Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

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226and since p Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) is an epimorphism, we get thatζ ′ ζ = 1 (Q•B Q ′ )• C Q ′′.Therefore, (Q • B Q ′ ) • C Q ′′ ≃ Q • B (Q ′ • C Q ′′ ) via ζ. Moreover, by Proposition 1<strong>1.</strong>5,we know that (Q • B Q ′ ) • C Q ′′ and Q • B (Q ′ • C Q ′′ ) are A-D-bimodules. We nowwant to prove that ζ is a morphism of left A-modules and right D-modules. Let uscomputeζλ (Q•B Q ′ )• C Q ′′ (1 A · p Q•B Q ′ ,Q ′′) (1 A · p Q,Q ′ · 1 Q ′′)defλ (Q•B Q ′ )• C Q ′′= ζp Q•B Q ′ ,Q ′′ (λ(Q•B Q ′ ) · 1 Q ′′)(1A · p Q,Q ′ · 1 Q ′′)defλ (Q•B Q ′ )= ζp Q•B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) (λ Q · 1 Q ′ · 1 Q ′′)(241)= p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) (λ Q · 1 Q ′ · 1 Q ′′)λ Q=pQ,Q ′ • C Q ′′ (λ Q · 1 Q ′ • C Q ′′) (1 A · 1 Q · p Q ′ ,Q ′′)defλ (Q•B Q ′ )= λ Q•B (Q ′ • C Q ′′ ) (1 A · p Q,Q ′ • C Q ′′) (1 A · 1 Q · p Q ′ ,Q ′′)(241)= λ Q•B (Q ′ • C Q ′′ ) (1 A · ζ) (1 A · p Q•B Q ′ ,Q ′′) (1 A · p Q,Q ′ · 1 Q ′′)and since (1 A · p Q•B Q ′ ,Q ′′) (1 A · p Q,Q ′ · 1 Q ′′) is epi, we get thatζλ (Q•B Q ′ )• C Q ′′ = λ Q• B (Q ′ • C Q ′′ ) (1 A · ζ)i.e. ζ is a morphism of left A-modules. Similarly, one can prove that ζ is a morphismof right D-modules.□Notation 1<strong>1.</strong>8. In the setting of Proposition 1<strong>1.</strong>7, let us consider the isomorphismof bimodulesζ : (Q • B Q ′ ) • C Q ′′ → Q • B (Q ′ • C Q ′′ ) .In order to be more clear, in the following, we will denote it byζ Q,Q ′ ,Q ′′ : (Q • B Q ′ ) • C Q ′′ → Q • B (Q ′ • C Q ′′ )which is the unique satisfying the following(244) ζ Q,Q ′ ,Q ′′p Q• B Q ′ ,Q ′′ (p Q,Q ′ · 1 Q ′′) = p Q,Q ′ • C Q ′′ (1 Q · p Q ′ ,Q ′′) .Proposition 1<strong>1.</strong>9. Let (X, A), (Y, B), (Z, C), (W, D), (U, E) be monads in the 2-category C and let (Q, λ Q , ρ Q ) be an A-B-bimodule, (Q ′ , λ Q ′, ρ Q ′) be a B-C-bimodule,(Q ′′ , λ Q ′′, ρ Q ′′) be a C-D-bimodule and (Q ′′′ , λ Q ′′′, ρ Q ′′′) be a D-E-bimodule. Then thePentagon Axiom holds, i.e. the following diagram is commutative((Q • B Q ′ ) • C Q ′′ ) • D Q ′′′ ζ Q,Q ′ ,Q ′′• D 1 Q ′′′(Q • B (Q ′ • C Q ′′ )) • D Q ′′′ζ Q•B Q ′ ,Q ′′ ,Q ′′′ζ Q,Q ′ •C Q ′′ ,Q ′′′(Q • B Q ′ ) • C (Q ′′ • D Q ′′′ )Q • B ((Q ′ • C Q ′′ ) • D Q ′′′ )ζ Q,Q ′ ,Q ′′ •D Q ′′′1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′Q • B (Q ′ • C (Q ′′ • D Q ′′′ ))
