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 ...
212Definition 9.27. Let k be a commutative ring. An entwining structure (A, C, ψ)over k consists of• A = (A, m, u) a k-algebra• C = (C, ∆, ε) a k-coalgebra• ψ : C ⊗ A → A ⊗ C satisfying the following relations(222)(m ⊗ C)◦(A ⊗ ψ)◦(ψ ⊗ A) = ψ ◦(C ⊗ m) and ψ ◦(C ⊗ u)◦r −1Cand= (u ⊗ C)◦l−1C(223)(ψ ⊗ C) ◦ (C ⊗ ψ) ◦ (∆ ⊗ A) = (A ⊗ ∆) ◦ ψ and r A ◦ (A ⊗ ε) ◦ ψ = l A ◦ (ε ⊗ A) .Notation 9.28. Let (A, C, ψ) be an entwining structure over k. We will use sigmanotationψ (c ⊗ a) = ∑ a α ⊗ c αor with summation understoodψ (c ⊗ a) = a α ⊗ c α .Using this notation we can rewrite (222) and (223) as follows(224)(225)(ab) α⊗ c α = a α b β ⊗ c αβ , ψ (c ⊗ 1 A ) = (1 A ) α⊗ c α = 1 A ⊗ ca α ⊗ c α 1 ⊗ c α 2 = a αβ ⊗ c β 1 ⊗ c α 2 , a α ε C (c α ) = ε C (c) aMoreover we set, for every a, b, a ′ , b ′ ∈ A and c ∈ Ca (b ⊗ c) = ab ⊗ c and (b ⊗ c) b ′ = bψ (c ⊗ b ′ ) = bb ′ α ⊗ c α .We also define a map ∆ C : C = A ⊗ C → C ⊗ A C = A ⊗ C ⊗ A A ⊗ C, by settingand a map ε C : C → A, as follows∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2)ε C (a ⊗ c) = aε (c) .Definition 9.29. Let (A, C, ψ) be an entwining structure. An entwined (A, C, ψ)-module is a triple ( M, µ A M , ρC M)where(M, µAM)is a right A-module,(M, ρCM)is aright C-comodule such that the structures are compatible(µAM ⊗ C ) ◦ (M ⊗ ψ) ◦ ( ρ C M ⊗ A ) = ρ C M ◦ µ A Mi.e. for every m ∈ M and for every a ∈ A we have∑(226)(ma)0 ⊗ (ma) 1= ∑ m 0 a α ⊗ m α 1 .A morphism of entwined modules f : ( M, µ A M , (M) ρC → N, µAN , ρN) C is a morphism ofright A-modules and a morphism of right C-comodules. We denote by M C A (ψ) thecategory of entwined (A, C, ψ)-modules.Proposition 9.30 ([BrWi, 3
2) If A ⊗ C is an A-coring then (A, C, ψ) is an entwining structure whereψ (c ⊗ a) = (1 A ⊗ c) · a.3) If C = A ⊗ C is the A-coring associated to the entwining (A, C, ψ), thenM C A = (Mod-A)C ≃ M C A (ψ).Proof. 1) Let us define the A-bimodule structures on C = A ⊗ C. Set, for everya, b, a ′ , b ′ ∈ A and c ∈ Ca (b ⊗ c) = ab ⊗ c and (b ⊗ c) b ′ = bψ (c ⊗ b ′ ) = bb ′ α ⊗ c αi.e.a ′ (b ⊗ c) b ′ = a ′ bψ (c ⊗ b ′ ) = a ′ bb ′ α ⊗ c α .We check the right module structure. Let us compute(a ⊗ c) (bb ′ ) = aψ (c ⊗ bb ′ ) = a [ψ (C ⊗ m) (c ⊗ b ⊗ b ′ )]ψentw= a [((m ⊗ C) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)) (c ⊗ b ⊗ b ′ )][= a [((m ⊗ C) ◦ (A ⊗ ψ)) (b α ⊗ c α ⊗ b ′ )] = a (m ⊗ C)(b α ⊗ b ′ β ⊗ (c α ) β)](= a b α b ′ β ⊗ (c α ) β) = ab α b ′ β ⊗ (c α ) β = (ab α ⊗ c α ) b ′ = ((a ⊗ c) b) b ′ .Let us calculate(a ⊗ c) 1 A = aψ (c ⊗ 1 A ) = a [(ψ ◦ (C ⊗ u)) (c ⊗ 1 k )]= a [( ) ]ψ ◦ (C ⊗ u) ◦ r −1 ψentwC (c) = a [( ) ](u ⊗ C) ◦ l −1C (c)= a [(u ⊗ C) (1 k ⊗ c)] = a (1 A ⊗ c) = a ⊗ c.Now, let us check that it is a bimodule(a ′ (a ⊗ c)) b ′ = a ′ aψ (c ⊗ b ′ ) = a ′ (aψ (c ⊗ b ′ )) = a ′ ((a ⊗ c) b ′ ) .We define the coproduct on C = A⊗C, ∆ C : C = A⊗C → C ⊗ A C = A⊗C ⊗ A A⊗C,by setting∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2)where we denote ∆ (c) = c (1) ⊗ c (2) . It is straightforward to check that it is leftA-linear. Let us check it is also right A-linear. Let us compute∆ C ((a ⊗ c) b ′ ) = ∆ C (aψ (c ⊗ b ′ )) = ∆ C (ab ′ α ⊗ c α )(225)= ab ′ α ⊗ c α (1) ⊗ A 1 A ⊗ c α (2) = a (b ′ α) β⊗ c β (1) ⊗ A 1 A ⊗ c α (2)= aψ ( c (1) ⊗ b α) ′ ⊗A 1 A ⊗ c α (224)(2) = aψ ( )c (1) ⊗ b ′ α ⊗A ψ ( )c α (2) ⊗ 1 A= ( )a ⊗ c (1) b′α ⊗ A ψ ( ) ( )c α (2) ⊗ 1 A = a ⊗ c(1) ⊗A b ′ αψ ( )c α (2) ⊗ 1 A= a ⊗ c (1) ⊗ A(b′α ⊗ c α (2))1A = a ⊗ c (1) ⊗ A b ′ α ⊗ c α (2)= a ⊗ c (1) ⊗ A ψ ( c (2) ⊗ b ′) = a ⊗ c (1) ⊗ A ψ ( c (2) ⊗ b ′)= a ⊗ c (1) ⊗ A(1A ⊗ c (2))b ′ = ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))b ′ = ( ∆ C (a ⊗ c) ) b ′Let us check the coassociativity(∆ C ⊗ C ) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= a ⊗ c(1)(1) ⊗ A 1 A ⊗ c (1)(2) ⊗ A 1 A ⊗ c (2)213
- 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 ⊗
- Page 210 and 211: 210Now, we consider a particular ca
- 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 (
- Page 234 and 235: 234Since we have(B • B (Q · A) ,
- Page 236 and 237: 2362-cells. This means that a comon
- Page 238 and 239: 238defined by settingu Q·A = ( u (
- Page 240 and 241: 240the unique A-bimodule morphism s
- Page 242 and 243: 242Let F be a finite subset of Hom
- Page 244 and 245: 244Lemma A.4. Let A be an abelian c
- Page 246 and 247: 246We haveT (ζ) ◦ ξ ◦ T H (p)
- Page 248 and 249: 248where k : Ker (Coker (f ◦ p))
- Page 250 and 251: 250be the codiagonal map of the ρ
- Page 252 and 253: 252Proposition A.12 ([ELGO2, Propos
- Page 254 and 255: 254(⇒) Let {A i } i∈Ibe a famil
- Page 256 and 257: 256We will prove that h : ∐ B i
