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 ...
136and hence we get(160) x ◦ (χP ) ◦ ( QP w l) = x ◦ (χP ) ◦ (QP w r ) .We observe thatν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ ( BP w l) ◦ (yP QP C)= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (yP QP ) ◦ ( P QP w l)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ ( P QP w l)(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ ( P QP w l)(149)= l ◦ (P x) ◦ (P χP ) ◦ ( P QP w l)(160)= l ◦ (P x) ◦ (P χP ) ◦ (P QP w r )(149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P QP w r )(109)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P QP w r )= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (yP QP ) ◦ (P QP w r )= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (BP w r ) ◦ (yP QP C)since yP QP C is an epimorphism, we obtain thatν 0 ′ ◦ (m B P ) ◦ (ByP ) ◦ ( BP w l) = ν 0 ′ ◦ (m B P ) ◦ (ByP ) ◦ (BP w r ) .(Since B preserves coequalizers, we have that B ̂Q,)Bν 0′ = Coequ Fun ((ByP ) ◦( ) BP wl, (ByP ) ◦ (BP w r )) so that there exists a unique functorial morphism B µb Q:( )̂Q, B µb QB ̂Q → ̂Q which satisfies (151).Now we want to show thatB-module functor. First let us prove that B µb Qis associative that is) ( )(B B µb Qm B ̂Q .B µb Q◦= B µb Q◦is a leftWe have)B µb Q◦(B B µb Q◦ (BBν ′ 0) (151)= B µb Q◦ (Bν ′ 0) ◦ (Bm B P )(151)= ν ′ 0 ◦ (m B P ) ◦ (Bm B P ) m Bass(151)= B µb Q◦ (Bν ′ 0) ◦ (m B BP ) m B= B µb Q◦= ν 0 ′ ◦ (m B P ) ◦ (m B BP )( )m B ̂Q ◦ (BBν 0) ′ .Since BBν 0 ′ is an epimorphism, we get that B µb Qis associative. Let us prove thatB µb Qis unital that is( )B µb Q◦ u B ̂Q = ̂Q.We calculate( )B µb Q◦ u B ̂Q◦ ν ′ 0 = B µb Q◦ (Bν ′ 0) ◦ (u B BP ) (151)= ν ′ 0 ◦ (m B P ) ◦ (u B BP ) = ν ′ 0.
Since ν 0 ′ is an epimorphism, we get that B µb Qis unital. Finally we have to prove thecompatibility condition) ( )(Bµ A Q b = µ A Q b ◦B µb QA .B µb Q◦137We haveB µb Q◦) (Bµ A Q b ◦ (BlA) ◦ (BP xx) ◦ (yP QP QP )(150)= B µb Q◦ (Bl) ◦ (BP m A ) ◦ (BP xx) ◦ (yP QP QP )(102)= B µb Q◦ (Bl) ◦ (BP x) ◦ (BP χP ) ◦ (yP QP QP )(149)= B µb Q◦ (Bν ′ 0) ◦ (ByP ) ◦ (BP χP ) ◦ (yP QP QP )y= B µb Q◦ (Bν ′ 0) ◦ (ByP ) ◦ (yP QP ) ◦ (P QP χP )= B µb Q◦ (Bν ′ 0) ◦ (yyP ) ◦ (P QP χP )(151)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P QP χP )(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P QP χP )(98)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P χP QP )(149)= l ◦ (P x) ◦ (P χP ) ◦ (P χP QP )(102)= l ◦ (P m A ) ◦ (P xx) ◦ (P χP QP )= l ◦ (P m A ) ◦ (P xA) ◦ (P