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 ...
108andsatisfying(B, y) = Coequ Fun((P χ) ◦ (δD P Q) , ε D P Q ) .A µ Q ◦ (xQ) = χ and µ B Q ◦ (Qy) = χ.Furthermore1) The morphisms cocan 1 := (A µ Q P ) ◦ (Aδ C ) : AC → QP is an isomorphism.2) The morphism cocan 1 := ( P µ B Q)◦ (δD B) : DB → P Q is an isomorphisms.3)4)5)6)(χP ) ◦ (QP δ C ) = cocan 1 ◦ (xC) and (P χ) ◦ (δ D P Q) = cocan 1 ◦ (Dy)x = ( Aε C) ◦ (cocan 1 ) −1y = ( ε D B ) ◦ (cocan 1 ) −1δ C = (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)δ D = ( P µ B Q)◦ (δD B) ◦ (Du B ) .From the last equalities, we deduce that if ε C A is a regular epimorphism, so is σ Aand if Bε D is a regular epimorphism, so is σ B .Proof. Note that we are in the setting of Theorem 6.29.1) Let us check that cocan 1 is an isomorphism.( The) inverse of the functorial morphism cocan 1 is given by cocan −11 = (xC) ◦QρCP : QP → AC. Indeed we compute(xC) ◦ ( Qρ C P)◦ cocan1 ◦ (xC) = (xC) ◦ ( Qρ C P)◦( Aµ Q P ) ◦ (Aδ C ) ◦ (xC)= (xC) ◦ ( Qρ C P)◦( Aµ Q P ) ◦ (xQP ) ◦ (QP δ C )(115)= (xC) ◦ ( Qρ C P)◦ (χP ) ◦ (QP δC ) = (xC) ◦ (χP C) ◦ ( QP Qρ C P)◦ (QP δC )P rightCcol= (xC) ◦ (χP C) ◦ (QP δ C C) ◦ ( QP ∆ C) = (xC) ◦ ( w l C ) ◦ ( QP ∆ C)x coequ= (xC) ◦ (w r C) ◦ ( QP ∆ C) = (xC) ◦ ( QP ε C C ) ◦ ( QP ∆ C) C comonad= (xC) .Since xC is an epimorphism, we obtain that(xC) ◦ ( Qρ C P)◦ cocan1 = AC.On the other hand, we havecocan 1 ◦ (xC) ◦ ( ) (Qρ C P = Aµ Q P ) ◦ (Aδ C ) ◦ (xC) ◦ ( )Qρ C P= (A µ Q P ) ◦ (xQP ) ◦ (QP δ C ) ◦ ( )Qρ C (115)P = (χP ) ◦ (QP δ C ) ◦ ( )Qρ C P(118)= (χP ) ◦ (Qδ D P ) ◦ ( )Q D ρ P(113)= ( Qε D P ) ◦ ( Q D ρ P) D ρ P counital= QP
so we obtain thatcocan −11 = (xC) ◦ ( Qρ C P).2) Similarly we prove that we prove that cocan 1 := ( P µ B Q)◦ (δD B) : DB → P Q isan isomorphism with (Dy) ◦ (D ρ P Q ) its inverse. In fact we have(Dy) ◦ (D ρ P Q ) ◦ cocan 1 ◦ (Dy) = (Dy) ◦ (D ρ P Q ) ◦ ( P µ B Q)◦ (δD B) ◦ (Dy)D ρ P= (Dy) ◦(DP µBQ)◦( Dρ P QB ) ◦ (δ D B) ◦ (Dy)δ D leftDcol= (Dy) ◦ ( DP µ Q) B ◦ (DδD B) ◦ ( ∆ D B ) ◦ (Dy)∆= D(Dy) ◦ ( )DP µ B Q ◦ (DδD B) ◦ (DDy) ◦ ( ∆ D P Q )δ=D(Dy) ◦ ( DP µ Q) B ◦ (DP Qy) ◦ (DδD P Q) ◦ ( ∆ D P Q )(117)= (Dy) ◦ (DP χ) ◦ (Dδ D P Q) ◦ ( ∆ D P Q )ycoequ= (Dy) ◦ ( Dε D P Q ) ◦ ( ∆ D P Q ) Dcomonad= Dyand since D preserves coequalizers, Dy is an epimorphism, so that we get(Dy) ◦ (D ρ P Q ) ◦ cocan 1 = DB.On the other hand we havecocan 1 ◦ (Dy) ◦ (D ρ P Q ) = ( )P µ B Q ◦ (δD B) ◦ (Dy) ◦ (D ρ P Q )δ= ( DP µ Q) B ◦ (P Qy) ◦ (δD P Q) ◦ (D ρ P Q ) (117)= (P χ) ◦ (δ D P Q) ◦ (D ρ P Q )so that we get3) We haveso that(118)= (P χ) ◦ (P δ C Q) ◦ ( ρ C P Q ) (112)= ( P ε C Q ) ◦ ( ρ C P Q )P com= P Qcocan 1 ◦ (Dy) ◦ (D ρ P Q ) = P Q.