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

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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
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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 ◦
Page 15 and 16: f ↦→ Rfis bijective for every X
Page 17 and 18: Remark 3.10. Let A = (A, m A , u A
Page 19 and 20: 19and fromµ A P ◦ ( µ A P A ) =
Page 21 and 22: and thusk ◦ (u A QZ) ◦ (Qz) = h
Page 23 and 24: and since A preserves equalizers, A
Page 25 and 26: Conversely, let Φ be a functorial
Page 27 and 28: Proof. Apply Proposition 3.24 to th
Page 29 and 30: Since Q is a left A-module functor,
Page 31 and 32: (Q BB F, p QB F ) = Coequ Fun(µBQ
Page 33 and 34: Theorem 3.37. Let B = (B, m B , u B
Page 35 and 36: where A UG B F : B → A is such th
Page 37 and 38: Proposition 3.44. Let A = (A, m A ,
Page 39 and 40: Note that, since f and g are A-bili
Page 41 and 42: Proposition 3.54. Let (L, R) be an
Page 43 and 44: 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 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 96 and 97: 96defΦ= ( QP A µ Q)◦(QP σ A Q
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
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116= ( Aε C Q ) ◦ ( cocan1 −1
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118so that we getχ= (Cx) ◦ (C ρ
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120We want to prove that Γ is an o
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122and since Dε D is an epimorphis
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124χ= (Cχ) ◦ (C ρ Q P Q ) ◦
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126Now, since cocan 1 : AC → QP i
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1287. Herds and Coherds7.1.
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130◦ ( σ A QQQ ) ◦ (A µ Q P Q
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132= µ B Q ◦ (A µ Q B ) ◦ ( A
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134Assume now that there is another
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136and hence we get(160) x ◦ (χP
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138Proposition 7.7. In the setting
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140We calculateA µ Q ◦ ( σ A Q
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142x=and=δ C=(l= QlQ ̂QQ)◦ (QP
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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
Page 178 and 179:
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
Page 188 and 189:
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 ∈
Page 220 and 221:
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,
Page 230 and 231:
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
Page 252 and 253:
252Proposition A.12 ([ELGO2, Propos
Page 254 and 255:
254(⇒) Let {A i } i∈Ibe a famil
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256We will prove that h : ∐ B i
Page 258 and 259:
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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