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

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156) ( )l=(pb Q AQ B ◦ ̂QA µ QBp bQ coeq=(150)=def A µ QB( )= pb Q AQ B ◦m=A◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P u A QP Q B U)( )̂QA Uλ AA Q B ◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U)◦ (P u A QP Q B U)) )(pb Q AQ B ◦(µ A Q bQ B ◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P u A QP Q B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P m A Q B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P u A QP Q B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P m A Q B U) ◦ (P u A AQ B U) ◦ (P xQ B U))Amonad=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P xQ B U)Now we want to prove that(pb Q AQ B)◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ ( z l BU ))=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (z r BU) .We have(pb Q AQ B)◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ ( z l BU ))=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U) ◦ (δ D P Q B U))(178)=(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (δ D P Q B U))x=(pb Q AQ B ◦ (lQ B ) ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (δ D P Q B U))δ=D(pb Q AQ B ◦ (lQ B ) ◦ (P xQ B ) ◦ (δ D P Q B ) ◦ (DP p Q ))(149)=(pb Q AQ B ◦ (ν 0Q ′ B ) ◦ (yP Q B ) ◦ (δ D P Q B ) ◦ (DP p Q ))(110)=(pb Q AQ B ◦ (ν 0Q ′ B ) ◦ (u B P Q B ) ◦ ( )ε D P Q B ◦ (DP pQ ))(176)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ ( )ε D P Q B ◦ (DP pQ )) (pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ ( ε D P Q B U )ε D =u A=Since, by Lemma <strong>2.</strong>10)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (z r BU)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (z r BU)(B B U, y B U) = Coequ Fun(zlB U, z r BU ) ,there exists a functorial morphism α : B B U → ̂Q AA Q B such that(179) α ◦ (y B U) = (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) .
157Now we want to prove that(̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ).Let us show the fork property for α, that is(180) α ◦ (m BB U) = α ◦ (B B Uλ B ) .We have(179)=α ◦ (B B Uλ B ) ◦ (yy B U) = α ◦ (B B Uλ B ) ◦ (yB B U) ◦ (P Qy B U)y= α ◦ (y B U) ◦ (P Q B Uλ B ) ◦ (P Qy B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P Q B Uλ B ) ◦ (P Qy B U))=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (P Q B Uλ B ) ◦ (P Qy B U)( )defp Q= pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ ( P µ B QBU ) ◦ (P Qy B U))(107)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (P χ B U))u=A(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U)(179)= α ◦ (y B U) ◦ (P χ B U) (109)= α ◦ (m BB U) ◦ (yy B U)and, since yy B U is an epimorphism, we conclude. Now, let us consider a functorialmorphism h : B B U → X such that h ◦ (m BB U) = h ◦ (B B Uλ B ) . We have to showthat there exists a unique functorial morphism ĥ : ̂Q AA Q B → X such thatĥ ◦ α = h.First we will show that there exists a functorial morphism ĥ such that ĥ and h fulfill(177) i.e.)ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h ◦ (y B U) ◦ ( P A µ QB U ) .To do this, we need a series of equalities. First of all, let us show that(181) y ◦ ( P A µ Q)◦(P AµBQ)◦ (P AQy) ◦ (P xQP Q) = mB ◦ (yy) ◦ (P χP Q) .In fact, we haveNow let us prove thaty ◦ ( P A µ Q)◦(P AµBQ)◦ (P AQy) ◦ (P xQP Q)(107)= y ◦ ( P A µ Q)◦ (P Aχ) ◦ (P xQP Q)x= y ◦ ( )P A (101)µ Q ◦ (P xQ) ◦ (P QP χ) = y ◦ (P χ) ◦ (P QP χ)(98)= y ◦ (P χ) ◦ (P χP Q) (109)= m B ◦ (yy) ◦ (P χP Q) .(182) (yy) ◦ (P χP Q) = (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .
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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 ◦
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
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 ◦
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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 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
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
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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 ∈
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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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