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 ...
34where A Q : B → A A is the functor defined in Lemma 3.29.Proof. Let ( Q : B → A, A µ Q)be a left A-module functor. Then, by Lemma 3.29,there exists a unique functor A Q : B → A A such thatAU ◦ A Q = Q and A Uλ AA Q = A µ Q .Note that, since AQ preserves coequalizers, by Lemma 3.19, Q = A U ◦ A Q preservescoequalizers. Then, by Lemma 3.20, also A Q preserves coequalizers. Conversely, ifH : B → A A is a functor preserving coequalizers, we get that A U ◦ H : B → A.Moreover, by Lemma 3.21, A U preserves coequalizers and thus also A U ◦H preservescoequalizers. Now, let us prove that A ν and A κ determine a bijective correspondencebetween A F and ( A A ← B). We compute( A κ ◦ A ν) (( Q, A µ Q))= A κ ( A Q) = ( A U A Q, A Uλ AA Q) = ( Q, A µ Q).On the other hand we have( A ν ◦ A κ) (H) = A ν (( A U ◦ H, A Uλ A H)) = A ( A U ◦ H) (17)= H.Theorem 3.39. Let A = (A, m A , u A ) be a monad on a category A with coequalizerssuch that A preserves coequalizers. Let B = (B, m B , u B ) be a monad on a categoryB with coequalizers such that B preserves coequalizers. Then there exists a bijectivecorrespondence between the following collections of data:AF B A-B-bimodule functors Q : B → A such that AQ and QB preserve coequalizers( A A ← B B) functors G : B B → A A preserving coequalizersgiven byAν B : AF B → ( A A ← B B) where A ν B((Q, A µ Q , µ B Q))= A Q BAκ B : ( A A ← B B) → A F B where A κ B (G) = ( A U ◦ G ◦ B F , A Uλ A G B F , A UGλ BB F ) .Proof. Let us consider an A-B-bimodule functor ( Q : B → A, A µ Q , µ Q) B such that AQand QB preserve coequalizers. In particular, ( Q, µ Q) B is a right B-module functor,so that we (( can apply )) the map ν B : F B → (A ← B B) of Theorem 3.37 and we get afunctor ν B Q, µBQ = QB : B B → A which preserves coequalizers. By Proposition3.30, ( )Q B , A µ QB is a left A-module functor so that we can also apply the mapAν : A F → ( A A ← B) of Theorem 3.38 where the category B is B B. The map A νis defined by A ν (( ))Q B , A µ QB = A (Q B ) = A Q B : B B → A A and A Q B preservescoequalizers. Conversely, let us consider a functor G : B B → A A which preservescoequalizers. By Theorem 3.38, we get a left A-module functor given byAκ (G) = ( A U ◦ G, A Uλ A G)where A U ◦ G : B B → A and A A UG preserves coequalizers. By Lemma 3.19, alsoAU ◦ G : B B → A preserves coequalizers. Thus, we can apply Theorem 3.37 and weget a right B-module functorκ B ( A UG) = ( A UG B F, A UGλ BB F )□
where A UG B F : B → A is such that A UG B F B preserves coequalizers. Clearly, sinceAUG preserves coequalizers, B F is a left adjoint and A preserves coequalizers byassumption, we deduce that also A A UG B F preserves coequalizers. Now, we wantto prove that A ν B : A F B → ( A A ← B B) and A κ B : ( A A ← B B) → A F B determine abijection. We haveand( A κ B ◦ A ν B ) (( Q, A µ Q , µ B Q))= A κ B ( A Q B )= ( A U ◦ A Q B ◦ B F , A Uλ AA Q BB F , A U A Q B λ BB F )= (Q, A Uλ AA Q, Q B λ BB F ) = ( Q, A µ Q , µ B Q B B F)=(Q, A µ Q , µ B Q( A ν B ◦ A κ B ) (G) = A ν B (( A U ◦ G ◦ B F , A Uλ A G B F , A UGλ BB F ))= A ( A U ◦ G ◦ B F ) B = A (( A U ◦ G ◦ B F ) B )(16)= A ( A U ◦ G) (17)= G.Proposition 3.40. Let A = (A, m A , u A ) be a monad over a category A with coequalizersand assume that A preserves coequalizers. Let Q : A → A be an A-bimodulefunctor. Then there exists a unique lifted functor A Q A : A A → A A such thatAU A Q AA F = Q.Proof. By Proposition 3.31 there exists a unique functor A Q A : A A → A A such thatAU A Q A = Q A . Now, by Proposition 3.34 we also get that Q AA F = Q so that weobtainAU A Q AA F = Q.□Proposition 3.4
- Page 1 and 2: Contents1.
