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 ...
74we have that Ld ϕ K ϕ Y is mono and hence we obtainso that L̂η is a functorial isomorphism.(L̂ηY ) ◦ (ɛLY ) ◦ (Ld ϕ K ϕ Y ) = LD ϕ K ϕ YDefinition 4.56. Let ( L, C ρ L)be a left comodule functor for a comonad C =(C, ∆ C , ε C) such that L has a right adjoint R. Then we can consider a canonicalcomonad morphismcan := (Cɛ) ◦ (C ρ L R ) : LR → Cwhere ɛ denotes the counit of the adjunction (L, R) . A left C-Galois functor is aleft C-comodule functor ( L, C ρ L)with a right adjoint R such that can is a comonadisomorphism.Corollary 4.57. Let ( L, C ρ L)be a left C-Galois comodule functor such that Lpreserves equalizers, L reflects isomorphisms and let C = ( C, ∆ C , ε C) be a comonadon A. Assume that, for every ( X, C ρ X)∈ C A, there exists Equ B(αX, R C ρ X)whereα = (Rcan) ◦ (ηR) where R is the right adjoint of L and η is the unit of theadjunction. Then we can consider the functor K can : B → C A. Then the functorK can is an equivalence of categories.Proof. We can apply Theorem 4.55 to the case ϕ = can.Theorem 4.58 (Beck’s Theorem). Let (L, R) be an adjunction where L : B → A andR : A → B. Let α = Θ (Id LR ) = ηR and assume that, for every ( X, LR ρ X)∈ LR A,there exists Equ B(ηRX, R LR ρ X). Then we can consider the functor K = Υ (IdLR ) :B → LR A and its right adjoint D : LR A → B. The functor K is an equivalence ofcategories if and only if1) L preserves the equalizer2) L reflects isomorphisms.(D, d) = Equ Fun(ηR LR U, R LR Uγ LR) .Proof. Apply Theorem 4.55 taking ϕ = Id LR and thus α = Θ (Id LR ) = ηR.Definition 4.59. Let C = ( C, ∆ C , ε C) be a comonad on the category A and letL : B → A. The functor L is called ϕ-comonadic if it has a right adjoint R : A → Bfor which there exists ϕ : LR → C a comonad morphism such that the functorK ϕ = Υ (ϕ) : B → C A is an equivalence of categories with D ϕ : C A → B which isright adjoint.Definition 4.60. Let L : B → A be a functor. The functor L is called comonadicif it has a right adjoint R : A → B such that the functor K = Υ (Id LR ) : B → LR Ais an equivalence of categories with right adjoint D : LR A → B.Lemma 4.6
e a L-contractible equalizer pair in B. Then (53) has an equalizer d : X ′′ → X ′ inB andLX ′′ Ld LX ′ Ld 0 LXLd 1is an equalizer in A.Proof. Since L is a ϕ-comonadic functor we know that K ϕ = Υ (ϕ) : B → C A is anequivalence of categories. Then, instead of consideringin the category B, we can considerX ′ d 0 Xd 1K ϕ X ′ K ϕd 0 K ϕ XK ϕd 1in C A which is a C U-contractible equalizer pair. Let us denote by ( Z ′ , C ρ Z ′):= Kϕ X ′and ( )Z, C ρ Z := Kϕ X so that we can rewrite the C U-contractible equalizer pair asfollows(Z ′ , C K ϕd 0ρ Z ′) ( )Z, C ρ ZK ϕd 1We want to prove that this pair has an equalizer in C A. Since the pair (K ϕ d 0 , K ϕ d 1 )is a C U-contractible equalizer in C A, we have thatZ ′′ d Z ′sis a contractible equalizer and thus, by Proposition
- 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 and 34: Theorem 3.37. Let B = (B, m B , u B
- Page 35 and 36: where A UG B F : B → A is such th
