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

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244Lemma A.4. Let A be an abelian category and let U ∈ A. Then, for every exactsequence in Athe sequence0 → Hom A (U, K) Hom A(U,k)−→0 → K k−→ Xis an exact sequence of abelian groups.f−→ Y,Hom A (U, X) Hom A(U,f)−→ Hom A (U, Y )Proof. Let h ∈ Hom A (U, K). Then Hom A (U, k) (h) = kh. Since k is a monomorphismit follows that kh = 0 if and only if h = 0. Hence Hom A (U, k) is also amonomorphism. Also (Hom A (U, f) ◦ Hom A (U, k)) (h) = fkh = 0. This impliesthat Im (Hom A (U, k)) ⊆ Ker (Hom A (U, f)). Let now g ∈ Hom A (U, X) and assumethat Hom A (U, f) (g) = 0 i.e. fg = 0. Since (K, k) = Ker (f) there exists a morphismg ′ : U → K such that g = kg ′ = Hom A (U, k) (g ′ ) ∈ Im (Hom A (U, k)). Thereforewe get that Ker (Hom A (U, f)) ⊆ Im (Hom A (U, k)) and hence Ker (Hom A (U, f)) =Im (Hom A (U, k)).□Lemma A.5. In the assumptions and notations of A.1, let (T, H) be an adjunctionwhere T : B → A and H : A → B and let f : X → Y be a morphism in A. Then fis a monomorphism (resp. epimorphism) if and only if T H (f) is a monomorphism(resp. epimorphism).Proof. First of all, for every X ∈ A let ɛX : T H (X) → X be the counit of theadjunction (T, H). Then, in view of Proposition A.3 and Proposition <strong>2.</strong>32, ɛX is anisomorphism and for every morphism f : X → Y in A we havef ◦ ɛX = ɛY ◦ T H (f) .Thus f is mono (resp. epi) if and only if T H (f) is mono (resp. epi).Lemma A.6. In the assumptions and notations of A.1, let m, n ∈ N, m, n ≥ 1,let f : B m → B n be a morphism of right B-modules, let X = Coim ( f ) and letj : X → B n be the canonical injection. Let T : Mod-B → A be a left adjoint of thefunctor Hom A (U, −) : A → Mod-B. Then T (j) is a monomorphism.Proof. Let m, n ∈ N, m, n ≥ 1, let f : U m → U n be a morphism in A. Let usconsider the diagram□0 Ker (f)k U m fU n iCoim (f)00where p is the canonical epimorphism and i is the canonical monomorphism. Byapplying to it the functor H = Hom A (U, −), in view of Lemma A.4, we obtain thep
245diagramH(k)H(f)0 Ker (H (f)) = H (Ker (f)) H (U m ) H (U n )H(p)H(i)H (Coim (f))0Since Coim (H (f)) = Coker (H (k)) and H (p) ◦ H (k) = 0 there exists a uniquemorphism ζ : Coim (H (f)) → H (Coim (f)) such that(259) H (p) = ζ ◦ qwhere q : H (U m ) → Coim (H (f)) is the canonical epimorphism. Let j : Coim (H (f)) →H (U n ) be the canonical monomorphism such that j ◦ q = H (f). Then fromj ◦ q ◦ H (k) = H (f ◦ k) = 0 we get that(260) q ◦ H (k) = 0.H(k)0 Ker (H (f)) = H (Ker (f)) H (U m ) H (UqH(p)n )H(i) jH (Coim (f)) Coim (H (f)) = Coker (H (k))ζ000From H (i) ◦ ζ ◦ q (259)= H (i) ◦ H (p) = H (f) = j ◦ q, since q is an epimorphism, weget that(261) H (i) ◦ ζ = j.Let us apply T to it having in mind that T is right exactH(f)0 T H (Ker (f))T H(k) T H (U m T H(f))T H (U n )T (q)T H(p)T H(i) T (j)ξT H (Coim (f)) T (Coim (H (f)))T (ζ)Let us prove that T (j) is mono. From formula (260) we obtain that T (q) ◦T H (k) = T (q ◦ H (k)) = 0. SinceT H (Coim (f)) = Coim (T H (f)) = Coker (Ker (T H (f)))= Coker (T H (Ker (f))) = Coker (T H (k))there exists a unique ξ : T H (Coim (f)) → T (Coim (H (f))) such that(262) ξ ◦ T H (p) = T (q) .
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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
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
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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 ′
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 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 262 and 263: 262andP (J ′ )e f ′−→ P (I
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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