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

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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
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Contents1.
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linearity and compatibility conditi
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5and since g ◦ f is an epimorphis
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Proof. Clearly (qP )◦(αP ) = (qP
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Lemma 2.13 ([BM, L
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i.e. Hom B (Y, iX) equalizes Hom B
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13such thatd 0 ◦ v = Id Yd 1 ◦
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f ↦→ Rfis bijective for every X
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Remark 3.10. Let A = (A, m A , u A
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19and fromµ A P ◦ ( µ A P A ) =
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and thusk ◦ (u A QZ) ◦ (Qz) = h
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and since A preserves equalizers, A
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Conversely, let Φ be a functorial
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Proof. Apply Proposition 3.24 to th
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Since Q is a left A-module functor,
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(Q BB F, p QB F ) = Coequ Fun(µBQ
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Theorem 3.37. Let B = (B, m B , u B
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where A UG B F : B → A is such th
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Proposition 3.44. Let A = (A, m A ,
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Note that, since f and g are A-bili
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Proposition 3.54. Let (L, R) be an
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Corollary 3.58. Let (L, R) be an ad
Page 45 and 46: Definition 4.2. A
Page 47 and 48: Proposition 4.13. Let C = ( C, ∆
Page 49 and 50: Then we have(P Cx) ◦ ( ρ C P X )
Page 51 and 52: and since C preserves coequalizers,
Page 53 and 54: Proof. Apply Corollary 4.24 to the
Page 55 and 56: Let( (CQ ) ()D, ι Q) C = Equ Fun
Page 58 and 59: 58F D right D-comodule functors Q :
Page 60 and 61: 60prove that C ν D : C F D → (C
Page 62 and 63: 624.2. The compari
Page 64 and 65: 64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY,
Page 66 and 67: 66for every ( X, C ρ X)∈ C A, th
Page 68 and 69: 68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂η
Page 70 and 71: 70In particular(49) d ϕ(CX, ∆ C
Page 72 and 73: 72We have to prove that (LD ϕ , Ld
Page 74 and 75: 74we have that Ld ϕ K ϕ Y is mono
Page 76 and 77: and since d is mono we get that(ε
Page 78 and 79: 78Corollary 4.63 (Beck’s Precise
Page 80 and 81: 80We compute(LRɛLY ′ ) ◦ ( LR
Page 82 and 83: 82Proof. First of all we prove that
Page 84 and 85: 84i.e. Aα is a functorial morphism
Page 86 and 87: 86Then we haveA µ CCX ◦ ( A∆ C
Page 88 and 89: 884.23) is a functor Ã : C A → C
Page 90 and 91: 90Let θ l = ( σ B P Q ) ◦ (P τ
Page 92 and 93: 925)σ A = ( ε C A ) ◦ ( Cσ A)
Page 94 and 95: 94(ii) the functorial morphism can
Page 98 and 99: 98AU A can AA F = can AA F = ( CσA
Page 100 and 101: 100Similarly, one can prove the sta
Page 102 and 103: 102(b) A comonad C = ( C, ∆ C ,
Page 104 and 105: 104We calculateso that we getx ◦
Page 106 and 107: 106There exist functorial morphisms
Page 108 and 109: 108andsatisfying(B, y) = Coequ Fun(
Page 110 and 111: 1104) With notations of Theorem 6.2
Page 112 and 113: 112Then ν : Y → D is the unique
Page 114 and 115: 114= A µ Q ◦ ( Aε C Q ) ◦ (AC
Page 116 and 117: 116= ( Aε C Q ) ◦ ( cocan1 −1
Page 118 and 119: 118so that we getχ= (Cx) ◦ (C ρ
Page 120 and 121: 120We want to prove that Γ is an o
Page 122 and 123: 122and since Dε D is an epimorphis
Page 124 and 125: 124χ= (Cχ) ◦ (C ρ Q P Q ) ◦
Page 126 and 127: 126Now, since cocan 1 : AC → QP i
Page 128 and 129: 1287. Herds and Coherds7.1.
