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

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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
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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
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Definition 4.2. A
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Proposition 4.13. Let C = ( C, ∆
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Then we have(P Cx) ◦ ( ρ C P X )
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and since C preserves coequalizers,
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Proof. Apply Corollary 4.24 to the
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Let( (CQ ) ()D, ι Q) C = Equ Fun
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58F D right D-comodule functors Q :
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60prove that C ν D : C F D → (C
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624.2. The compari
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64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY,
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66for every ( X, C ρ X)∈ C A, th
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68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂η
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70In particular(49) d ϕ(CX, ∆ C
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72We have to prove that (LD ϕ , Ld
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74we have that Ld ϕ K ϕ Y is mono
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and since d is mono we get that(ε
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78Corollary 4.63 (Beck’s Precise
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80We compute(LRɛLY ′ ) ◦ ( LR
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82Proof. First of all we prove that
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84i.e. Aα is a functorial morphism
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86Then we haveA µ CCX ◦ ( A∆ C
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884.23) is a functor Ã : C A → C
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90Let θ l = ( σ B P Q ) ◦ (P τ
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925)σ A = ( ε C A ) ◦ ( Cσ A)
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94(ii) the functorial morphism can
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96defΦ= ( QP A µ Q)◦(QP σ A Q
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98AU A can AA F = can AA F = ( CσA
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100Similarly, one can prove the sta
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102(b) A comonad C = ( C, ∆ C ,
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104We calculateso that we getx ◦
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106There exist functorial morphisms
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108andsatisfying(B, y) = Coequ Fun(
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1104) With notations of Theorem 6.2
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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
Page 146 and 147: 146given byWe computeσ B = m B ◦
Page 148 and 149: 148andy= ′m B ◦ (ν B B) ◦ (y
Page 150 and 151: 150Now we compute(hQ) ◦ ( Qχ )
Page 152 and 153: 152Thus we obtainσ B ◦ ( ) (P µ
Page 154 and 155: 154Thus hQ is an isomorphism with i
Page 156 and 157: 156) ( )l=(pb Q AQ B ◦ ̂QA µ QB
Page 158 and 159: 158In fact we haveTherefore we dedu
Page 160 and 161: 160χ= h 1 ◦ (P xQ B ) ◦ (P QP
Page 162 and 163: 162so that we obtain:(190)We comput
Page 166 and 167: 166= Ξ ◦ (A A U A λ) ◦ (xx A
Page 168 and 169: 168)(155)= k 2 ◦(Qpb Q◦ (Ql A U
Page 170 and 171: 170) ) (χ= ρ ◦(p QB ̂QA ◦(Qp
Page 172 and 173: 172Theorem 8.13. Let A and B be cat
Page 174 and 175: 174l = eC ρ L : L = − ⊗ B A
Page 176 and 177: and[µBQ ◦ ( Qσ B)] (− ⊗ T x
Page 178 and 179: 178so that− ⊗ R 1 A ⊗ R c = (
Page 180 and 181: 180− ⊗ T x ⊗ R 1 A ⊗ A f
Page 182 and 183: 182(208)(209)(210)(211)(h1 ) 0 ⊗
Page 184 and 185: 184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d
Page 186 and 187: 186so that h 1 ⊗ h 2 ⊗ a ∈ A
Page 188 and 189: 188= 〈( h (1) y (1))εH ( h (2) y
Page 190 and 191: 190H C is faithfully coflat. Assume
Page 192 and 193: 192=(Qε C H C) ( ∑ )−□ C k i
Page 194 and 195: 194Following Theorem 6.29, we now c
Page 196 and 197: 196)û E(ε C H C (h) = û E (π (h
Page 198 and 199: 198Letandα l = (ϕ ⊗ H) ( (x ⊗
Page 200 and 201: 200This map is well-defined, in fac
Page 202 and 203: 202We now have to prove that this m
Page 204 and 205: 2041) − ⊗ B Σ A preserves the
Page 206 and 207: 206functorial isomorphism. In parti
Page 208 and 209: 208coaction ρ C Σ : Σ → Σ ⊗
Page 210 and 211: 210Now, we consider a particular ca
Page 212 and 213: 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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