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 ...
120We want to prove that Γ is an opposite mixed distributive law. We compute(m B D) ◦ (BΓ) ◦ (ΓB) ◦ (Dyy) y = (m B D) ◦ (BΓ) ◦ (ΓB) ◦ (DyB) ◦ (DP Qy)defγ= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P χB) ◦ (δ D P QB) ◦ (DP Qy)δ D= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P χB) ◦ (P QP Qy) ◦ (δ D P QP Q)χ= (m B D) ◦ (BΓ) ◦ (yDB) ◦ ( P ρ D QB ) ◦ (P Qy) ◦ (P χP Q) ◦ (δ D P QP Q)ρ D Q= (m B D) ◦ (BΓ) ◦ (yDB) ◦ (P QDy) ◦ ( P ρ D QP Q ) ◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (BΓ) ◦ (BDy) ◦ (yDP Q) ◦ ( P ρ D QP Q ) ◦ (P χP Q) ◦ (δ D P QP Q)defγ= (m B D) ◦ (ByD) ◦ ( )BP ρ D Q ◦ (BP χ) ◦ (BδD P Q) ◦ (yDP Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (yDP Q) ◦ (P QyD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)y= (m B D) ◦ (yyD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)(109)= (yD) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q) ◦ ( P ρ D QP Q )◦ (P χP Q) ◦ (δ D P QP Q)(127)= (yD) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P δC QP Q) ◦ ( P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)δ=C(yD) ◦ (P χD) ◦ (P δ C QD) ◦ ( (P CρQ) D ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)(112)= (yD) ◦ ( P ε C QD ) ◦ ( ) (P Cρ D Q ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (P χP Q) ◦ (δ D P QP Q)ε= C (yD) ◦ ( ) (P ρ D Q ◦ (P χ) ◦ P ε C QP Q ) ◦ ( P C ρ Q P Q ) ◦ (P χP Q) ◦ (δ D P QP Q)Qcomfun= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (P χP Q) ◦ (δD P QP Q)(111)= (yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (P QP χ) ◦ (δD P QP Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) ◦ (DP χ)defγ= Γ ◦ (Dy) ◦ (DP χ) (109)= Γ ◦ (Dm B ) ◦ (Dyy)and since Dyy is an epimorphism we deduce thatLet us compute(m B D) ◦ (BΓ) ◦ (ΓB) = Γ ◦ (Dm B ) .(ΓD) ◦ (DΓ) ◦ ( ∆ D B ) ◦ (Dy) ∆D= (ΓD) ◦ (DΓ) ◦ (DDy) ◦ ( ∆ D P Q )
defγ= (ΓD) ◦ (DyD) ◦ ( )DP ρ D Q ◦ (DP χ) ◦ (DδD P Q) ◦ ( ∆ D P Q )defγ= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ (δ D P QD) ◦ ( DP ρ D Q)◦ (DP χ)◦ (Dδ D P Q) ◦ ( ∆ D P Q )121δ=D(yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q)◦ (δ D DP Q) ◦ ( ∆ D P Q )(126)= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( P QP ρ D Q)◦ (P QP χ) ◦ (P QδD P Q)◦ ( P ρ D QP Q ) ◦ (δ D P Q)(127)= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( P QP ρ D Q)◦ (P QP χ) ◦ (P δC QP Q)◦ ( P C ρ Q P Q ) ◦ (δ D P Q)δ C= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ (P δ C QD) ◦ ( P Cρ D Q)◦ (P Cχ) ◦(P C ρ Q P Q )◦ (δ D P Q)(112)= (yDD) ◦ ( P ρ D QD ) ◦ ( P ε C QD ) ◦ ( ) (P Cρ D Q ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (δ D P Q)ε= C (yDD) ◦ ( P ρ D QD ) ◦ ( ) (P ρ D Q ◦ (P χ) ◦ P ε C QP Q ) ◦ ( P C ρ Q P Q ) ◦ (δ D P Q)Qcomfun= (yDD) ◦ ( P ρ D QD ) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q)Qcomfun= (yDD) ◦ ( P Q∆ D) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q)y= ( B∆ D) ◦ (yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) defγ= ( B∆ D) ◦ Γ ◦ (Dy)and since Dy is an epimorphism we get thatNow we compute(ΓD) ◦ (DΓ) ◦ ( ∆ D B ) = ( B∆ D) ◦ Γ.