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 ...
168)(155)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP b A U) ◦ (QP QP x A U))(154)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xAU A ) ◦ (QP QP xU A ))= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xx A U))(102)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U))(149)= k 2 ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (QyP A U) ◦ (QP χP A U))(109)= k 2 ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (Qm B P A U) ◦ (QyyP A U)) ( )(151)= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0P ′ A U) ◦ (QyyP A U)) ( )= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0P ′ A U) ◦ (QByP A U) ◦ (QyP QP A U)) ( )(149)= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBlP A U) ◦ (QBP x A U) ◦ (QyP QP A U)) ( ) (y= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U ◦ (QP QlP A U) ◦ (QP QP x A U)) ) ((175)= k 2 ◦(Q B µb QA◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)def A µ QB) ) (= k2 ◦(Q B Uλ BB ̂QA ◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)) (Since(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U) is epi, we get))k 2 ◦(µ B QBU B ̂QA = k 2 ◦(Q B Uλ BB ̂QA .)()From(Q BB ̂QA , p QB ̂QA = Coequ Fun µ B QBU B ̂QA , Q B Uλ BB ̂QA we deduce thatthere exists a unique functorial morphism ̂k : Q BB ̂QA → Y such that(201) ̂k ◦(p QB ̂QA)= k 2 .Moreover we have( ) ))̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) = k 2 ◦(Qpb Q◦ (Ql A U)i.e.(202) ̂k ◦(p QB ̂QA)◦Now we compute(200)= k 1 ◦ (Ql A U) (199)= k ◦ (m AA U) ◦ (xA A U)̂k ◦ Ξ ◦ (xA U) (191)= ̂k ◦) (Qpb Q◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)(202)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP u AA U) x = k ◦ (m AA U) ◦ (Au AA U) ◦ (x A U)= k ◦ (x A U) .
Since x A U is epi we get that̂k ◦ Ξ = k.Let now ̂k ′ : Q BB ̂QA → Y be another functorial morphism such that ̂k ′ ◦ Ξ = k.Then we have( ) )̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)(191)= ̂k ◦ Ξ ◦ (x A U) = k ◦ (x A U) = ̂k ′ ◦ Ξ ◦ (x A U)(191)= ̂k) )′ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)) )Since we already observed from (196) that(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)is an epimorphism, we have ̂k ′ = ̂k.( )Hence we have proved that Q BB ̂QA , Ξ =Coequ Fun (m AA U, A A U A λ) and, in view of (202) , we get that (197)holds.Theorem 8.9. Within the assumptions and notations of Theorem 6.29, we have afunctorial isomorphism A A ∼ = A Q BB ̂QA .Proof. In Proposition 8.8 we ( have proved)the existence of a functorial morphismΞ : AU A → Q BB ̂QA such that Q BB ̂QA , Ξ = Coequ Fun (m AA U, A A Uλ A ). By Proposition3.13 and Proposition 3.14 also ( A U, A Uλ A ) = Coequ Fun ((m AA U, A A Uλ A )) , inview of uniqueness of a coequalizer up to isomorphisms, there exists a functorialisomorphismρ : A U A Q BB ̂QA = Q BB ̂QA → A U such that ρ ◦ Ξ = A Uλ A .Now since( A Uλ A ) ◦ (m AA U) = ( A Uλ A ) ◦ (A A Uλ A )and since ρ ◦ Ξ = A Uλ A , by Proposition 8.8, we deduce that) )(203) ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) = ( A Uλ A ) ◦ (m AA U) ◦ (xA A U)Now we want to prove that ρ lifts to a functorial morphism A Q BB ̂QA → A A of A-leftmodule functors. First we observe that ( A U, A Uλ A ) is an A-left module functor inview of Proposition 3.13. Also(Q BB ̂QA , A µ ̂Q)QB B A is an A-left module functor(see proof of Proposition 3.30) where A µ QB = A Uλ AA Q B : AQ B → Q B . To showthat ρ is morphism of A-left module functors we have to prove( A Uλ A ) ◦ (Aρ) = ρ ◦ ( A µ QB B ̂Q A ).We have(A ρ ◦ µ ̂Q))) )QB B A ◦(Ap QB ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U))(174)= ρ ◦(p (A )) )QB ̂QA ◦ µ Q BU B ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U)◦ (QP QP x A U))) )(101)= ρ ◦(p QB ̂QA ◦(χ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U)169□
- 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 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 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 ⊗
- Page 210 and 211: 210Now, we consider a particular ca
- Page 212 and 213: 212Definition 9.27. Let k be a comm
- Page 214 and 215: 214∆coass= a ⊗ c (1) ⊗ A 1 A
- Page 216 and 217: 216Definition 9.32.</strong
168)(155)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP b A U) ◦ (QP QP x A U))(154)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xAU A ) ◦ (QP QP xU A ))= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP m AA U) ◦ (QP xx A U))(102)= k 2 ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U))(149)= k 2 ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (QyP A U) ◦ (QP χP A U))(109)= k 2 ◦(Qpb Q◦ (Qν 0AU) ′ ◦ (Qm B P A U) ◦ (QyyP A U)) ( )(151)= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0P ′ A U) ◦ (QyyP A U)) ( )= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBν 0P ′ A U) ◦ (QByP A U) ◦ (QyP QP A U)) ( )(149)= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ (QBlP A U) ◦ (QBP x A U) ◦ (QyP QP A U)) ( ) (y= k 2 ◦(Qpb Q◦ Q B µb Q AU ◦ Qy ̂Q)A U ◦ (QP QlP A U) ◦ (QP QP x A U)) ) ((175)= k 2 ◦(Q B µb QA◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)def A µ QB) ) (= k2 ◦(Q B Uλ BB ̂QA ◦(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U)) (Since(QBpb Q◦ Qy ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP x A U) is epi, we get))k 2 ◦(µ B QBU B ̂QA = k 2 ◦(Q B Uλ BB ̂QA .)()From(Q BB ̂QA , p QB ̂QA = Coequ Fun µ B QBU B ̂QA , Q B Uλ BB ̂QA we deduce thatthere exists a unique functorial morphism ̂k : Q BB ̂QA → Y such that(201) ̂k ◦(p QB ̂QA)= k 2 .Moreover we have( ) ))̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) = k 2 ◦(Qpb Q◦ (Ql A U)i.e.(202) ̂k ◦(p QB ̂QA)◦Now we compute(200)= k 1 ◦ (Ql A U) (199)= k ◦ (m AA U) ◦ (xA A U)̂k ◦ Ξ ◦ (xA U) (191)= ̂k ◦) (Qpb Q◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)(202)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP u AA U) x = k ◦ (m AA U) ◦ (Au AA U) ◦ (x A U)= k ◦ (x A U) .
Change language