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 ...
104We calculateso that we getx ◦ δ C ◦ ( Cε C) δ C= x ◦ ( QP ε C) ◦ (δ C C) = x ◦ w r ◦ (δ C C)xcoequ= x ◦ w l ◦ (δ C C) = x ◦ (χP ) ◦ (QP δ C ) ◦ (δ C C)δ C= x ◦ (χP ) ◦ (δ C QP ) ◦ (Cδ C ) (99)= x ◦ ( ε C QP ) ◦ (Cδ C )ε C = x ◦ δ C ◦ (ε C C)x ◦ δ C ◦ (Cε C ) = x ◦ δ C ◦ (ε C C).Since ( A, ε C) = Coequ Fun(Cε C , ε C C ) there exists a unique functorial morphismu A : A → A such that (103) is fulfilled. Now we want to show that A = (A, m A , u A )is a monad over A that isWe calculatem A ◦ (m A A) = m A ◦ (Am A )m A ◦ (Au A ) = A = m A ◦ (u A A) .m A ◦ (m A A) ◦ (xxx) = m A ◦ (m A A) ◦ (xxA) ◦ (QP QP x)(102)= m A ◦ (xA) ◦ (χP A) ◦ (QP QP x)χ= m A ◦ (xA) ◦ (QP x) ◦ (χP QP ) = m A ◦ (xx) ◦ (χP QP )(102)= x ◦ (χP ) ◦ (χP QP ) (98)= x ◦ (χP ) ◦ (QP χP )(102)= m A ◦ (xx) ◦ (QP χP ) = m A ◦ (xA) ◦ (QP x) ◦ (QP χP )(102)= m A ◦ (xA) ◦ (QP m A ) ◦ (QP xx) x = m A ◦ (Am A ) ◦ (xAA) ◦ (QP xx)Thus we get that= m A ◦ (Am A ) ◦ (xxx) .m A ◦ (m A A) ◦ (xxx) = m A ◦ (Am A ) ◦ (xxx)and since xxx is an epimorphism, we deduce that m A is associative. We computeThus we get thatm A ◦ (Au A ) ◦ ( Aε C) ◦ (xC) (103)= m A ◦ (Ax) ◦ (Aδ C ) ◦ (xC)x= m A ◦ (Ax) ◦ (xQP ) ◦ (QP δ C ) = m A ◦ (xx) ◦ (QP δ C )(102)= x ◦ (χP ) ◦ (QP δ C ) = x ◦ w l= x ◦ w r = x ◦ (QP ε C ) x = ( Aε C) ◦ (xC) .m A ◦ (Au A ) ◦ ( Aε C) ◦ (xC) = ( Aε C) ◦ (xC) .and since ( Aε C) ◦ (xC) is epimorphism we deduce thatm A ◦ (Au A ) = A.
105We computeso that we getm A ◦ (u A A) ◦ ( ε C A ) ◦ (Cx) (103)= m A ◦ (xA) ◦ (δ C A) ◦ (Cx)δ C= m A ◦ (xA) ◦ (QP x) ◦ (δ C QP ) = m A ◦ (xx) ◦ (δ C QP )(102)= x ◦ (χP ) ◦ (δ C QP ) (99)= x ◦ (ε C QP ) εC = ( ε C A ) ◦ (Cx)m A ◦ (u A A) ◦ ( ε C A ) ◦ (Cx) = ( ε C A ) ◦ (Cx)and since ( ε C A ) ◦ (Cx) is an epimorphism we deduce thatm A ◦ (u A A) = A.Therefore we obtain that m A is unital. We computeA µ Q ◦ ( A A µ Q)◦ (AxQ) ◦ (xQP Q)(101)= A µ Q ◦ (Aχ) ◦ (xQP Q) x = A µ Q ◦ (xQ) ◦ (QP χ)(101)= χ ◦ (QP χ) (98)= χ ◦ (χP Q) (101)= A µ Q ◦ (xQ) ◦ (χP Q)(102)= A µ Q ◦ (m A Q) ◦ (xxQ) = A µ Q ◦ (m A Q) ◦ (AxQ) ◦ (xQP Q) .Since (AxQ) ◦ (xQP Q) is an epimorphism we getWe calculateA µ Q ◦ ( A A µ Q)= A µ Q ◦ (m A Q) .A µ Q ◦ (u A Q) ◦ ( ε C Q ) (103)= A µ Q ◦ (xQ) ◦ (δ C Q)(101)= χ ◦ (δ C Q) (99)= ( ε C Q ) .Since ( ε C Q ) is an epimorphism we obtainA µ Q ◦ (u A Q) = Q.Proposition 6.26. Let A and B be categories with coequalizers and let P : A → B,Q : B → A, and D : B → B be functors. Assume that all the functors P, Q and Dpreserve coequalizers. Let ε D : D → B be a functorial morphism and assume that(B, εD ) = Coequ Fun(Dε D , ε D D ) . Let χ : QP Q → Q be a functorial morphism suchthatχ ◦ (QP χ) = χ ◦ (χP Q) .Let δ D : D → P Q be a functorial morphism such that(105) χ ◦ (Qδ D ) = Qε D .Let z l = (P χ) ◦ (δ D P Q) and z r = ε D P Q : DP Q → P Q. Set(106) (B, y) = Coequ Fun(z l , z r) .There exists a functorial morphism µ B Q : QB → Q such that(107) µ B Q ◦ (Qy) = χ.□
- 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 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 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
- 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
104We calculateso that we getx ◦ δ C ◦ ( Cε C) δ C= x ◦ ( QP ε C) ◦ (δ C C) = x ◦ w r ◦ (δ C C)xcoequ= x ◦ w l ◦ (δ C C) = x ◦ (χP ) ◦ (QP δ C ) ◦ (δ C C)δ C= x ◦ (χP ) ◦ (δ C QP ) ◦ (Cδ C ) (99)= x ◦ ( ε C QP ) ◦ (Cδ C )ε C = x ◦ δ C ◦ (ε C C)x ◦ δ C ◦ (Cε C ) = x ◦ δ C ◦ (ε C C).Since ( A, ε C) = Coequ Fun(Cε C , ε C C ) there exists a unique functorial morphismu A : A → A such that (103) is fulfilled. Now we want to show that A = (A, m A , u A )is a m<strong>on</strong>ad over A that isWe calculatem A ◦ (m A A) = m A ◦ (Am A )m A ◦ (Au A ) = A = m A ◦ (u A A) .m A ◦ (m A A) ◦ (xxx) = m A ◦ (m A A) ◦ (xxA) ◦ (QP QP x)(102)= m A ◦ (xA) ◦ (χP A) ◦ (QP QP x)χ= m A ◦ (xA) ◦ (QP x) ◦ (χP QP ) = m A ◦ (xx) ◦ (χP QP )(102)= x ◦ (χP ) ◦ (χP QP ) (98)= x ◦ (χP ) ◦ (QP χP )(102)= m A ◦ (xx) ◦ (QP χP ) = m A ◦ (xA) ◦ (QP x) ◦ (QP χP )(102)= m A ◦ (xA) ◦ (QP m A ) ◦ (QP xx) x = m A ◦ (Am A ) ◦ (xAA) ◦ (QP xx)Thus we get that= m A ◦ (Am A ) ◦ (xxx) .m A ◦ (m A A) ◦ (xxx) = m A ◦ (Am A ) ◦ (xxx)and since xxx is an epimorphism, we deduce that m A is associative. We computeThus we get thatm A ◦ (Au A ) ◦ ( Aε C) ◦ (xC) (103)= m A ◦ (Ax) ◦ (Aδ C ) ◦ (xC)x= m A ◦ (Ax) ◦ (xQP ) ◦ (QP δ C ) = m A ◦ (xx) ◦ (QP δ C )(102)= x ◦ (χP ) ◦ (QP δ C ) = x ◦ w l= x ◦ w r = x ◦ (QP ε C ) x = ( Aε C) ◦ (xC) .m A ◦ (Au A ) ◦ ( Aε C) ◦ (xC) = ( Aε C) ◦ (xC) .and since ( Aε C) ◦ (xC) is epimorphism we deduce thatm A ◦ (Au A ) = A.
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