- Page 3 and 4: linearity and compatibility conditi
- Page 5 and 6: 5and since g ◦ f is an epimorphis
- Page 7 and 8: Proof. Clearly (qP )◦(αP ) = (qP
- Page 9 and 10: Lemma 2.13 ([BM, L
- Page 11 and 12: i.e. Hom B (Y, iX) equalizes Hom B
- Page 13 and 14: 13such thatd 0 ◦ v = Id Yd 1 ◦
- Page 15 and 16: f ↦→ Rfis bijective for every X
- Page 17 and 18: Remark 3.10. Let A = (A, m A , u A
- Page 19 and 20: 19and fromµ A P ◦ ( µ A P A ) =
- Page 21 and 22: and thusk ◦ (u A QZ) ◦ (Qz) = h
- Page 23 and 24: and since A preserves equalizers, A
- Page 25 and 26: Conversely, let Φ be a functorial
- Page 27 and 28: Proof. Apply Proposition 3.24 to th
- Page 29 and 30: Since Q is a left A-module functor,
- Page 31 and 32: (Q BB F, p QB F ) = Coequ Fun(µBQ
- Page 33: Theorem 3.37. Let B = (B, m B , u B
- Page 37 and 38: Proposition 3.44. Let A = (A, m A ,
- Page 39 and 40: Note that, since f and g are A-bili
- Page 41 and 42: Proposition 3.54. Let (L, R) be an
- Page 43 and 44: Corollary 3.58. Let (L, R) be an ad
- Page 45 and 46: Definition 4.2. A
- Page 47 and 48: Proposition 4.13. Let C = ( C, ∆
- Page 49 and 50: Then we have(P Cx) ◦ ( ρ C P X )
- Page 51 and 52: and since C preserves coequalizers,
- Page 53 and 54: Proof. Apply Corollary 4.24 to the
- Page 55 and 56: Let( (CQ ) ()D, ι Q) C = Equ Fun
- Page 58 and 59: 58F D right D-comodule functors Q :
- Page 60 and 61: 60prove that C ν D : C F D → (C
- Page 62 and 63: 624.2. The compari
- Page 64 and 65: 64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY,
- Page 66 and 67: 66for every ( X, C ρ X)∈ C A, th
- Page 68 and 69: 68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂η
- Page 70 and 71: 70In particular(49) d ϕ(CX, ∆ C
- Page 72 and 73: 72We have to prove that (LD ϕ , Ld
- Page 74 and 75: 74we have that Ld ϕ K ϕ Y is mono
- Page 76 and 77: and since d is mono we get that(ε
- Page 78 and 79: 78Corollary 4.63 (Beck’s Precise
- Page 80 and 81: 80We compute(LRɛLY ′ ) ◦ ( LR
- Page 82 and 83: 82Proof. First of all we prove that
where A UG B F : B → A is such that A UG B F B preserves coequalizers. Clearly, sinceAUG preserves coequalizers, B F is a left adjoint and A preserves coequalizers byassumpti<strong>on</strong>, we deduce that also A A UG B F preserves coequalizers. Now, we wantto prove that A ν B : A F B → ( A A ← B B) and A κ B : ( A A ← B B) → A F B determine abijecti<strong>on</strong>. We haveand( A κ B ◦ A ν B ) (( Q, A µ Q , µ B Q))= A κ B ( A Q B )= ( A U ◦ A Q B ◦ B F , A Uλ AA Q BB F , A U A Q B λ BB F )= (Q, A Uλ AA Q, Q B λ BB F ) = ( Q, A µ Q , µ B Q B B F)=(Q, A µ Q , µ B Q( A ν B ◦ A κ B ) (G) = A ν B (( A U ◦ G ◦ B F , A Uλ A G B F , A UGλ BB F ))= A ( A U ◦ G ◦ B F ) B = A (( A U ◦ G ◦ B F ) B )(16)= A ( A U ◦ G) (17)= G.Propositi<strong>on</strong> 3.40. Let A = (A, m A , u A ) be a m<strong>on</strong>ad over a category A with coequalizersand assume that A preserves coequalizers. Let Q : A → A be an A-bimodulefunctor. Then there exists a unique lifted functor A Q A : A A → A A such thatAU A Q AA F = Q.Proof. By Propositi<strong>on</strong> 3.31 there exists a unique functor A Q A : A A → A A such thatAU A Q A = Q A . Now, by Propositi<strong>on</strong> 3.34 we also get that Q AA F = Q so that weobtainAU A Q AA F = Q.□Propositi<strong>on</strong> 3.4<str<strong>on</strong>g>1.</str<strong>on</strong>g> Let A = (A, m A , u A ) be a m<strong>on</strong>ad over a category A with coequalizersand assume that A preserves coequalizers. Let B = (B, m B , u B ) be a m<strong>on</strong>adover a category B with coequalizers and let Q : B → A be an A-B-bimodule functor.Then there exists a unique lifted functor A Q B : B B → A A such thatAU A Q BB F = Q.Proof. By Propositi<strong>on</strong> 3.31 there exists a unique functor A Q B : B B → A A such thatAU A Q B = Q B . Now, by Propositi<strong>on</strong> 3.34 we also get that Q BB F = Q so that weobtainAU A Q BB F = Q.□Propositi<strong>on</strong> 3.4<str<strong>on</strong>g>2.</str<strong>on</strong>g> Let A = (A, m A , u A ) be a m<strong>on</strong>ad over a category A with coequalizersand assume that A preserves coequalizers. Let B = (B, m B , u B ) be am<strong>on</strong>ad over a category B with coequalizers and let P, Q : B → A be A-B-bimodulefunctors. Let f : P → Q be a functorial morphism of left A-module functors andof right B-module functors. Then there exists a unique functorial morphism of leftA-module functorsf B : P B → Q B)35□
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