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

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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
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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
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 164 and 165: 164(194)=) )(p QB ̂QA ◦(Qpb Q◦
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
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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
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 ′
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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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