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 ...
118so that we getχ= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (χP C)defλ= Λ ◦ (xC) ◦ (χP C) (102)= Λ ◦ (m A C) ◦ (xxC)(Cm A ) ◦ (ΛA) ◦ (AΛ) ◦ (xxC) = Λ ◦ (m A C) ◦ (xxC)and since xxC is an epimorphism we obtainLet now compute(Cm A ) ◦ (ΛA) ◦ (AΛ) = Λ ◦ (m A C) .(CΛ) ◦ (ΛC) ◦ ( A∆ C) ◦ (xC) x = (CΛ) ◦ (ΛC) ◦ (xCC) ◦ ( QP ∆ C)defλ= (CΛ) ◦ (CxC) ◦ (C ρ Q P C ) ◦ (χP C) ◦ (QP δ C C) ◦ ( QP ∆ C)defλ= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (CQP δ C ) ◦ (C ρ Q P C ) ◦ (χP C) ◦ (QP δ C C)◦ ( QP ∆ C)C ρ Q ,(125)= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (QP δ C ) ◦ (χP C)◦ ( QP Qρ C P)◦ (QP δC )χ= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (QP δ C ) ◦ ( Qρ C P)◦ (χP ) ◦ (QP δC )C ρ Q= (CCx) ◦(C C ρ Q P ) ◦ (CχP ) ◦ (CQP δ C ) ◦ ( CQρ C P)◦( Cρ Q P ) ◦ (χP ) ◦ (QP δ C )(127)= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (CQδ D P ) ◦ ( CQ D ρ P)◦( Cρ Q P ) ◦ (χP )◦ (QP δ C )(113)= (CCx) ◦ ( C C ρ Q P ) ◦ ( CQε D P ) ◦ ( CQ D ρ P)◦( Cρ Q P ) ◦ (χP ) ◦ (QP δ C )so that we getQcomfun= (CCx) ◦ ( C C ρ Q P ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Qcomfun= (CCx) ◦ ( ∆ C QP ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )∆ C= ( ∆ C A ) ◦ (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) defλ= ( ∆ C A ) ◦ Λ ◦ (xC)(CΛ) ◦ (ΛC) ◦ ( A∆ C) ◦ (xC) = ( ∆ C A ) ◦ Λ ◦ (xC)and since xC is an epimorphism we deduce thatNow we compute(CΛ) ◦ (ΛC) ◦ ( A∆ C) = ( ∆ C A ) ◦ Λ.Λ ◦ (u A C) ◦ ( ε C C ) (103)= Λ ◦ (xC) ◦ (δ C C)= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (δ C C)δ=C(Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (δ C QP ) ◦ (Cδ C )(112)= (Cx) ◦ (C ρ Q P ) ◦ ( ε C QP ) ◦ (Cδ C ) = (Cx) ◦ (C ρ Q P ) ◦ δ C ◦ ( ε C C )
(112)= (Cx) ◦ (Cδ C ) ◦ ∆ C ◦ ( ε C C ) (103)= (Cu A ) ◦ ( Cε C) ◦ ∆ C ◦ ( ε C C )Ccomonad= (Cu A ) ◦ ( ε C C )and since ε C C is an epimorphism we get thatΛ ◦ (u A C) = (Cu A ) .Finally we compute(ε C A ) ◦ Λ ◦ (xC) defλ= ( ε C A ) ◦ (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )119ε C = x ◦ ( ε C QP ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Qcomfun= x ◦ (χP ) ◦ (QP δ C ) (102)= m A ◦ (xx) ◦ (QP δ C )x= m A ◦ (xA) ◦ (QP x) ◦ (QP δ C ) (103)= m A ◦ (xA) ◦ (QP u A ) ◦ ( QP ε C)x= m A ◦ (Au A ) ◦ x ◦ ( QP ε C) Acomonad,x=and since xC is an epimorphism we get that(ε C A ) ◦ Λ = Aε C .2) Consider the functorial morphism given by(AεC ) ◦ (xC)DP Q δ DP Q−→ P QP Q P χ−→ P Q P ρD Q−→ P QD yD−→ BDγ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .Recall that z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q and let us computethat isLet us computeγ ◦ ( Dz l) ? = γ ◦ (Dz r )(yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) ◦ (DP χ) ◦ (Dδ D P Q)?= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) ◦ ( Dε D P Q ) .(yD) ◦ ( P ρQ) D ◦ (P χ) ◦ (δD P Q) ◦ (DP χ) ◦ (Dδ D P Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (P QP χ) ◦ (δD P QP Q) ◦ (Dδ D P Q)δ D ,(111)= (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (P χP Q) ◦ (P QδD P Q) ◦ (δ D DP Q)(113)= (yD) ◦ ( (P ρQ) D ◦ (P χ) ◦ P Qε D P Q ) ◦ (δ D DP Q)δ=D(yD) ◦ ( )P ρ D Q ◦ (P χ) ◦ (δD P Q) ◦ ( Dε D P Q ) .Since (DB, Dy) = Coequ Fun(Dz l , Dz r) , there exists a functorial morphism Γ :DB → BD such thatΓ ◦ (Dy) = γ = (yD) ◦ ( P ρ D Q)◦ (P χ) ◦ (δD P Q) .
