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 ...
106There exist functorial morphisms m B : BB → B and u B : B → B such thatB = (B, m B , u B ) is a monad over B that preserves coequalizers. Moreover m B andu B are uniquely determined by(108) m B ◦ (yB) = y ◦ ( )P µ B Qor equivalently(109) m B ◦ (yy) = y ◦ (P χ)and(110) y ◦ δ D = u B ◦ ε D .Moreover ( Q, µ B Q)is a right B-module functor.Proof. By left-right symmetric argument of those used in proof of Proposition 6.25,one can prove this Proposition.□Definition 6.27. Let A and B be categories. A preformal codual structure is aeightuple Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) where C : A → A, D : B → B, P : A → Band Q : B → A are functors, δ C : C → QP, δ D : D → P Q, ε C : C → A, ε D : D → Bare functorial morphisms. A copretorsor χ for Θ is a functorial morphism χ :QP Q → Q satisfying the following conditions:1) Coassociativity, in the sense that(111) χ ◦ (χP Q) = χ ◦ (QP χ)2) Counitality, in the sense that(112) χ ◦ (δ C Q) = ε C Qand(113) χ ◦ (Qδ D ) = Qε D .Definition 6.28. A preformal codual structure Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D)will be called regular whenever ( A, ε C) (= Coequ Fun Cε C , ε C C ) and ( (B, ε D) =Coequ Fun Dε D , ε D D ) . In this case a copretorsor for Θ will be called a regularcopretorsor.Theorem 6.29. Let A and B be categories with coequalizers and let χ : QP Q →Q be a regular copretorsor for Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) . Assume that theunderlying functors P, Q, C and D preserve coequalizers. Let w l = (χP ) ◦ (QP δ C )and w r = QP ε C : QP C → QP . Set(114) (A, x) = Coequ Fun(w l , w r) .There exists a functorial morphism A µ Q : AQ → Q such that(115)A µ Q ◦ (xQ) = χ.There exist functorial morphisms m A : AA → A and u A : A → A such that A =(A, m A , u A ) is a monad over A that preserves coequalizers. Moreover m A and u Aare uniquely determined byx ◦ (χP ) = m A ◦ (xx) and x ◦ δ C = u A ◦ ε C .
Moreover ( Q, A µ Q)is a left A-module functor.Let z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q. Set(116) (B, y) = Coequ Fun(z l , z r) .There exists a functorial morphism µ B Q : QB → Q such that(117) µ B Q ◦ (Qy) = χ.There exist functorial morphisms m B : BB → B and u B : B → B such thatB = (B, m B , u B ) is a monad over B that preserves coequalizers. Moreover m B andu B are uniquely determined bym B ◦ (yy) = y ◦ (P χ) and y ◦ δ D = u B ◦ ε D .Moreover ( Q, µ B Q)is a right B-module functor.Finally ( Q, A µ Q , µ B Q)is an A-B-module functor.Proof. Within these assumption, we can apply Proposition 6.25 to get the monad Aand the functorial morphism A µ Q : AQ → Q satisfying 115 such that ( )Q, A µ Q is aleft A-module functor and Proposition 6.26 to get the monad B and the functorialmorphism µ B Q : QB → Q satisfying 117 such that ( Q, µ Q) B is a right B-modulefunctor. Let us check the compatibility condition. We calculateA µ Q ◦ ( )Aµ B Q ◦ (AQy) ◦ (xQP Q) = x A µ Q ◦ (xQ) ◦ ( )QP µ B Q ◦ (QP Qy)(115),(117)= χ ◦ (QP χ) (98)= χ ◦ (χP Q)(117),(115)= µ B Q ◦ (Qy) ◦ (A µ Q P Q ) ◦ (xQP Q)A µ Q= µBQ ◦ (A µ Q B ) ◦ (AQy) ◦ (xQP Q) .Since (AQy) ◦ (xQP Q) is an epimorphism we get thatA µ Q ◦ ( )Aµ B Q = µBQ ◦ (A µ Q B ) .Therefore (Q, A µ Q , µ B Q ) is an A-B-bimodule functor.□Theorem 6.30. Let χ : QP Q → Q be a regular copretorsor for a preformal codualstructure Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) on categories A and B such that the underlyingfunctors P, Q, C and D preserve coequalizers. Assume that C and D arecomonads, ( P, D ρ P)is a left D-comodule functor and(P, ρCP)is a right C-comodulefunctor. Moreover assume that the functorial morphism δ C is right C-colinear, thatis ( Qρ C P)◦ δC = (δ C C) ◦ ∆ C and the functorial morphism δ D is left D-colinear thatis (D ρ P Q ) ◦ δ D = (Dδ D ) ◦ ∆ D and that they are compatible in the sense that(118) (δ D P ) ◦ D ρ P = (P δ C ) ◦ ρ C P .Then there exists a monad A = (A, m A , u A ) on the category A together with afunctorial morphism A µ Q : AQ → Q such that ( Q, A µ Q)is a left A-module functorand a monad B = (B, m B , u B ) together with a functorial morphism µ B Q : QB → Qsuch that ( Q, µ B Q)is a right B-module functor. The underlying functors are definedas follows(A, x) = Coequ Fun((χP ) ◦ (QP δC ) , QP ε C) .107
- 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
- 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 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
- Page 124 and 125: 124χ= (Cχ) ◦ (C ρ Q P Q ) ◦
- Page 126 and 127: 126Now, since cocan 1 : AC → QP i
- Page 128 and 129: 1287. Herds and Coherds7.1.