Γ ◦ (Du B ) ◦ ( Dε D) (110)= Γ ◦ (Dy) ◦ (Dδ D )defγ= (yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q) ◦ (Dδ D )δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (P QδD ) ◦ (δ D D)(113)= (yD) ◦ ( P ρ D Q)◦(P QεD ) ◦ (δ D D)ρ D Q= (yD) ◦ ( P QDε D) ◦ ( P ρ D QD ) ◦ (δ D D)(126)= (yD) ◦ ( P QDε D) ◦ (δ D DD) ◦ ( ∆ D D )δ D= (yD) ◦ (δ D D) ◦ ( DDε D) ◦ ( ∆ D D ) ∆ D= (yD) ◦ (δ D D) ◦ ∆ D ◦ ( Dε D)(110)= (u B D) ◦ ( ε D D ) ◦ ∆ D ◦ ( Dε D) Dcomonad= (u B D) ◦ ( Dε D)
- 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 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 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
- Page 156 and 157: 156) ( )l=(pb Q AQ B ◦ ̂QA µ QB
- Page 158 and 159: 158In fact we haveTherefore we dedu
- Page 160 and 161: 160χ= h 1 ◦ (P xQ B ) ◦ (P QP
- Page 162 and 163: 162so that we obtain:(190)We comput
- Page 164 and 165: 164(194)=) )(p QB ̂QA ◦(Qpb Q◦
- Page 166 and 167: 166= Ξ ◦ (A A U A λ) ◦ (xx A
- Page 168 and 169: 168)(155)= k 2 ◦(Qpb Q◦ (Ql A U
defγ= (ΓD) ◦ (DyD) ◦ ( )DP ρ D Q ◦ (DP χ) ◦ (DδD P Q) ◦ ( ∆ D P Q )defγ= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ (δ D P QD) ◦ ( DP ρ D Q)◦ (DP χ)◦ (Dδ D P Q) ◦ ( ∆ D P Q )121δ=D(yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( )P QP ρ D Q ◦ (P QP χ) ◦ (P QδD P Q)◦ (δ D DP Q) ◦ ( ∆ D P Q )(126)= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( P QP ρ D Q)◦ (P QP χ) ◦ (P QδD P Q)◦ ( P ρ D QP Q ) ◦ (δ D P Q)(127)= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ ( P QP ρ D Q)◦ (P QP χ) ◦ (P δC QP Q)◦ ( P C ρ Q P Q ) ◦ (δ D P Q)δ C= (yDD) ◦ ( P ρ D QD ) ◦ (P χD) ◦ (P δ C QD) ◦ ( P Cρ D Q)◦ (P Cχ) ◦(P C ρ Q P Q )◦ (δ D P Q)(112)= (yDD) ◦ ( P ρ D QD ) ◦ ( P ε C QD ) ◦ ( ) (P Cρ D Q ◦ (P Cχ) ◦ P C ρ Q P Q )◦ (δ D P Q)ε= C (yDD) ◦ ( P ρ D QD ) ◦ ( ) (P ρ D Q ◦ (P χ) ◦ P ε C QP Q ) ◦ ( P C ρ Q P Q ) ◦ (δ D P Q)Qcomfun= (yDD) ◦ ( P ρ D QD ) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q)Qcomfun= (yDD) ◦ ( P Q∆ D) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q)y= ( B∆ D) ◦ (yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) defγ= ( B∆ D) ◦ Γ ◦ (Dy)and since Dy is an epimorphism we get thatNow we compute(ΓD) ◦ (DΓ) ◦ ( ∆ D B ) = ( B∆ D) ◦ Γ.Γ ◦ (Du B ) ◦ ( Dε D) (110)= Γ ◦ (Dy) ◦ (Dδ D )defγ= (yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q) ◦ (Dδ D )δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (P QδD ) ◦ (δ D D)(113)= (yD) ◦ ( P ρ D Q)◦(P QεD ) ◦ (δ D D)ρ D Q= (yD) ◦ ( P QDε D) ◦ ( P ρ D QD ) ◦ (δ D D)(126)= (yD) ◦ ( P QDε D) ◦ (δ D DD) ◦ ( ∆ D D )δ D= (yD) ◦ (δ D D) ◦ ( DDε D) ◦ ( ∆ D D ) ∆ D= (yD) ◦ (δ D D) ◦ ∆ D ◦ ( Dε D)(110)= (u B D) ◦ ( ε D D ) ◦ ∆ D ◦ ( Dε D) Dcom<strong>on</strong>ad= (u B D) ◦ ( Dε D)
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