- 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 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 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
- 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
118so that we getχ= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (χP C)defλ= Λ ◦ (xC) ◦ (χP C) (102)= Λ ◦ (m A C) ◦ (xxC)(Cm A ) ◦ (ΛA) ◦ (AΛ) ◦ (xxC) = Λ ◦ (m A C) ◦ (xxC)and since xxC is an epimorphism we obtainLet now compute(Cm A ) ◦ (ΛA) ◦ (AΛ) = Λ ◦ (m A C) .(CΛ) ◦ (ΛC) ◦ ( A∆ C) ◦ (xC) x = (CΛ) ◦ (ΛC) ◦ (xCC) ◦ ( QP ∆ C)defλ= (CΛ) ◦ (CxC) ◦ (C ρ Q P C ) ◦ (χP C) ◦ (QP δ C C) ◦ ( QP ∆ C)defλ= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (CQP δ C ) ◦ (C ρ Q P C ) ◦ (χP C) ◦ (QP δ C C)◦ ( QP ∆ C)C ρ Q ,(125)= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (QP δ C ) ◦ (χP C)◦ ( QP Qρ C P)◦ (QP δC )χ= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (C ρ Q P QP ) ◦ (QP δ C ) ◦ ( Qρ C P)◦ (χP ) ◦ (QP δC )C ρ Q= (CCx) ◦(C C ρ Q P ) ◦ (CχP ) ◦ (CQP δ C ) ◦ ( CQρ C P)◦( Cρ Q P ) ◦ (χP ) ◦ (QP δ C )(127)= (CCx) ◦ ( C C ρ Q P ) ◦ (CχP ) ◦ (CQδ D P ) ◦ ( CQ D ρ P)◦( Cρ Q P ) ◦ (χP )◦ (QP δ C )(113)= (CCx) ◦ ( C C ρ Q P ) ◦ ( CQε D P ) ◦ ( CQ D ρ P)◦( Cρ Q P ) ◦ (χP ) ◦ (QP δ C )so that we getQcomfun= (CCx) ◦ ( C C ρ Q P ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )Qcomfun= (CCx) ◦ ( ∆ C QP ) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C )∆ C= ( ∆ C A ) ◦ (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) defλ= ( ∆ C A ) ◦ Λ ◦ (xC)(CΛ) ◦ (ΛC) ◦ ( A∆ C) ◦ (xC) = ( ∆ C A ) ◦ Λ ◦ (xC)and since xC is an epimorphism we deduce thatNow we compute(CΛ) ◦ (ΛC) ◦ ( A∆ C) = ( ∆ C A ) ◦ Λ.Λ ◦ (u A C) ◦ ( ε C C ) (103)= Λ ◦ (xC) ◦ (δ C C)= (Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (QP δ C ) ◦ (δ C C)δ=C(Cx) ◦ (C ρ Q P ) ◦ (χP ) ◦ (δ C QP ) ◦ (Cδ C )(112)= (Cx) ◦ (C ρ Q P ) ◦ ( ε C QP ) ◦ (Cδ C ) = (Cx) ◦ (C ρ Q P ) ◦ δ C ◦ ( ε C C )
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