- Page 130 and 131: 130◦ ( σ A QQQ ) ◦ (A µ Q P Q
- Page 132 and 133: 132= µ B Q ◦ (A µ Q B ) ◦ ( A
- Page 134 and 135: 134Assume now that there is another
- Page 136 and 137: 136and hence we get(160) x ◦ (χP
- Page 138 and 139: 138Proposition 7.7. In the setting
- Page 140 and 141: 140We calculateA µ Q ◦ ( σ A Q
- Page 142 and 143: 142x=and=δ C=(l= QlQ ̂QQ)◦ (QP
- Page 144 and 145: 144(◦ ρ D ̂QQ)Q◦ (QlQ) ◦ (Q
- Page 146 and 147: 146given byWe computeσ B = m B ◦
- Page 148 and 149: 148andy= ′m B ◦ (ν B B) ◦ (y
- Page 150 and 151: 150Now we compute(hQ) ◦ ( Qχ )
- Page 152 and 153: 152Thus we obtainσ B ◦ ( ) (P µ
- Page 154 and 155: 154Thus hQ is an isomorphism with i
Moreover ( Q, A µ Q)is a left A-module functor.Let z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q. Set(116) (B, y) = Coequ Fun(z l , z r) .There exists a functorial morphism µ B Q : QB → Q such that(117) µ B Q ◦ (Qy) = χ.There exist functorial morphisms m B : BB → B and u B : B → B such thatB = (B, m B , u B ) is a m<strong>on</strong>ad over B that preserves coequalizers. Moreover m B andu B are uniquely determined bym B ◦ (yy) = y ◦ (P χ) and y ◦ δ D = u B ◦ ε D .Moreover ( Q, µ B Q)is a right B-module functor.Finally ( Q, A µ Q , µ B Q)is an A-B-module functor.Proof. Within these assumpti<strong>on</strong>, we can apply Propositi<strong>on</strong> 6.25 to get the m<strong>on</strong>ad Aand the functorial morphism A µ Q : AQ → Q satisfying 115 such that ( )Q, A µ Q is aleft A-module functor and Propositi<strong>on</strong> 6.26 to get the m<strong>on</strong>ad B and the functorialmorphism µ B Q : QB → Q satisfying 117 such that ( Q, µ Q) B is a right B-modulefunctor. Let us check the compatibility c<strong>on</strong>diti<strong>on</strong>. We calculateA µ Q ◦ ( )Aµ B Q ◦ (AQy) ◦ (xQP Q) = x A µ Q ◦ (xQ) ◦ ( )QP µ B Q ◦ (QP Qy)(115),(117)= χ ◦ (QP χ) (98)= χ ◦ (χP Q)(117),(115)= µ B Q ◦ (Qy) ◦ (A µ Q P Q ) ◦ (xQP Q)A µ Q= µBQ ◦ (A µ Q B ) ◦ (AQy) ◦ (xQP Q) .Since (AQy) ◦ (xQP Q) is an epimorphism we get thatA µ Q ◦ ( )Aµ B Q = µBQ ◦ (A µ Q B ) .Therefore (Q, A µ Q , µ B Q ) is an A-B-bimodule functor.□Theorem 6.30. Let χ : QP Q → Q be a regular copretorsor for a preformal codualstructure Θ = ( C, D, P, Q, δ C , δ D , ε C , ε D) <strong>on</strong> categories A and B such that the underlyingfunctors P, Q, C and D preserve coequalizers. Assume that C and D arecom<strong>on</strong>ads, ( P, D ρ P)is a left D-comodule functor and(P, ρCP)is a right C-comodulefunctor. Moreover assume that the functorial morphism δ C is right C-colinear, thatis ( Qρ C P)◦ δC = (δ C C) ◦ ∆ C and the functorial morphism δ D is left D-colinear thatis (D ρ P Q ) ◦ δ D = (Dδ D ) ◦ ∆ D and that they are compatible in the sense that(118) (δ D P ) ◦ D ρ P = (P δ C ) ◦ ρ C P .Then there exists a m<strong>on</strong>ad A = (A, m A , u A ) <strong>on</strong> the category A together with afunctorial morphism A µ Q : AQ → Q such that ( Q, A µ Q)is a left A-module functorand a m<strong>on</strong>ad B = (B, m B , u B ) together with a functorial morphism µ B Q : QB → Qsuch that ( Q, µ B Q)is a right B-module functor. The underlying functors are definedas follows(A, x) = Coequ Fun((χP ) ◦ (QP δC ) , QP ε C) .107
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