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 ...
148andy= ′m B ◦ (ν B B) ◦ (y ′ B) ◦ ( ) (QQν )B ◦ QQy′(166)= m B ◦ (m B B) ◦ ( σ B σ B B ) ◦ (P iQB) ◦ (qQB) ◦ ( QQm B)◦(QQσ B σ B)◦ ( QQP iQ ) ◦ ( QQqQ )(144)= m B ◦ (m B B) ◦ ( σ B σ B B ) ◦ (jP QB) ◦ (κ ′ 0QB) ◦ ( QQm B)◦(QQσ B σ B)◦ ( QQjP Q ) ◦ ( QQκ ′ 0Q )= m B ◦ (m B B) ◦ ( Bσ B B ) ◦ ( σ B P QB ) ◦ (jP QB) ◦ (κ ′ 0QB) ◦ ( QQm B)◦ ( QQBσ B) ◦ ( QQσ B P Q ) ◦ ( QQjP Q ) ◦ ( QQκ ′ 0Q )(67)= m B ◦ (m B B) ◦ ( Bσ B B ) ◦ (u B P QB) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQm B)◦ ( QQBσ B) ◦ ( QQu B P Q ) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )u B= m B ◦ (m B B) ◦ (u B BB) ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQm B)◦ ( QQu B B ) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )Bmonad= m B ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )ν B ◦ m B ′ ◦ (y ′ y ′ ) (163)= ν B ◦ y ′ ◦ ( Qχ ) = m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) ◦ ( Qχ )(144)= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) ◦ (κ ′ 0Q) ◦ ( Qχ )(67)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qχ )u B= m B ◦ (u B B) ◦ σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qχ )Bmonad= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qχ )defχ= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ B Q ◦ Q A µ Q B ) ◦ ( QAQσ B) ◦ ( Qσ A QP Q )◦ ( QQP iQ ) ◦ ( QQqQ )A µ Q= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ ) B Q ◦ QQσB◦ ( Q A µ Q P Q ) ◦ ( Qσ A QP Q )◦ ( QQP iQ ) ◦ ( QQqQ )(82)= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ ) B Q ◦ QQσB◦ ( Qµ B QP Q )◦ ( QQσ B P Q ) ◦ ( QQP iQ ) ◦ ( QQqQ )(144)= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( ) (Qµ ) B Q ◦ QQσB◦ ( Qµ B QP Q )◦ ( QQσ B P Q ) ◦ ( QQjP Q ) ◦ ( QQκ ′ 0Q )(67)= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qµ B Q◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )Qmodulefunctor= σ B ◦ ( ε D P Q ) ◦ (κ ′ 0Q) ◦ ( Qµ B Q)◦(QQσB ) ◦ ( Qµ B QP Q ) ◦ ( QQu B P Q ))◦(QQσB ) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )
κ ′ 0= σ B ◦ ( ε D P Q ) ◦ ( )DP µ B Q ◦ (κ′0 QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )ε= D σ B ◦ ( (P µ Q) B ◦ ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )(81)= m B ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )so that we obtainm B ◦ (ν B ν B ) ◦ (y ′ y ′ ) = ν B ◦ m B ′ ◦ (y ′ y ′ )and since y ′ is an epimorphism we deduce thatNow, let us calculatem B ◦ (ν B ν B ) = ν B ◦ m B ′.ν B ◦ u B ′ ◦ ε D (163)= ν B ◦ y ′ ◦ δ D(166)= m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) ◦ δ D= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ δ D(144)= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) ◦ (κ ′ 0Q) ◦ δ D(67),(148)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ (ε D P Q) ◦ (Dj) ◦ ∆ Du B ,ε D= m B ◦ (u B B) ◦ σ B ◦ j ◦ (ε D D) ◦ ∆ DBmonadDcomonad= σ B ◦ j (67)= u B ◦ ε D .Since the tame condition is assumed, BA σ B ABAσ B A(97)= A σ B ABBF (94)= B U BA σ B ABAFis also an isomorphism. By (95) we have thatAσ B A ◦ (p P A Q) = σ Bis a functorial isomorphism and thusand Lemma