- 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 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
- Page 84 and 85: 84i.e. Aα is a functorial morphism
- Page 86 and 87: 86Then we haveA µ CCX ◦ ( A∆ C
- Page 88 and 89: 884.23) is a functor à : C A → C
- Page 90 and 91: 90Let θ l = ( σ B P Q ) ◦ (P τ
- Page 92 and 93: 925)σ A = ( ε C A ) ◦ ( Cσ A)
- Page 94 and 95: 94(ii) the functorial morphism can
- Page 96 and 97: 96defΦ= ( QP A µ Q)◦(QP σ A Q
- Page 98 and 99: 98AU A can AA F = can AA F = ( CσA
- Page 100 and 101: 100Similarly, one can prove the sta
- Page 102 and 103: 102(b) A comonad C = ( C, ∆ C ,
- Page 104 and 105: 104We calculateso that we getx ◦
- Page 106 and 107: 106There exist functorial morphisms
- Page 108 and 109: 108andsatisfying(B, y) = Coequ Fun(
- Page 110 and 111: 1104) With notations of Theorem 6.2
- Page 112 and 113: 112Then ν : Y → D is the unique
- Page 114 and 115: 114= A µ Q ◦ ( Aε C Q ) ◦ (AC
- Page 116 and 117: 116= ( Aε C Q ) ◦ ( cocan1 −1
- Page 118 and 119: 118so that we getχ= (Cx) ◦ (C ρ
- Page 120 and 121: 120We want to prove that Γ is an o
- Page 122 and 123: 122and since Dε D is an epimorphis
e a L-c<strong>on</strong>tractible equalizer pair in B. Then (53) has an equalizer d : X ′′ → X ′ inB andLX ′′ Ld LX ′ Ld 0 LXLd 1is an equalizer in A.Proof. Since L is a ϕ-com<strong>on</strong>adic functor we know that K ϕ = Υ (ϕ) : B → C A is anequivalence of categories. Then, instead of c<strong>on</strong>sideringin the category B, we can c<strong>on</strong>siderX ′ d 0 Xd 1K ϕ X ′ K ϕd 0 K ϕ XK ϕd 1in C A which is a C U-c<strong>on</strong>tractible equalizer pair. Let us denote by ( Z ′ , C ρ Z ′):= Kϕ X ′and ( )Z, C ρ Z := Kϕ X so that we can rewrite the C U-c<strong>on</strong>tractible equalizer pair asfollows(Z ′ , C K ϕd 0ρ Z ′) ( )Z, C ρ ZK ϕd 1We want to prove that this pair has an equalizer in C A. Since the pair (K ϕ d 0 , K ϕ d 1 )is a C U-c<strong>on</strong>tractible equalizer in C A, we have thatZ ′′ d Z ′sis a c<strong>on</strong>tractible equalizer and thus, by Propositi<strong>on</strong> <str<strong>on</strong>g>2.</str<strong>on</strong>g>19, an equalizer in A. Let usc<strong>on</strong>sider the following diagramC ρ Z ′′C C ρ Z ′′Ld 0tLd 1Z ′′ d Z ′C ρ Z ′CZ ′′ Cd CZ ′ ∆ C Z ′∆ C Z ′′C C ρ Z ′ ZLd 0Ld 1CLd 0CLd 1CCZ ′′ CCd CCZ ′ CCLd 0 CCLd 1 Z CZC C ρ ZC ρ Z CCZBy Propositi<strong>on</strong> <str<strong>on</strong>g>2.</str<strong>on</strong>g>20, all the rows are c<strong>on</strong>tractible equalizers. Since Ld 0 = C UK ϕ d 0and Ld 1 = C UK ϕ d 1 where K ϕ d 0 and K ϕ d 1 are morphisms in C A, we have that theupper right square serially commutes. Moreover, since we also have that ∆ C is afunctorial morphism, the lower right square serially commutes. We also have thatC ρ Z ′◦d is a fork for (CLd 0 , CLd 1 ) and, since (CZ ′′ , CZ ′ , CZ, Cd, CLd 0 , CLd 1 , Cs, Ct)is a c<strong>on</strong>tractible equalizer, in particular (CZ ′′ , Cd) = Equ A (CLd 0 , CLd 1 ); by theuniversal property of the equalizer, there exists a unique morphism C ρ Z ′′ : Z ′′ →CZ ′′ such that(54)C ρ Z ′ ◦ d = (Cd) ◦ C ρ Z ′′.∆ C Z75