Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...
Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ... Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...
134Assume now that there is another morphism t : ̂Q → X such that ξ = t ◦ ν ′ 0. Thenwe havet ◦ l ◦ (P x) (149)= t ◦ ν ′ 0 ◦ (yP ) = ξ ◦ (yP ) = ν ◦ ν ′ 0 ◦ (yP ) (149)= ν ◦ l ◦ (P x) .Since l ◦ (P x) is an epimorphism, we deduce that t = ν.(2) We want to equip ̂Q with the structure of a B-A-bimodule functor. To beginwith, let us prove a number of equalities. Let us calculateso thatχ ◦ ( Qz l) = χ ◦ (QP χ) ◦ (Qδ D P Q) (98)= χ ◦ (χP Q) ◦ (Qδ D P Q)(105)= χ ◦ ( Qε D P Q ) = χ ◦ (Qz r )(153) χ ◦ ( Qz l) = χ ◦ (Qz r ) .Let(154) b = m A ◦ (xA) .Thenso thatx ◦ (χP ) (102)= m A ◦ (xx) = m A ◦ (xA) ◦ (QP x)(155) x ◦ (χP ) = b ◦ (QP x) .We haveand hence(P χ) ◦ ( z l P Q ) = (P χ) ◦ (P χP Q) ◦ (δ D P QP Q)(98)= (P χ) ◦ (P QP χ) ◦ (δ D P QP Q)δ D= (P χ) ◦ (δ D P Q) ◦ (DP χ) = z l ◦ (DP χ)y ◦ (P χ) ◦ ( z l P Q ) = y ◦ z l ◦ (DP χ) ycoequ= y ◦ z r ◦ (DP χ) = y ◦ ( ε D P Q ) ◦ (DP χ)so that we getε D = y ◦ (P χ) ◦ ( ε D P QP Q ) = y ◦ (P χ) ◦ (z r P Q)(156) y ◦ (P χ) ◦ ( z l P Q ) = y ◦ (P χ) ◦ (z r P Q) .From previous equalities, it follows thatl ◦ (P b) ◦ ( z l P A ) ◦ (DP QP x) = zll ◦ (P b) ◦ (P QP x) ◦ ( z l P QP )(155)= l ◦ (P x) ◦ (P χP ) ◦ ( z l P QP ) (149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ ( z l P QP )(156)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (z r P QP ) (149)= l ◦ (P x) ◦ (P χP ) ◦ (z r P QP )(155)= l ◦ (P b) ◦ (P QP x) ◦ (z r P QP ) zr= l ◦ (P b) ◦ (z r P A) ◦ (DP QP x) .Since DP QP x is an epimorphism, we obtain(157) l ◦ (P b) ◦ ( z l P A ) = l ◦ (P b) ◦ (z r P A)
135that isl ◦ (P m A ) ◦ (P xA) ◦ ( z l P A ) = l ◦ (P m A ) ◦ (P xA) ◦ (z r P A) .From
- Page 84 and 85: 84i.e. Aα is a functorial morphism
- Page 86 and 87: 86Then we haveA µ CCX ◦ ( A∆ C
- Page 88 and 89: 884.23) is a functor à : C A → C
- Page 90 and 91: 90Let θ l = ( σ B P Q ) ◦ (P τ
- Page 92 and 93: 925)σ A = ( ε C A ) ◦ ( Cσ A)
- Page 94 and 95: 94(ii) the functorial morphism can
- Page 96 and 97: 96defΦ= ( QP A µ Q)◦(QP σ A Q
- Page 98 and 99: 98AU A can AA F = can AA F = ( CσA
- Page 100 and 101: 100Similarly, one can prove the sta
- Page 102 and 103: 102(b) A comonad C = ( C, ∆ C ,
- Page 104 and 105: 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 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
- 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 ⊗
135that isl ◦ (P m A ) ◦ (P xA) ◦ ( z l P A ) = l ◦ (P m A ) ◦ (P xA) ◦ (z r P A) .From <str<strong>on</strong>g>2.</str<strong>on</strong>g>9 we have that(̂QA, lA)= Coequ Fun((P xA) ◦(z l P A ) , (P xA) ◦ (z r P A) ) .Hence there exists a unique functorial morphism µ A Q b : ̂QA → ̂Q which satisfies (150).( )Now we want to prove that ̂Q, µ A Q b is a right A-module functor. First let us provethat µ A b Qis associative that isµ A b Q◦( ) ( )µ A Q bA= µ A Q b ◦ ̂QmA .We computeµ A b Q◦( )µ A Q bA◦ (lAA) (150)= µ A Q b ◦ (lA) ◦ (P m A A)(150)= l ◦ (P m A ) ◦ (P m A A) m Aass= l ◦ (P m A ) ◦ (P Am A )(150)= µ A b Q◦ (lA) ◦ (P Am A ) l = µ A b Q◦( )̂QmA ◦ (lAA) .Since lAA is an epimorphism, we get that µ A b Qis associative. Let us prove that µ A b Qis unital that isin fact( )µ A Q b ◦ ̂QuA( )µ A Q b ◦ ̂QuA = ̂Q◦ l l = µ A b Q◦ (lA) ◦ (P Au A ) (150)= l ◦ (P m A ) ◦ (P Au A ) Am<strong>on</strong>ad= land since l is an epimorphism we c<strong>on</strong>clude. We want to prove a series of equalities.First of all, let us prove that(158)(159)In fact, we haveand(χP ) ◦ ( QP w l) = w l ◦ (χP C)(χP ) ◦ (QP w r ) = w r ◦ (χP C)(χP ) ◦ ( QP w l) = (χP ) ◦ (QP χP ) ◦ (QP QP δ C )(98)= (χP ) ◦ (χP QP ) ◦ (QP QP δ C )χ= (χP ) ◦ (QP δ C ) ◦ (χP C) = w l ◦ (χP C)(χP ) ◦ (QP w r ) = (χP ) ◦ ( QP QP ε C) χ=(QP εC ) ◦ (χP ) = w r ◦ (χP C) .From (158) we deduce thatx ◦ (χP ) ◦ ( QP w l) (158)= x ◦ w l ◦ (χP C)xcoequ= x ◦ w r ◦ (χP C) (159)= x ◦ (χP ) ◦ (QP w r )
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