- 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 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 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
- Page 170 and 171: 170) ) (χ= ρ ◦(p QB ̂QA ◦(Qp
- Page 172 and 173: 172Theorem 8.13. Let A and B be cat
- Page 174 and 175: 174l = eC ρ L : L = − ⊗ B A
- Page 176 and 177: and[µBQ ◦ ( Qσ B)] (− ⊗ T x
- Page 178 and 179: 178so that− ⊗ R 1 A ⊗ R c = (
- Page 180 and 181: 180− ⊗ T x ⊗ R 1 A ⊗ A f
- Page 182 and 183: 182(208)(209)(210)(211)(h1 ) 0 ⊗
- Page 184 and 185: 184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d
- Page 186 and 187: 186so that h 1 ⊗ h 2 ⊗ a ∈ A
- Page 188 and 189: 188= 〈( h (1) y (1))εH ( h (2) y
- Page 190 and 191: 190H C is faithfully coflat. Assume
- Page 192 and 193: 192=(Qε C H C) ( ∑ )−□ C k i
- Page 194 and 195: 194Following Theorem 6.29, we now c
- Page 196 and 197: 196)û E(ε C H C (h) = û E (π (h
κ ′ 0= σ B ◦ ( ε D P Q ) ◦ ( )DP µ B Q ◦ (κ′0 QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )ε= D σ B ◦ ( (P µ Q) B ◦ ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )(81)= m B ◦ ( σ B B ) ◦ ( ε D P QB ) ◦ (κ ′ 0QB) ◦ ( QQσ B) ◦ ( QQε D P Q ) ◦ ( QQκ ′ 0Q )so that we obtainm B ◦ (ν B ν B ) ◦ (y ′ y ′ ) = ν B ◦ m B ′ ◦ (y ′ y ′ )and since y ′ is an epimorphism we deduce thatNow, let us calculatem B ◦ (ν B ν B ) = ν B ◦ m B ′.ν B ◦ u B ′ ◦ ε D (163)= ν B ◦ y ′ ◦ δ D(166)= m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ) ◦ δ D= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (P iQ) ◦ (qQ) ◦ δ D(144)= m B ◦ ( Bσ B) ◦ ( σ B P Q ) ◦ (jP Q) ◦ (κ ′ 0Q) ◦ δ D(67),(148)= m B ◦ ( Bσ B) ◦ (u B P Q) ◦ (ε D P Q) ◦ (Dj) ◦ ∆ Du B ,ε D= m B ◦ (u B B) ◦ σ B ◦ j ◦ (ε D D) ◦ ∆ DBm<strong>on</strong>adDcom<strong>on</strong>ad= σ B ◦ j (67)= u B ◦ ε D .Since the tame c<strong>on</strong>diti<strong>on</strong> is assumed, BA σ B ABAσ B A(97)= A σ B ABBF (94)= B U BA σ B ABAFis also an isomorphism. By (95) we have thatAσ B A ◦ (p P A Q) = σ Bis a functorial isomorphism and thusand Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>6, since p P A Q is a regular epimorphism by c<strong>on</strong>structi<strong>on</strong>, we deducethat σ B is also a regular epimorphism. Thus, by Theorem 6.6, so is Bε D . Since Breflects coequalizers we get that ε D is an epimorphism and therefore we obtainHence ν B is a morphism of m<strong>on</strong>ads.2) C<strong>on</strong>sider the following diagramν B ◦ u B ′ = u B .149DQQDhQDP Q(Qχ)◦(δ D QQ)(ε D QQ)(P µ S Q )◦(jS)◦(Dσ S) (ε D P Q)QQ y′ hQ P Q π Z ZS ′ ν Zwhere (Z, π Z ) = Coequ Fun((P µBQ)◦ (jB) ◦(DσB ) , ε D P Q ) . Note that(ε D P Q ) ◦ (DhQ) εD = (hQ) ◦ ( ε D QQ ) .
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