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

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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 ′ )) ,
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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
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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
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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
Page 212 and 213: 212Definition 9.27. Let k be a comm
Page 214 and 215: 214∆coass= a ⊗ c (1) ⊗ A 1 A
Page 216 and 217: 216Definition 9.32.</strong
Page 218 and 219: 218Let us compute, for every d ∈
Page 220 and 221: 220• 2-cells: monad functor trans
Page 222 and 223: 222We now want to prove that ρ Q·
Page 224 and 225: 224Proof. Let us consider the follo
Page 226 and 227: 226and since p Q•B Q ′ ,Q ′
Page 228 and 229: 228(241)= (1 Q • B l Q ′) ζ Q,
Page 230 and 231: 230On the other hand, we can first
Page 232 and 233: 232so that we define the map φ F (
Page 234 and 235: 234Since we have(B • B (Q · A) ,
Page 236 and 237: 2362-cells. This means that a comon
Page 238 and 239: 238defined by settingu Q·A = ( u (
Page 240 and 241: 240the unique A-bimodule morphism s
Page 242 and 243: 242Let F be a finite subset of Hom
Page 244 and 245: 244Lemma A.4. Let A be an abelian c
Page 246 and 247: 246We haveT (ζ) ◦ ξ ◦ T H (p)
Page 248 and 249: 248where k : Ker (Coker (f ◦ p))
Page 250 and 251: 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
Page 256 and 257: 256We will prove that h : ∐ B i
Page 258 and 259: 258Proposition A.19. Let (T, H) be
Page 260 and 261: 260Since P is finite Hom A (P, P )
Page 264 and 265: 264hence there exists a unique morp
Page 266: 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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