(χP ) ◦ (QP δ C ) (115)= (A µ Q P ) ◦ (xQP ) ◦ (QP δ C )x= (A µ Q P ) ◦ (Aδ C ) ◦ (xC) defcocan 1= cocan 1 ◦ (xC)(119) (χP ) ◦ (QP δ C ) = cocan 1 ◦ (xC) .Similarly we haveso that(P χ) ◦ (δ D P Q) (117)= ( P µ B Q)◦ (P Qy) ◦ (δD P Q)δ= ( DP µ Q) B ◦ (δD B) ◦ (Dy) defcocan 1= cocan 1 ◦ (Dy)(120) (P χ) ◦ (δ D P Q) = cocan 1 ◦ (Dy) .109
- 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 106 and 107: 106There exist functorial morphisms
- 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
- Page 156 and 157: 156) ( )l=(pb Q AQ B ◦ ̂QA µ QB
so we obtain thatcocan −11 = (xC) ◦ ( Qρ C P).2) Similarly we prove that we prove that cocan 1 := ( P µ B Q)◦ (δD B) : DB → P Q isan isomorphism with (Dy) ◦ (D ρ P Q ) its inverse. In fact we have(Dy) ◦ (D ρ P Q ) ◦ cocan 1 ◦ (Dy) = (Dy) ◦ (D ρ P Q ) ◦ ( P µ B Q)◦ (δD B) ◦ (Dy)D ρ P= (Dy) ◦(DP µBQ)◦( Dρ P QB ) ◦ (δ D B) ◦ (Dy)δ D leftDcol= (Dy) ◦ ( DP µ Q) B ◦ (DδD B) ◦ ( ∆ D B ) ◦ (Dy)∆= D(Dy) ◦ ( )DP µ B Q ◦ (DδD B) ◦ (DDy) ◦ ( ∆ D P Q )δ=D(Dy) ◦ ( DP µ Q) B ◦ (DP Qy) ◦ (DδD P Q) ◦ ( ∆ D P Q )(117)= (Dy) ◦ (DP χ) ◦ (Dδ D P Q) ◦ ( ∆ D P Q )ycoequ= (Dy) ◦ ( Dε D P Q ) ◦ ( ∆ D P Q ) Dcom<strong>on</strong>ad= Dyand since D preserves coequalizers, Dy is an epimorphism, so that we get(Dy) ◦ (D ρ P Q ) ◦ cocan 1 = DB.On the other hand we havecocan 1 ◦ (Dy) ◦ (D ρ P Q ) = ( )P µ B Q ◦ (δD B) ◦ (Dy) ◦ (D ρ P Q )δ= ( DP µ Q) B ◦ (P Qy) ◦ (δD P Q) ◦ (D ρ P Q ) (117)= (P χ) ◦ (δ D P Q) ◦ (D ρ P Q )so that we get3) We haveso that(118)= (P χ) ◦ (P δ C Q) ◦ ( ρ C P Q ) (112)= ( P ε C Q ) ◦ ( ρ C P Q )P com= P Qcocan 1 ◦ (Dy) ◦ (D ρ P Q ) = P Q.(χP ) ◦ (QP δ C ) (115)= (A µ Q P ) ◦ (xQP ) ◦ (QP δ C )x= (A µ Q P ) ◦ (Aδ C ) ◦ (xC) defcocan 1= cocan 1 ◦ (xC)(119) (χP ) ◦ (QP δ C ) = cocan 1 ◦ (xC) .Similarly we haveso that(P χ) ◦ (δ D P Q) (117)= ( P µ B Q)◦ (P Qy) ◦ (δD P Q)δ= ( DP µ Q) B ◦ (δD B) ◦ (Dy) defcocan 1= cocan 1 ◦ (Dy)(120) (P χ) ◦ (δ D P Q) = cocan 1 ◦ (Dy) .109
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