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 ...
184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d∈C= a ⊗ b ⊗ d.Therefore, A ⊗ ϕ is an isomorphism and since A/k is faithfully flat, also ϕ is anisomorphism, i.e. Ĉ ∼ = H.Now we want to compute the comonad D. We haveθ l = ( A ⊗ A ⊗ σ B) ◦ (τ ⊗ A) : A ⊗ A → A ⊗ A ⊗ Aθ l (a ⊗ b) = a 0 ⊗ a 1 1 ⊗ a 2 1bθ r = A ⊗ A ⊗ u B : A ⊗ A → A ⊗ A ⊗ Aθ r (a ⊗ b) = a ⊗ b ⊗ 1 A .((Since (D, j) = Equ ) )Fun A ⊗ A ⊗ σB◦ (τ ⊗ A) , A ⊗ A ⊗ u B , we have{∑D = a i ⊗ b i | ∑ ( ) ai ⊗ ( a i) 1⊗ ( a i) 20 1 1 bi = ∑ }a i ⊗ b i ⊗ 1 A .By applying A ⊗ can to θ l and θ r , for every ∑ a i ⊗ b i ∈ D, we get that[ ] ( ∑ )(∑ ((A ⊗ can) ◦ θl a i ⊗ b i = (A ⊗ can)) ai ⊗ ( a i) 1⊗ ( a i) )20 1 1 bi= ∑ ( ) ai ⊗ ( a i) (1 (a ) (i 2 (a0 1 1)0 bi i⊗) 21)1bi= ∑ ( ) ai ⊗ ( a i) (1 (a ) )i 2 ( ) (0 1 1 bi (a ⊗ ) )i 2 ( )00 1 bi11(206)= ∑ ( ) ai ⊗ ( b i) ⊗ ( a i) ( ) (∑ )0 0 1 bi = 1 ρH A⊗A a i ⊗ b iand(∑ )[(A ⊗ can) ◦ θ r ] a i ⊗ b i(∑ )= (A ⊗ can) a i ⊗ b i ⊗ 1 A = ∑ a i ⊗ b i ⊗ 1 H .Since (A ⊗ can) is an isomorphism, we get D = Equ ( ρ H A⊗A , A ⊗ A ⊗ u H)= (A ⊗ A) co(H) .By Theorem 6.5, ∆ D and ε D are uniquely determined by(P τ) ◦ j = (jj) ◦ ∆ D and σ B ◦ j = u B ◦ ε D .Let ∑ a i ⊗ b i ∈ D = (A ⊗ A) co(H) . Then we have(∑ ) (∑ )(jj) ◦ ∆ D a i ⊗ b i = [(τ ⊗ A) ◦ j] a i ⊗ b iand also(uA ◦ ε D) ( ∑a i ⊗ b i )= ∑ (ai ) 0 ⊗ ( a i) 11 ⊗ ( a i) 21 ⊗ bi(∑ )= (τ ⊗ A) a i ⊗ b i(∑ )= (m A ◦ j) a i ⊗ b i = ∑ a i · b iSince ∑ a i ⊗ b i ∈ (A ⊗ A) co(H) , we have∑ ( ) ai ⊗ ( b i) ⊗ ( a i) ( )0 0 1 bi = ∑ a i ⊗ b i ⊗ 11 Hso that, by applying m A ⊗ H, since A is an H-comodule algebra, we get∑ (a i b i) ⊗ ( a i b i) = ∑ ( ) ( )0 1 ai0 bi ⊗ ( a i) ( )0 1 bi = ∑ a i · b i ⊗ 11 H
so that (θ l P ) ◦ (P i) : H ⊗ A → A ⊗ A ⊗ A∑i.e. ai · b i ∈ A co(H) = k1 A∼ = k. Note that from mA ◦ (A ⊗ u A ) ◦ (r A ) −1 = Id Awe get that A ⊗ u A is a monomorphism and hence, since A is faithfully flat overk, also u A is a monomorphism. We denote by ν = u |k1 AA: k → k1 A the obviousisomorphism. Thus from(uA ◦ ε D) ( ∑ )a i ⊗ b i = ∑ a i · b i185we getLet us computei.e.ε D (∑a i ⊗ b i )= v −1 (∑ai · b i ).θ l P = A ⊗ θ l : a ⊗ b ⊗ c ↦→ a ⊗ b 0 ⊗ b 1 1 ⊗ b 2 1cθ r P = A ⊗ θ r : a ⊗ b ⊗ c ↦→ a ⊗ b ⊗ c ⊗ 1 AA ⊗ θ l : A ⊗ A ⊗ AP i : C ⊗ A → A ⊗ A ⊗ Ah ⊗ a ↦→ h 1 ⊗ h 2 ⊗ ah ⊗ a ↦→ h 1 ⊗ ( h 2) ⊗ ( h 2) 1h 2) 20 1 1 h1 1 ⊗ h 2 1 ⊗ h 1 2 ⊗ h 2 2a(θ r P ) ◦ (P i) : H ⊗ A → A ⊗ A ⊗ Ah ⊗ a ↦→ h 1 ⊗ h 2 ⊗ a ⊗ 1 A .Recall that H ⊆ C ⊆ A ⊗ A. Such Q = (C ⊗ A) ∩ [A ⊗ D] ∼[= ” (H ⊗ A) ∩A ⊗ (A ⊗ A) co(H)] ”. Note that, assuming that A preserves equalizers, forevery h ⊗ a ∈ H ⊗ A, i.e. h 1 ⊗ h 2 ⊗ a ∈ C ⊗ A, h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) ifand only if h 1 ⊗ h 2 ⊗ a ∈ Equ ( A ⊗ ρ H A⊗A , A ⊗ A ⊗ A ⊗ u H)whereEqu ( )A ⊗ ρ H A⊗A, A ⊗ A ⊗ A ⊗ u H= Equ ( (A ⊗ A ⊗ A ⊗ m H ) ◦ (A ⊗ A ⊗ f ⊗ H) ◦ ( ) )A ⊗ ρ H A ⊗ ρ H A , A ⊗ A ⊗ A ⊗ uH= {a ⊗ b ⊗ c | a ⊗ b 0 ⊗ c 0 ⊗ b 1 c 1 = a ⊗ b ⊗ c ⊗ 1 H }so that h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) = Equ ( A ⊗ ρ H A⊗A , A ⊗ A ⊗ A ⊗ u H)if andonly ifh 1 ⊗ ( h 2) 0 ⊗ a 0 ⊗ ( h 2) 1 a 1 = h 1 ⊗ h 2 ⊗ a ⊗ 1 H .Let us prove that Q = (H ⊗ A) co(H) . Now,(H ⊗ A) co(H) = {h ⊗ a | h 1 ⊗ a 0 ⊗ h 2 a 1 = h ⊗ a ⊗ 1 H }where h ∈ H, h ϕ↦→ h 1 ⊗ h 2 ∈ C.1) Let h ⊗ a ∈ (H ⊗ A) co(H) and let us prove that h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) .Since h ⊗ a ∈ (H ⊗ A) co(H) , we computeh 1 ⊗ ( h 2) 0 ⊗ a 0 ⊗ ( h 2) 1 a 1(207)= (h 1 ) 1 ⊗ (h 1 ) 2 ⊗ a 0 ⊗ h 2 a 1 = h 1 ⊗ h 2 ⊗ a ⊗ 1 H
- 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 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 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
- Page 218 and 219: 218Let us compute, for every d ∈
- Page 220 and 221: 220• 2-cells: monad functor trans
- Page 222 and 223: 222We now want to prove that ρ Q·
- Page 224 and 225: 224Proof. Let us consider the follo
- Page 226 and 227: 226and since p Q•B Q ′ ,Q ′
- Page 228 and 229: 228(241)= (1 Q • B l Q ′) ζ Q,
- Page 230 and 231: 230On the other hand, we can first
- Page 232 and 233: 232so that we define the map φ F (
184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d∈C= a ⊗ b ⊗ d.Therefore, A ⊗ ϕ is an isomorphism and since A/k is faithfully flat, also ϕ is anisomorphism, i.e. Ĉ ∼ = H.Now we want to compute the com<strong>on</strong>ad D. We haveθ l = ( A ⊗ A ⊗ σ B) ◦ (τ ⊗ A) : A ⊗ A → A ⊗ A ⊗ Aθ l (a ⊗ b) = a 0 ⊗ a 1 1 ⊗ a 2 1bθ r = A ⊗ A ⊗ u B : A ⊗ A → A ⊗ A ⊗ Aθ r (a ⊗ b) = a ⊗ b ⊗ 1 A .((Since (D, j) = Equ ) )Fun A ⊗ A ⊗ σB◦ (τ ⊗ A) , A ⊗ A ⊗ u B , we have{∑D = a i ⊗ b i | ∑ ( ) ai ⊗ ( a i) 1⊗ ( a i) 20 1 1 bi = ∑ }a i ⊗ b i ⊗ 1 A .By applying A ⊗ can to θ l and θ r , for every ∑ a i ⊗ b i ∈ D, we get that[ ] ( ∑ )(∑ ((A ⊗ can) ◦ θl a i ⊗ b i = (A ⊗ can)) ai ⊗ ( a i) 1⊗ ( a i) )20 1 1 bi= ∑ ( ) ai ⊗ ( a i) (1 (a ) (i 2 (a0 1 1)0 bi i⊗) 21)1bi= ∑ ( ) ai ⊗ ( a i) (1 (a ) )i 2 ( ) (0 1 1 bi (a ⊗ ) )i 2 ( )00 1 bi11(206)= ∑ ( ) ai ⊗ ( b i) ⊗ ( a i) ( ) (∑ )0 0 1 bi = 1 ρH A⊗A a i ⊗ b iand(∑ )[(A ⊗ can) ◦ θ r ] a i ⊗ b i(∑ )= (A ⊗ can) a i ⊗ b i ⊗ 1 A = ∑ a i ⊗ b i ⊗ 1 H .Since (A ⊗ can) is an isomorphism, we get D = Equ ( ρ H A⊗A , A ⊗ A ⊗ u H)= (A ⊗ A) co(H) .By Theorem 6.5, ∆ D and ε D are uniquely determined by(P τ) ◦ j = (jj) ◦ ∆ D and σ B ◦ j = u B ◦ ε D .Let ∑ a i ⊗ b i ∈ D = (A ⊗ A) co(H) . Then we have(∑ ) (∑ )(jj) ◦ ∆ D a i ⊗ b i = [(τ ⊗ A) ◦ j] a i ⊗ b iand also(uA ◦ ε D) ( ∑a i ⊗ b i )= ∑ (ai ) 0 ⊗ ( a i) 11 ⊗ ( a i) 21 ⊗ bi(∑ )= (τ ⊗ A) a i ⊗ b i(∑ )= (m A ◦ j) a i ⊗ b i = ∑ a i · b iSince ∑ a i ⊗ b i ∈ (A ⊗ A) co(H) , we have∑ ( ) ai ⊗ ( b i) ⊗ ( a i) ( )0 0 1 bi = ∑ a i ⊗ b i ⊗ 11 Hso that, by applying m A ⊗ H, since A is an H-comodule algebra, we get∑ (a i b i) ⊗ ( a i b i) = ∑ ( ) ( )0 1 ai0 bi ⊗ ( a i) ( )0 1 bi = ∑ a i · b i ⊗ 11 H
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