Hom Mod-B (Hom A (P, A) , Hom A (P, A ′ ))( )HomA (P, f) : Homf ↦→A (P, A) → Hom A (P, A ′ )ξ ↦→ f ◦ ξis bijective for every A, A ′ ∈ A. Note that Hom A (P, −) induces an isomorphismφ P,P : Hom A (P, P ) −→ Hom Mod-B (Hom A (P, P ) , Hom A (P, P )) .In fact, for every ζ ∈ Hom A (P, P ) such that Hom A (P, ζ) = 0 we have that, forevery ξ ∈ Hom A (P, P ) , 0 = Hom A (P, ζ) (ξ) = ζ ◦ ξ. Since P is a generator, wededuce that ζ = 0. Now, let f : Hom A (P, P ) → Hom A (P, P ) be a morphism inMod-B and set f (Id P ) = χ. Then, for every ξ ∈ Hom A (P, P ), we haveand thusf (ξ) = f (Id P ◦ ξ) f∈Mod-B= f (Id P ) ◦ ξ = χ ◦ ξ = Hom A (P, χ) (ξ)f = Hom A (P, χ) = Hom A (P, −) (χ)so that φ P,P is an epimorphism. Let us consider families (P i ) i∈Iand (P j ) j∈JwhereP i ≃ P ≃ P j for every i ∈ I and j ∈ J. Set B i = Hom A (P, P i ) and B j =Hom A (P, P j ). Then B i = Hom A (P, P i ) ≃ Hom A (P, P ) = B and similarly B j ≃ B.Let us compute( ∐Hom A P j , ∐ )coprodP i ≃∏ (Hom A P j , ∐ )P finiteP i ≃∏ ∐Hom A (P j , P i )j∈J i∈Ij∈Ji∈Ij∈J i∈Iφ P,P≃ ∏ ∐Hom Mod-B (Hom A (P, P j ) , Hom A (P, P i )) = ∏ ∐Hom Mod-B (B j , B i )j∈J i∈I j∈J i∈I(B i ≃Bfinite∏≃ Hom Mod-B B j , ∐ )(coprod∐B i ≃ Hom Mod-B B j , ∐ )B ij∈Ji∈Ij∈J i∈Ihence(275) Hom A( ∐j∈JP j , ∐ i∈IP i)≃ Hom Mod-B( ∐j∈JB j , ∐ i∈IB i)which says that Hom A (P, −) induces a bijection between the full subcategory of thecoproducts of copies of P in A and the full subcategory of coproducts of copies ofB in Mod-B, i.e. Hom A (P, −) is full and faithful on the full subcategory of thecoproducts of copies of P in A. Let us denote by ε P i : P → P (I) , p P i : P (I) → P andε j : B = Hom A (P, P ) → B (I) = Hom A (P, P ) (I) , p j : B (I) = Hom A (P, P ) (I) → B =Hom A (P, P ) the canonical maps. Now, let A, A ′ ∈ A. Since P is a generator, wehaveP (J)e f−→ P (I)h−→ A = Coker (h) → 0261

262andP (J ′ )e f ′−→ P (I′ ) h ′−→ A ′ = Coker (h ′ ) → 0.Let z : Hom A (P, A) → Hom A (P, A ′ ) be a morphism in Mod-B. We have to provethat there exists a morphism a : A → A ′ such that z = Hom A (P, a). Since P isprojective, Hom A (P, −) is exact so that we get the exact sequencesand(Hom ) A P, P(J)A(P,e Hom f)−→Hom A(P, P(I) ) Hom A (P,h)−→ Hom A (P, A) → 0( )Hom A P, P (J ′ ) HomA(P,e f ′) ( )−→ Hom A P, P (I′ ) HomA (P,h ′ )−→ Hom A (P, A ′ ) → 0Then we can considerHom A(P, P(J) ) Hom A(P,e f) Hom A(P, P(I) ) Hom A (P,h) Hom A (P, A)0Hom A(P, P(J ′ ) ) Hom A(P,e f ′) Hom A(P, P(I ′ ) ) Hom A (P,h ′ ) HomA (P, A ′ ) 0(Since Hom ) A P, P(I)(≃ B (I) it is projective, so that there exists a morphismy : Hom ) (A P, P(I)→ Hom ) (A P, P(I ′ )and a morphism x : Hom ) A P, P(J)(→Hom ) A P, P(J ′ ). Thus we have the following diagramzHom A(P, P(J) )A(P,e Hom f) ( Hom )A P, P(I)Hom A (P,h)Hom A (P, A)0xyHom A(P, P(J ′ ) ) Hom A(P,e f ′) Hom A(P, P(I ′ ) ) Hom A (P,h ′ ) HomA (P, A ′ ) 0Since Hom A (P, −) is full and faithful on coproducts of copies of P, every morphismx : Hom A(P, P(J) ) → Hom A(P, P(J ′ ) ) is of the form x = Hom A (P, ˜x) for ˜x :P (J) → P (J ′) , so that we can rewrite the diagram as followszHom A(P, P(J) )A(P,e Hom f) ( Hom )A P, P(I)Hom A (P,h)Hom A (P, A)0Hom A (P,ex)Hom A (P,ey)Hom A(P, P(J ′ ) ) Hom A(P,e f ′) Hom A(P, P(I ′ ) ) Hom A (P,h ′ ) HomA (P, A ′ ) 0zThus we haveHom A(P, ˜f) ′ ◦ ˜x = Hom A(P, ˜f)′= Hom A (P, ỹ) ◦ Hom A(P, ˜f◦ Hom A (P, ˜x)) (= Hom A P, ỹ ◦ ˜f).Since P is a generator we already now that Hom A (P, −) is faithful, so that wededuce that˜f ′ ◦ ˜x = ỹ ◦ ˜f