- Page 258 and 259: 258Proposition A.19. Let (T, H) be
- Page 260 and 261: 260Since P is finite Hom A (P, P )
2) If A ⊗ C is an A-coring then (A, C, ψ) is an entwining structure whereψ (c ⊗ a) = (1 A ⊗ c) · a.3) If C = A ⊗ C is the A-coring associated to the entwining (A, C, ψ), thenM C A = (Mod-A)C ≃ M C A (ψ).Proof. 1) Let us define the A-bimodule structures <strong>on</strong> C = A ⊗ C. Set, for everya, b, a ′ , b ′ ∈ A and c ∈ Ca (b ⊗ c) = ab ⊗ c and (b ⊗ c) b ′ = bψ (c ⊗ b ′ ) = bb ′ α ⊗ c αi.e.a ′ (b ⊗ c) b ′ = a ′ bψ (c ⊗ b ′ ) = a ′ bb ′ α ⊗ c α .We check the right module structure. Let us compute(a ⊗ c) (bb ′ ) = aψ (c ⊗ bb ′ ) = a [ψ (C ⊗ m) (c ⊗ b ⊗ b ′ )]ψentw= a [((m ⊗ C) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)) (c ⊗ b ⊗ b ′ )][= a [((m ⊗ C) ◦ (A ⊗ ψ)) (b α ⊗ c α ⊗ b ′ )] = a (m ⊗ C)(b α ⊗ b ′ β ⊗ (c α ) β)](= a b α b ′ β ⊗ (c α ) β) = ab α b ′ β ⊗ (c α ) β = (ab α ⊗ c α ) b ′ = ((a ⊗ c) b) b ′ .Let us calculate(a ⊗ c) 1 A = aψ (c ⊗ 1 A ) = a [(ψ ◦ (C ⊗ u)) (c ⊗ 1 k )]= a [( ) ]ψ ◦ (C ⊗ u) ◦ r −1 ψentwC (c) = a [( ) ](u ⊗ C) ◦ l −1C (c)= a [(u ⊗ C) (1 k ⊗ c)] = a (1 A ⊗ c) = a ⊗ c.Now, let us check that it is a bimodule(a ′ (a ⊗ c)) b ′ = a ′ aψ (c ⊗ b ′ ) = a ′ (aψ (c ⊗ b ′ )) = a ′ ((a ⊗ c) b ′ ) .We define the coproduct <strong>on</strong> C = A⊗C, ∆ C : C = A⊗C → C ⊗ A C = A⊗C ⊗ A A⊗C,by setting∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2)where we denote ∆ (c) = c (1) ⊗ c (2) . It is straightforward to check that it is leftA-linear. Let us check it is also right A-linear. Let us compute∆ C ((a ⊗ c) b ′ ) = ∆ C (aψ (c ⊗ b ′ )) = ∆ C (ab ′ α ⊗ c α )(225)= ab ′ α ⊗ c α (1) ⊗ A 1 A ⊗ c α (2) = a (b ′ α) β⊗ c β (1) ⊗ A 1 A ⊗ c α (2)= aψ ( c (1) ⊗ b α) ′ ⊗A 1 A ⊗ c α (224)(2) = aψ ( )c (1) ⊗ b ′ α ⊗A ψ ( )c α (2) ⊗ 1 A= ( )a ⊗ c (1) b′α ⊗ A ψ ( ) ( )c α (2) ⊗ 1 A = a ⊗ c(1) ⊗A b ′ αψ ( )c α (2) ⊗ 1 A= a ⊗ c (1) ⊗ A(b′α ⊗ c α (2))1A = a ⊗ c (1) ⊗ A b ′ α ⊗ c α (2)= a ⊗ c (1) ⊗ A ψ ( c (2) ⊗ b ′) = a ⊗ c (1) ⊗ A ψ ( c (2) ⊗ b ′)= a ⊗ c (1) ⊗ A(1A ⊗ c (2))b ′ = ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))b ′ = ( ∆ C (a ⊗ c) ) b ′Let us check the coassociativity(∆ C ⊗ C ) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= a ⊗ c(1)(1) ⊗ A 1 A ⊗ c (1)(2) ⊗ A 1 A ⊗ c (2)213
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