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 ...
138Proposition 7.7. In the setting of Theorem 6.29 and Proposition 7.6, there existtwo functorial morphisms σ A : Q ̂Q → A and σ B : ̂QQ → B where σ A is A-bilinearand σ B is B-bilinear and they fulfill(161) σ A ◦ (Ql) = m A ◦ (xA)and(162) σ B ◦ (ν ′ 0Q) = m B ◦ (By) .Moreover the associative conditions hold, that isA µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B) and B µb Q◦Proof. First we want to prove thatIn fact we have( ) (σ B ̂Q = µ A Q b ◦ ̂QσA).m A ◦ (xA) ◦ (QP x) ◦ ( Qz l P ) = m A ◦ (xA) ◦ (QP x) ◦ (Qz r P ) .m A ◦ (xA) ◦ (QP x) ◦ ( Qz l P ) (102)= x ◦ (χP ) ◦ ( Qz l P )= x ◦ (χP ) ◦ (QP χP ) ◦ (Qδ D P QP ) (128)= x ◦ (χP ) ◦ (χP QP ) ◦ (Qδ D P QP )(130)= x ◦ (χP ) ◦ ( Qε D P QP ) = x ◦ (χP ) ◦ (Qz r P )(102)= m A ◦ (xx) ◦ (Qz r P ) = m A ◦ (xA) ◦ (QP x) ◦ (Qz r P ) .Since Q preserves coequalizers we have(Q ̂Q, Ql) = Coequ Fun((QP x) ◦(Qz l P ) , (QP x) ◦ (Qz r P ) )so that there exists a functorial morphism σ A : Q ̂Q → A which satisfies (161). Nowwe want to show that σ A is A-bilinear that is the following equalities hold( )σ A ◦A µ Q ̂Q = m A ◦ ( Aσ A))σ A ◦(Qµ A Q b = m A ◦ ( σ A A ) .We computem A ◦ ( Aσ A) ◦ (AQl) (161)= m A ◦ (Am A ) ◦ (AxA)m A ass= m A ◦ (m A A) ◦ (AxA) (104)= m A ◦ (xA) ◦ (A µ Q P A )(161)= σ A ◦ (Ql) ◦ (A µ Q P A ) A µ Q= σ A ◦Since AQl is an epimorphism, we get that σ A ◦(A µ Q ̂Q( )A µ Q ̂Q)◦ (AQl) .m A ◦ ( σ A A ) ◦ (QlA) (161)= m A ◦ (m A A) ◦ (xAA)= m A ◦ (Am A ) ◦ (xAA) = x m A ◦ (xA) ◦ (QP m A ))(161)= σ A ◦ (Ql) ◦ (QP m A ) (150)= σ A ◦(Qµ A Q b ◦ (QlA)= m A ◦ ( Aσ A) . We compute
)Since QlA is an epimorphism, we obtain that σ A ◦(Qµ A Q b = m A ◦ ( σ A A ) . Symmetrically,wewant to define σ B . We prove thatIn fact, we havem B ◦ (By) ◦ (yP Q) ◦ ( P w l Q ) = m B ◦ (By) ◦ (yP Q) ◦ (P w r Q) .m B ◦ (By) ◦ (yP Q) ◦ ( P w l Q ) = m B ◦ (yy) ◦ ( P w l Q )(109)= y ◦ (P χ) ◦ ( P w l Q ) = y ◦ (P χ) ◦ (P χP Q) ◦ (P QP δ C Q)(128)= y ◦ (P χ) ◦ (P QP χ) ◦ (P QP δ C Q)(129)= y ◦ (P χ) ◦ ( P QP ε C Q ) = y ◦ (P χ) ◦ (P w r Q)(109)= m B ◦ (yy) ◦ (P w r Q) = m B ◦ (By) ◦ (yP Q) ◦ (P w r Q) .By Lemma
- 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 134 and 135: 134Assume now that there is another
- Page 136 and 137: 136and hence we get(160) x ◦ (χP
- 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 ⊗
- 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
)Since QlA is an epimorphism, we obtain that σ A ◦(Qµ A Q b = m A ◦ ( σ A A ) . Symmetrically,wewant to define σ B . We prove thatIn fact, we havem B ◦ (By) ◦ (yP Q) ◦ ( P w l Q ) = m B ◦ (By) ◦ (yP Q) ◦ (P w r Q) .m B ◦ (By) ◦ (yP Q) ◦ ( P w l Q ) = m B ◦ (yy) ◦ ( P w l Q )(109)= y ◦ (P χ) ◦ ( P w l Q ) = y ◦ (P χ) ◦ (P χP Q) ◦ (P QP δ C Q)(128)= y ◦ (P χ) ◦ (P QP χ) ◦ (P QP δ C Q)(129)= y ◦ (P χ) ◦ ( P QP ε C Q ) = y ◦ (P χ) ◦ (P w r Q)(109)= m B ◦ (yy) ◦ (P w r Q) = m B ◦ (By) ◦ (yP Q) ◦ (P w r Q) .By Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>9, we have(̂QQ, ν′0 Q)= Coequ Fun((yP Q) ◦(P w l Q ) , (yP Q) ◦ (P w r Q) )139so that there exists a functorial morphism σ B : ̂QQ → B which satisfies (162). Nowwe want to show that σ B is B-bicolinear that is the following equalities hold( )σ B ◦B µb QQ = m B ◦ ( Bσ B)m B ◦ ( σ B B ) ( )= σ B ◦ ̂QµBQWe calculatem B ◦ ( Bσ B) ◦ (Bν ′ 0Q) (162)= m B ◦ (Bm B ) ◦ (BBy)m B ass= m B ◦ (m B B) ◦ (BBy) m =Bm B ◦ (By) ◦ (m B P Q)( )(162)= σ B ◦ (ν 0Q) ′ ◦ (m B P Q) (151)= σ B ◦B µb QQ ◦ (Bν 0Q) ′ .( )Since Bν 0Q ′ is an epimorphism, we deduce that σ B ◦B µb QQ = m B ◦ ( Bσ B) . Wecomputem B ◦ ( σ B B ) ◦ (ν ′ 0QB) (162)= m B ◦ (m B B) ◦ (ByB)m B ass= m B ◦ (Bm B ) ◦ (ByB) (108)(162)= σ B ◦ (ν ′ 0Q) ◦ ( BP µ B Q= m B ◦ (By) ◦ ( )BP µ B Q( )̂QµBQ ◦ (ν 0QB) ′ .) ν ′0= σ B ◦Since ν 0QB ′ is an epimorphism, we get that m B ◦ ( σ B B ) = σ B ◦have to prove the associative c<strong>on</strong>diti<strong>on</strong>sA µ Q ◦ ( σ A Q ) = µ B Q ◦ ( Qσ B)( ) (σ B ̂Q = µ A Q b ◦ ̂QσA).B µb Q◦(̂QµBQ). Finally we
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