QP x) ◦ (P χP QP )χ= l ◦ (P m A ) ◦ (P xA) ◦ (P χP A) ◦ (P QP QP x)(150)= µ A b Q◦ (lA) ◦ (P xA) ◦ (P χP A) ◦ (P QP QP x)(149)= µ A b Q◦ (ν ′ 0A) ◦ (yP A) ◦ (P χP A) ◦ (P QP QP x)(109)= µ A b Q◦ (ν ′ 0A) ◦ (m B P A) ◦ (yyP A) ◦ (P QP QP x)= µ A b Q◦ (ν ′ 0A) ◦ (m B P A) ◦ (ByP A) ◦ (yP QP A) ◦ (P QP QP x)y= µ A Q b ◦ (ν 0A) ′ ◦ (m B P A) ◦ (ByP A) ◦ (BP QP x) ◦ (yP QP QP )( )(151)= µ A Q b ◦B µb QA ◦ (Bν 0A) ′ ◦ (ByP A) ◦ (BP QP x) ◦ (yP QP QP )( )(149)= µ A Q b ◦B µb QA ◦ (BlA) ◦ (BP xA) ◦ (BP QP x) ◦ (yP QP QP )( )= µ A Q b ◦B µb QA ◦ (BlA) ◦ (BP xx) ◦ (yP QP QP )Since (BlA) ◦ (BP xx) ◦ (yP QP QP ) is an epimorphism, we conclude. Then(̂Q, B µb Q, µ A b Q)is a B-A-bimodule functor.□
- 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 134 and 135: 134Assume now that there is another
- 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 ⊗
- Page 184 and 185: 184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d
Since ν 0 ′ is an epimorphism, we get that B µb Qis unital. Finally we have to prove thecompatibility c<strong>on</strong>diti<strong>on</strong>) ( )(Bµ A Q b = µ A Q b ◦B µb QA .B µb Q◦137We haveB µb Q◦) (Bµ A Q b ◦ (BlA) ◦ (BP xx) ◦ (yP QP QP )(150)= B µb Q◦ (Bl) ◦ (BP m A ) ◦ (BP xx) ◦ (yP QP QP )(102)= B µb Q◦ (Bl) ◦ (BP x) ◦ (BP χP ) ◦ (yP QP QP )(149)= B µb Q◦ (Bν ′ 0) ◦ (ByP ) ◦ (BP χP ) ◦ (yP QP QP )y= B µb Q◦ (Bν ′ 0) ◦ (ByP ) ◦ (yP QP ) ◦ (P QP χP )= B µb Q◦ (Bν ′ 0) ◦ (yyP ) ◦ (P QP χP )(151)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P QP χP )(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P QP χP )(98)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P χP QP )(149)= l ◦ (P x) ◦ (P χP ) ◦ (P χP QP )(102)= l ◦ (P m A ) ◦ (P xx) ◦ (P χP QP )= l ◦ (P m A ) ◦ (P xA) ◦ (P QP x) ◦ (P χP QP )χ= l ◦ (P m A ) ◦ (P xA) ◦ (P χP A) ◦ (P QP QP x)(150)= µ A b Q◦ (lA) ◦ (P xA) ◦ (P χP A) ◦ (P QP QP x)(149)= µ A b Q◦ (ν ′ 0A) ◦ (yP A) ◦ (P χP A) ◦ (P QP QP x)(109)= µ A b Q◦ (ν ′ 0A) ◦ (m B P A) ◦ (yyP A) ◦ (P QP QP x)= µ A b Q◦ (ν ′ 0A) ◦ (m B P A) ◦ (ByP A) ◦ (yP QP A) ◦ (P QP QP x)y= µ A Q b ◦ (ν 0A) ′ ◦ (m B P A) ◦ (ByP A) ◦ (BP QP x) ◦ (yP QP QP )( )(151)= µ A Q b ◦B µb QA ◦ (Bν 0A) ′ ◦ (ByP A) ◦ (BP QP x) ◦ (yP QP QP )( )(149)= µ A Q b ◦B µb QA ◦ (BlA) ◦ (BP xA) ◦ (BP QP x) ◦ (yP QP QP )( )= µ A Q b ◦B µb QA ◦ (BlA) ◦ (BP xx) ◦ (yP QP QP )Since (BlA) ◦ (BP xx) ◦ (yP QP QP ) is an epimorphism, we c<strong>on</strong>clude. Then(̂Q, B µb Q, µ A b Q)is a B-A-bimodule functor.□
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