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 ...
228(241)= (1 Q • B l Q ′) ζ Q,B,Q ′p Q•B B,Q ′ (p Q,B · 1 Q ′)so that, since p Q•B B,Q ′ (p Q,B · 1 Q ′) is an epimorphism, we getr Q • B 1 Q ′ = (1 Q • B l Q ′) ζ Q,B,Q ′.Proposition 1
We now want to prove that f • B f ′ is a morphism of A-C-bimodules. Note that, byProposition 1
- 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
- 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 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 )
- 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
We now want to prove that f • B f ′ is a morphism of A-C-bimodules. Note that, byPropositi<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>5, P • B P ′ and Q • B Q ′ are A-C-bimodules. We computeλ Q•B Q ′ (1 A · f • B f ′ ) (1 A · p P,P ′) (247)= λ Q•B Q ′ (1 A · p Q,Q ′) (1 A · f · f ′ )(236)= p Q,Q ′ (λ Q · 1 Q ′) (1 A · f · f ′ )= p Q,Q ′ (λ Q · 1 Q ′) (1 A · f · 1 Q ′) (1 A · 1 P · f ′ )(245)= p Q,Q ′ (f · 1 Q ′) (λ P · 1 Q ′) (1 A · 1 P · f ′ )λ P= p Q,Q ′ (f · 1 Q ′) (1 P · f ′ ) (λ P · 1 P ′) = p Q,Q ′ (f · f ′ ) (λ P · 1 P ′)(247)= (f • B f ′ ) p P,P ′ (λ P · 1 P ′)(236)= (f • B f ′ ) λ P •B P ′ (1 A · p P,P ′)229and since 1 A · p P,P ′is an epimorphism, we get thatλ Q•B Q ′ (1 A · f • B f ′ ) = (f • B f ′ ) λ P •B P ′i.e. f • B f ′ is a morphism of left A-modules. Similarly, we also haveρ Q•B Q ′ (f • B f ′ · 1 C ) (p P,P ′ · 1 C ) (247)= ρ Q•B Q ′ (p Q,Q ′ · 1 C) (f · f ′ · 1 C )(237)= p Q,Q ′ (1 Q · ρ Q ′) (f · f ′ · 1 C )= p Q,Q ′ (1 Q · ρ Q ′) (1 Q · f ′ · 1 C ) (f · 1 P ′ · 1 C )(246)= p Q,Q ′ (1 Q · f ′ ) (1 Q · ρ P ′) (f · 1 P ′ · 1 C )f= p Q,Q ′ (1 Q · f ′ ) (f · 1 P ′) (1 P · ρ P ′) = p Q,Q ′ (f · f ′ ) (1 P · ρ P ′)and since p P,P ′ · 1 C is epi, we get that(247)= (f • B f ′ ) p P,P ′ (1 P · ρ P ′)(237)= (f • B f ′ ) ρ P •B P ′ (p P,P ′ · 1 C)ρ Q•B Q ′ (f • B f ′ · 1 C ) = (f • B f ′ ) ρ P •B P ′i.e. f • B f ′ is also a morphism of right C-modules.□Propositi<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>1<str<strong>on</strong>g>2.</str<strong>on</strong>g> For any m<strong>on</strong>ad (Y, B) in C, the compositi<strong>on</strong> denoted by • B iscompatible with the vertical can<strong>on</strong>ical compositi<strong>on</strong>.Proof. Let (X, A) , (Y, B) , (Z, C) be m<strong>on</strong>ads in C, let (P, λ P , ρ P ) , (Q, λ Q , ρ Q ) ,(W, λ W , ρ W ) be A-B-bimodules in C, let (P ′ , λ P ′, ρ P ′) , (Q ′ , λ Q ′, ρ Q ′) , (W ′ , λ W ′, ρ W ′)be B-C-bimodules in C and let f : P → Q, g : Q → W be A-B-bimodule morphisms,f ′ : P ′ → Q ′ , g ′ : Q ′ → W ′ be B-C-bimodule morphisms in C. By Propositi<strong>on</strong> 1<str<strong>on</strong>g>1.</str<strong>on</strong>g>11we can c<strong>on</strong>sider the A-C-bimodule morphisms f • B f ′ : P • B P ′ → Q • B Q ′ andg • B g ′ : Q • B Q ′ → W • B W ′ and we can compose them in order to get(g • B g ′ ) (f • B f ′ ) : P • B P ′ → W • B W ′ .