263i.e.f eP (J) P (I) h A 0exef ′eyP (J ′ h)P (I′ )′ A ′ 0( )Since A = Coker ˜f and by the commutative of the diagram, we deduce that thereexists a unique morphism a : A → A ′ in A such that the diagramf eP (J) P (I) h A0exP (J ′ )ef ′eyP (I′ )ah ′ A ′ 0is commutative. By applying the exact functor Hom A (P, −) thus we getHom A(P, P(J) )A(P,e Hom f) ( Hom )A P, P(I)Hom A (P,h)Hom A (P, A)0Hom A (P,ex)Hom A (P,ey)zHom A (P,a)Hom A(P, P(J ′ ) ) Hom A(P,e f ′) Hom A(P, P(I ′ ) ) Hom A (P,h ′ ) HomA (P, A ′ ) 0which saysz ◦ Hom A (P, h) = Hom A (P, h ′ ) ◦ Hom A (P, ỹ)= Hom A (P, a) ◦ Hom A (P, h)and since Hom A (P, h) is epi we deduce thatz = Hom A (P, a) .This proves that Hom A (P, −) is full. Since P is a generator, Hom A (P, −) is faithful.Then we only need to prove that Hom A (P, −) is surjective on objects. Let M ∈Mod-B. Then we have the following exact sequence in Mod-BB (X) → B (M) → M → 0.Since B = Hom A (P, P ) we can rewrite is asHom A (P, P ) (X) → Hom A (P, P ) (M) → M → 0.Since P is finite Hom A (P, P ) (X) ≃ Hom A(P, P(X) ) andHom A (P, P ) (M) ≃ Hom A(P, P(M) ) and then we have an exact sequence in Mod-B(276) Hom A(P, P(X) ) f−→ Hom A(P, P(M) ) → Q → 0where Q = Coker (f). Then Q ≃ M. Since Hom A (P, −) is full (and faithful) wehaveHom A (A, A ′ ) ≃ Hom Mod-B (Hom A (P, A) , Hom A (P, A ′ )) ,

264hence there exists a unique morphism g : P (X) → P (M) in A such that f =Hom A (P, −) (g). Let us consider in AP (X)g−→ P (M) → X → 0where X = Coker (g). Since P is projective, Hom A (P, −) is exact, and applying itwe get the exact sequenceHom A(P, P(X) ) f=Hom A (P,g)−→ Hom A(P, P(M) ) −→ Hom A (P, X) → 0.From this sequence and (276) we deduce that Q ≃ Hom A (P, X) where X =Coker (g) ∈ A.Conversely, let us assume that F : A → Mod-B is an equivalence of categories.Let G : Mod-B → A be its inverse equivalence. Since B is a progenerator and G isan equivalence of categories, by Proposition A.19 1) and 6), we deduce that G (B)is a progenerator in A. Moreover we haveB ≃ Hom Mod-B (B, B) ≃ Hom A (G (B) , G (B)) .Observe that G is a left adjoint to F and thus we haveHom A (G (B) , −) ≃ Hom Mod-B (B, F−) .Since Hom Mod-B (B, F−) ≃ F as functors, we deduce thatF ≃ Hom A (G (B) , −)where G (B) is a progenerator in A. Moreover, since G is an equivalence, G ( B (X)) ≃G (B) (X) .□

