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 ...
160χ= h 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (P χP Q B U)x= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (P χP Q B U)(184)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ (P χP Q B U)(187)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(184)= h 1 ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(186)= h 2 ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(188)= h 2 ◦ ( ̂Q A µ QB ) ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q ) .Since (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q ) is an epimorphism, we deduce thath 2 ◦ ( ̂Q)A µ QB ) = h 2 ◦(µ A Q bQ B .By Proposition 8.3 we have(̂QAA Q B , pb Q AQ B)= Coequ Fun(µ A b QAU A Q B , ̂Q A Uλ AA Q B)Coequ Fun(µ A b QQ B , ̂Q A µ QB)and hence we infer that there exists a functorial morphismĥ : ̂Q AA Q B → X such that)ĥ ◦(pb Q AQ B = h 2 .Hence we get)ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h 2 ◦ (l A U A Q B ) ◦ (P Ap Q )= h 2 ◦ (lQ B ) ◦ (P Ap Q ) (186)= h 1 ◦ (P Ap Q ) (184)= h ◦ (y B U) ◦ ( P A µ Q BU )and hence equality (177) is proven. Now we have( )ĥ ◦ α ◦ (y B U) (179)= ĥ ◦ pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)(177)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P u A Q B U) A µ Q is unital= h ◦ (y B U)so we getĥ ◦ α = h.Let now ĥ′ : ̂QAA Q B → X be another functorial morphism such that ĥ′ ◦ α = h.Then we haveĥ ◦ (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)(179)= ĥ ◦ α ◦ (y BU) = h ◦ (y B U) = ĥ′ ◦ α ◦ (y B U)(179)= ĥ′ ◦ (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) .)Note that, by (178) , since the second term is an epimorphism,(pb Q AQ B ◦(l A U A Q B )◦(P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U) is epi and so (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)is also ( an epimorphism. Therefore we deduce that ĥ′ = ĥ. Hence we have provedthat ̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ). □=
Theorem 8.6. Within the assumptions and notations of Theorem 6.29, we have afunctorial isomorphism B ̂QAA Q B∼ = B B.Proof. In Proposition 8.5 we have proved the existence of a functorial morphism α :B B U → ̂Q(AA Q B such that ̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B. ) . By Proposition3.13 and Proposition 3.14 also ( B U, ( B Uλ B )) = Coequ Fun (m BB U, B B Uλ B. ). Inview of uniqueness of a coequalizer up to isomorphisms, there exists a functorialisomorphismNow sinceβ : ̂Q AA Q B = B U B ̂QAA Q B → B U such that β ◦ α = B Uλ B .( B Uλ B ) ◦ (m BB U) = ( B Uλ B ) ◦ (B B Uλ B )and since β ◦ α = B Uλ B , by applying (177) where ”ĥ” = β and ”h” = BUλ B , wededuce that)β ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = ( B Uλ B ) ◦ (y B U) ◦ ( P A µ Q BU )equivalentlyi.e.(189) β ◦)β ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U)= ( B Uλ B ) ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U)(101)= ( B Uλ B ) ◦ (y B U) ◦ (P χ B U)(pb Q AQ B)◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U) = ( B Uλ B ) ◦ (y B U) ◦ (P χ B U) .Recall ( that, in view of Proposition 3.13, ( B U, B Uλ B ) is an B-left module functor.Also ̂QAA Q B , B µb QA A B)Q is an B-left module functor (see proof of Proposition3.30 and Lemma 3.17) where B µb QA= B Uλ BB ̂QA : B ̂Q A → ̂Q A .Now we want toprove that β lifts to a functorial morphism B ̂QAA Q B → B B i.e. thatβ :(̂QAA Q B , B µb QA AQ B)→ ( B U, B Uλ B )is a morphism of B-left module functors. Thus we have to proveWe calculate( B Uλ B ) ◦ (Bβ) = β ◦ ( B µb QA AQ B )(y ̂Q)◦ (P Ql) ◦ (P QP x)B µb Q◦ (Bl) ◦ (yP A) ◦ (P QP x) = B µb Q◦((149)= B µb Q◦ y ̂Q)◦ (P Qν 0) ′ ◦ (P QyP ) = y B µb Q◦ (Bν 0) ′ ◦ (yBP ) ◦ (P QyP )(151)= ν ′ 0 ◦ (m B P ) ◦ (yBP ) ◦ (P QyP ) = ν ′ 0 ◦ (m B P ) ◦ (yyP )(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) (149)= l ◦ (P x) ◦ (P χP )161
- 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 ◦
- 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 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
- Page 206 and 207: 206functorial isomorphism. In parti
- Page 208 and 209: 208coaction ρ C Σ : Σ → Σ ⊗
160χ= h 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (P χP Q B U)x= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (P χP Q B U)(184)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ (P χP Q B U)(187)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(184)= h 1 ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(186)= h 2 ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(188)= h 2 ◦ ( ̂Q A µ QB ) ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q ) .Since (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q ) is an epimorphism, we deduce thath 2 ◦ ( ̂Q)A µ QB ) = h 2 ◦(µ A Q bQ B .By Propositi<strong>on</strong> 8.3 we have(̂QAA Q B , pb Q AQ B)= Coequ Fun(µ A b QAU A Q B , ̂Q A Uλ AA Q B)Coequ Fun(µ A b QQ B , ̂Q A µ QB)and hence we infer that there exists a functorial morphismĥ : ̂Q AA Q B → X such that)ĥ ◦(pb Q AQ B = h 2 .Hence we get)ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h 2 ◦ (l A U A Q B ) ◦ (P Ap Q )= h 2 ◦ (lQ B ) ◦ (P Ap Q ) (186)= h 1 ◦ (P Ap Q ) (184)= h ◦ (y B U) ◦ ( P A µ Q BU )and hence equality (177) is proven. Now we have( )ĥ ◦ α ◦ (y B U) (179)= ĥ ◦ pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)(177)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P u A Q B U) A µ Q is unital= h ◦ (y B U)so we getĥ ◦ α = h.Let now ĥ′ : ̂QAA Q B → X be another functorial morphism such that ĥ′ ◦ α = h.Then we haveĥ ◦ (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)(179)= ĥ ◦ α ◦ (y BU) = h ◦ (y B U) = ĥ′ ◦ α ◦ (y B U)(179)= ĥ′ ◦ (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) .)Note that, by (178) , since the sec<strong>on</strong>d term is an epimorphism,(pb Q AQ B ◦(l A U A Q B )◦(P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U) is epi and so (pb Q AQ B ) ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U)is also ( an epimorphism. Therefore we deduce that ĥ′ = ĥ. Hence we have provedthat ̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ). □=
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