Page 130 and 131: 130◦ ( σ A QQQ ) ◦ (A µ Q P Q
Page 132 and 133: 132= µ B Q ◦ (A µ Q B ) ◦ ( A
Page 134 and 135: 134Assume now that there is another
Page 136 and 137: 136and hence we get(160) x ◦ (χP
Page 138 and 139: 138Proposition 7.7. In the setting
Page 140 and 141: 140We calculateA µ Q ◦ ( σ A Q
Page 142 and 143: 142x=and=δ C=(l= QlQ ̂QQ)◦ (QP
Page 144 and 145: 144(◦ ρ D ̂QQ)Q◦ (QlQ) ◦ (Q
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146given byWe computeσ B = m B ◦
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148andy= ′m B ◦ (ν B B) ◦ (y
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150Now we compute(hQ) ◦ ( Qχ )
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152Thus we obtainσ B ◦ ( ) (P µ
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154Thus hQ is an isomorphism with i
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156) ( )l=(pb Q AQ B ◦ ̂QA µ QB
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158In fact we haveTherefore we dedu
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160χ= h 1 ◦ (P xQ B ) ◦ (P QP
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162so that we obtain:(190)We comput
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164(194)=) )(p QB ̂QA ◦(Qpb Q◦
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166= Ξ ◦ (A A U A λ) ◦ (xx A
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168)(155)= k 2 ◦(Qpb Q◦ (Ql A U
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170) ) (χ= ρ ◦(p QB ̂QA ◦(Qp
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172Theorem 8.13. Let A and B be cat
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174l = eC ρ L : L = − ⊗ B A
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and[µBQ ◦ ( Qσ B)] (− ⊗ T x
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178so that− ⊗ R 1 A ⊗ R c = (
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180− ⊗ T x ⊗ R 1 A ⊗ A f
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182(208)(209)(210)(211)(h1 ) 0 ⊗
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184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d
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186so that h 1 ⊗ h 2 ⊗ a ∈ A
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188= 〈( h (1) y (1))εH ( h (2) y
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190H C is faithfully coflat. Assume
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192=(Qε C H C) ( ∑ )−□ C k i
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194Following Theorem 6.29, we now c
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196)û E(ε C H C (h) = û E (π (h
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198Letandα l = (ϕ ⊗ H) ( (x ⊗
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200This map is well-defined, in fac
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202We now have to prove that this m
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2041) − ⊗ B Σ A preserves the
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206functorial isomorphism. In parti
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208coaction ρ C Σ : Σ → Σ ⊗
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210Now, we consider a particular ca
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212Definition 9.27. Let k be a comm
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214∆coass= a ⊗ c (1) ⊗ A 1 A
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216Definition 9.32.</strong
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218Let us compute, for every d ∈
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220• 2-cells: monad functor trans
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222We now want to prove that ρ Q·
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224Proof. Let us consider the follo
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226and since p Q•B Q ′ ,Q ′
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228(241)= (1 Q • B l Q ′) ζ Q,
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230On the other hand, we can first
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232so that we define the map φ F (
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234Since we have(B • B (Q · A) ,
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2362-cells. This means that a comon
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238defined by settingu Q·A = ( u (
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240the unique A-bimodule morphism s
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242Let F be a finite subset of Hom
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244Lemma A.4. Let A be an abelian c
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246We haveT (ζ) ◦ ξ ◦ T H (p)
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248where k : Ker (Coker (f ◦ p))
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250be the codiagonal map of the ρ
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252Proposition A.12 ([ELGO2, Propos
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254(⇒) Let {A i } i∈Ibe a famil
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256We will prove that h : ∐ B i
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258Proposition A.19. Let (T, H) be
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260Since P is finite Hom A (P, P )
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262andP (J ′ )e f ′−→ P (I
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264hence there exists a unique morp
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266[RW] R. Rosebrugh, R.J. Wood, Di
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Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

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