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 ...
234Since we have(B • B (Q · A) , p B,Q·A ) = Coequ C (m B · 1 Q · 1 A , 1 B · λ Q·A )and we are assuming that the composition with 1-cells preserves coequalizer, we alsohave(P · (B • B (Q · A)) , 1 P · p B,Q·A ) = Coequ C (1 P · m B · 1 Q · 1 A , 1 P · 1 B · λ Q·A ) .Therefore, there exists a unique isomorphism h : (P · B)• B (Q · A) → P ·(B • B (Q · A))such that(248) h (p P ·B,Q·A ) = 1 P · p B,Q·A .Moreover, by Proposition 1
and thus we can rewrite the above relation([ρ (P ·B)•B (Q·A) (p P ·B,Q·A · 1 A ) = ρ (P ·B)•B (Q·A) h −1 (1 P · p B,Q·A ) ] )· 1 Aandso that= ρ (P ·B)•B (Q·A)(h−1 · 1 A)(1P · p B,Q·A · 1 A )p P ·B,Q·A (1 P · 1 B · ρ Q·A ) = h −1 (1 P · p B,Q·A ) (1 P · 1 B · ρ Q·A )(237)= h −1 ( 1 P · ρ B•B (Q·A))(1P · p B,Q·A · 1 A )ρ (P ·B)•B (Q·A)(h−1 · 1 A)(1P · p B,Q·A · 1 A ) = h −1 ( 1 P · ρ B•B (Q·A))(1P · p B,Q·A · 1 A ) .Since 1 P · p B,Q·A · 1 A is epi, we get( )ρ (P ·B)•B (Q·A) h−1 · 1 ( )A = h−11 P · ρ B•B (Q·A)and thusso that we get thatρ (P ·B)•B (Q·A) = h −1 ( 1 P · ρ B•B (Q·A))(h · 1A )ρ (P ·B)•B (Q·A) ≃ 1 P · ρ B•B (Q·A) ≃ 1 P · ρ Q·A = ρ P ·Q·A .1
- 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
- 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 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 )
- Page 262 and 263: 262andP (J ′ )e f ′−→ P (I
- Page 264 and 265: 264hence there exists a unique morp
- Page 266: 266[RW] R. Rosebrugh, R.J. Wood, Di
234Since we have(B • B (Q · A) , p B,Q·A ) = Coequ C (m B · 1 Q · 1 A , 1 B · λ Q·A )and we are assuming that the compositi<strong>on</strong> with 1-cells preserves coequalizer, we alsohave(P · (B • B (Q · A)) , 1 P · p B,Q·A ) = Coequ C (1 P · m B · 1 Q · 1 A , 1 P · 1 B · λ Q·A ) .Therefore, there exists a unique isomorphism h : (P · B)• B (Q · A) → P ·(B • B (Q · A))such that(248) h (p P ·B,Q·A ) = 1 P · p B,Q·A .Moreover, by Propositi<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>6, P · (B • B (Q · A)) ≃ P · (Q · A) so that we get(P · B) • B (Q · A) ≃ P · (Q · A) = P · Q · A.Now, the left C-module structure λ (P ·B)•B (Q·A) of (P · B) • B (Q · A), by (236) isuniquely determined byλ (P ·B)•B (Q·A) (1 C · p P ·B,Q·A ) = p P ·B,Q·A (λ P ·B · 1 Q · 1 A ) .By (248) we getp P ·B,Q·A = h −1 (1 P · p B,Q·A )and thus we can rewrite the above relati<strong>on</strong>(λ (P ·B)•B (Q·A) (1 C · p P ·B,Q·A ) = λ (P ·B)•B (Q·A) 1C · [h −1 (1 P · p B,Q·A ) ])andso that= λ (P ·B)•B (Q·A)(1C · h −1) (1 C · 1 P · p B,Q·A )p P ·B,Q·A (λ P ·B · 1 Q · 1 A ) = h −1 (1 P · p B,Q·A ) (λ P ·B · 1 Q · 1 A )defλ P ·B= h −1 (1 P · p B,Q·A ) (λ P · 1 B · 1 Q · 1 A )λ=Ph ( ) −1 λ P · 1 B•B (Q·A) (1C · 1 P · p B,Q·A )λ (P ·B)•B (Q·A)(1C · h −1) (1 C · 1 P · p B,Q·A ) = h −1 ( λ P · 1 B•B (Q·A))(1C · 1 P · p B,Q·A ) .Since 1 C · 1 P · p B,Q·A is epi, we get(λ (P ·B)•B (Q·A) 1C · h −1) = h ( )−1 λ P · 1 B•B (Q·A)and thusso that we get thatλ (P ·B)•B (Q·A) = h −1 ( λ P · 1 B•B (Q·A))(1C · h)λ (P ·B)•B (Q·A) ≃ λ P · 1 B•B (Q·A) ≃ λ P · 1 Q·A ≃ λ P ·Q·A .Similarly, the right A-module structure λ (P ·B)•B (Q·A) of (P · B) • B (Q · A), by (237)is uniquely determined byBy (248) we getρ (P ·B)•B (Q·A) (p P ·B,Q·A · 1 A ) = p P ·B,Q·A (1 P · 1 B · ρ Q·A ) .p P ·B,Q·A = h −1 (1 P · p B,Q·A )
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