References[Ap] H. Appelgate, Acyclic models and resolvent functors, Ph.D. thesis (Columbia University,1965).[Ba] R. Baer, Zur Einführung des Scharbegriffs, J. Reine Angew. Math. 160 (1929), 199-207.[Be] J. Beck, Distributive laws, in Seminar on Triples and Categorical Homology Theory, B.Eckmann (ed.), Springer LNM 80 (1969), 119-140.[Bo] G. Böhm, Private communication to the author.[BW] M. Barr and T. Wells, Toposes, Triples and Theories, Grundl. der math. Wiss. 278,Springer-Verlag, (1983), available athttp://www.cwru.edu/artsci/math/wells/pub/ttt.html.http://www.case.edu/artsci/math/wells/pub/ttt.html.[BB] G. Böhm and T. Brzeziński, Pre-torsors and equivalences, J. Algebra 317 (2007), 544-580. Corrigendum, J. Algebra 319 (2008), 1339-1340.[BM] G. Böhm and C. Menini, Pre-torsors and Galois comodules overmixed distributive laws, to appear in Appl. Cat. Str.,online versionhttp://www.springerlink.com/content/xtk716u4517t0m63/fulltext.pdf, preprintarXiv:RA0806.121<strong>2.</strong>[BRZ2002] T. Brzeziński, The structure of corings. Induction functors, Maschke-type theorem, andFrobenius and Galois-type properties, Algebra Represent. Theory 5 (2002), 389-410.[BrHaj] T. Brzeziński, P.M. Hajac, Coalgebra extensions and algebra coextensions of Galoistype, Commun. Algebra 27(3) (1999), 1347-1368.[BMV] T. Brzeziński, A. Vazquez Marquez, J. Vercruysse, The Eilenberg-Moore categoryand a Beck-type theorem for a Morita context, to appear in Appl. Cat. Str.,online version http://www.springerlink.com/content/y68l7724n12n5u18/fulltext.pdf,preprint arXiv:081<strong>1.</strong>4304.[BV] T. Brzeziński and J. Vercruysse, Bimodule herds, J. Algebra 321 (2009), 2670-2704.[BrWi] T. Brzeziński and R. Wisbauer, Corings and comodules, London Mathematical SocietyLecture Note Series 309, Cambridge University Press, Cambridge, (2003).[D] E. Dubuc, Kan extensions in enriched category theory, Lecture Notes in Mathematics145, Springer Verlag Berlin (1970).[ELGO2] L. El Kaoutit, J. Gómez-Torrecillas, F.J. Lobillo, Semisimple corings, Algebra Colloq.11 (2004), no. 4, 427-44<strong>2.</strong>[GT] J. Gómez-Torrecillas, Comonads and Galois corings, Appl. Categorical Structures 14(2006), 579-598. Available online at arXiv:math.RA/0607043v<strong>1.</strong>[Gra] John W. Gray, Formal category theory: adjointness for 2-categories, Lecture Notes inMathematics, Vol. 39<strong>1.</strong> Springer-Verlag, Berlin-New York, 1974. xii+282 pp. 18DXX[G] C. Grunspan, Quantum torsors, J. Pure Appl. Algebra 184 (2003), 229-255.[Ho] D. Hobst, Antipodes in the theory of noncommutative torsors, Ph.D. thesis Ludwig-Maximilians Universität München 2004, Logos Verlag, Berlin (2004).[H] P. Huber, Homotopy theory in general categories, Math. Ann. 144 (1961), 361-385.[J] P.T. Johnstone, Adjoint lifting theorems for categories of algebras, Bull. London Math.Soc., 7 (1975), 294-297.[KT] H.F. Kreimer, M. Takeuchi, Hopf algebras and Galois extensions of an algebra, IndianaUniv. Math. J. 30 (1981), 675-69<strong>2.</strong>[MeZu] C. Menini, M. Zuccoli, Equivalence Theorems and Hopf-Galois Extensions, J. Algebra194 (1997), 245-274.[Mesa] B. Mesablishvili, Entwining structures in monoidal categories, J. Algebra 319 (2008),[Po]2496-2517.N. Popescu, Abelian Categories with Application to Rings and Modules, AcademicPress, London & New York, (1973).[Pr] H. Prüfer, Theorie der Abelschen Gruppen. I. Grundeigenschaften, Math. Z. 20,(1924),165-187.265

266[RW] R. Rosebrugh, R.J. Wood, Distributive laws and factorization, J. Pure Appl. Algebra175 (2002), 327-353.[Sch1] P. Schauenburg, Quantum torsors with fewer axioms, preprint arXivmath.QA/0302003.[Scha4] P. Schauenburg, Hopf-Galois and bi-Galois extensions, in: Galois theory, Hopf algebras,and semiabelian categories, Fields Inst. Commun. 43, AMS 2004, 469-515.[Sch4] P. Schauenburg, Quantum torsors and Hopf-Galois objects, preprint arXiv[SS]math.QA/0208047.P. Schauenburg and H.-J. Schneider, On generalized Hopf Galois extensions, J. PureAppl. Algebra 202 (2005), 168-194.[Schn1] H.-J. Schneider, Principal homogeneous spaces for arbitrary Hopf algebras, Israel J.Math. 72 (1990), 167-195.[Schn2] H.-J. Schneider, Normal basis and transitivity of crossed products for Hopf algebras, J.Algebra 152 (1992), 289-31<strong>2.</strong>[St] R. Street, The formal theory of monads, J. Pure Appl. Algebra 2 (1972), 149-168.[W] R. Wisbauer, Algebras versus coalgebras, Applied Categorical Structures 16 (2008),255-295.[Wis] R. Wisbauer, On Galois Comodules, Comm. Algebra 34, (2006), 2683-271<strong>1.</strong>