227Proof. We computeand(1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′ (ζ Q,Q ′ ,Q ′′ • D 1 Q ′′′) p (Q•B Q ′ )• C Q ′′ ,Q ′′′(p Q•B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)= (1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′p Q• B (Q ′ • C Q ′′ ),Q ′′′ (ζ Q,Q ′ ,Q ′′ · 1 Q ′′′)(p Q•B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)(244)= (1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′p Q• B (Q ′ • C Q ′′ ),Q ′′′ (p Q,Q ′ • C Q ′′ · 1 Q ′′′)(1 Q · p Q ′ ,Q ′′ · 1 Q ′′′)= (1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) p Q,(Q ′ • C Q ′′ )• D Q ′′′ (1 Q · p Q ′ • C Q ′′ ,Q ′′′) (1 Q · p Q ′ ,Q ′′ · 1 Q ′′′)= p Q,Q ′ • C (Q ′′ • D Q ′′′ ) (1 Q · ζ Q ′ ,Q ′′ ,Q ′′′) (1 Q · p Q ′ • C Q ′′ ,Q ′′′) (1 Q · p Q ′ ,Q ′′ · 1 Q ′′′)(241)= p Q,Q ′ • C (Q ′′ • D Q ′′′ ) (1 Q · p Q ′ ,Q ′′ • D Q ′′′) (1 Q · 1 Q ′ · p Q ′′ ,Q ′′′)ζ Q,Q ′ ,Q ′′ • D Q ′′′ζ Q• B Q ′ ,Q ′′ ,Q ′′′p (Q• B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)(241)= ζ Q,Q ′ ,Q ′′ • D Q ′′′p Q• B Q ′ ,Q ′′ • D Q ′′′ (1 Q• B Q ′ · p Q ′′ ,Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)p Q,Q ′= ζ Q,Q ′ ,Q ′′ • D Q ′′′p Q• B Q ′ ,Q ′′ • D Q ′′′ (p Q,Q ′ · 1 Q ′′ • D Q ′′′) (1 Q · 1 Q ′ · p Q ′′ ,Q ′′′)so that we get that(241)= p Q,Q ′ • C (Q ′′ • D Q ′′′ ) (1 Q · p Q ′ ,Q ′′ • D Q ′′′) (1 Q · 1 Q ′ · p Q ′′ ,Q ′′′)(1 Q • B ζ Q ′ ,Q ′′ ,Q ′′′) ζ Q,Q ′ • C Q ′′ ,Q ′′′ (ζ Q,Q ′ ,Q ′′ • D 1 Q ′′′) p (Q•B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′)(p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)= ζ Q,Q ′ ,Q ′′ • D Q ′′′ζ Q• B Q ′ ,Q ′′ ,Q ′′′p (Q• B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′)and since p (Q•B Q ′ )• C Q ′′ ,Q ′′′ (p Q• B Q ′ ,Q ′′ · 1 Q ′′′) (p Q,Q ′ · 1 Q ′′ · 1 Q ′′′) is an epimorphism, wededuce that the Pentagon Axiom holds.□Proposition 1<strong>1.</strong>10. Let (X, A), (Y, B), (Z, C) be monads in the 2-category C andlet (Q, λ Q , ρ Q ) be an A-B-bimodule and (Q ′ , λ Q ′, ρ Q ′) be a B-C-bimodule. Then theTriangle Axiom holds, i.e. the following diagram is commutativeProof. We compute(Q • B B) • B Q ′ ζ Q,B,Q ′Q • B (B • B Q ′ )rQ • B 1 Q ′1 Q • B l Q ′Q • B Q ′(r Q • B 1 Q ′) (p Q•B B,Q ′) (p Q,B · 1 Q ) = p Q,Q ′ (r Q · 1 Q ′) (p Q,B · 1 Q )(239)= p Q,Q ′ (ρ Q · 1 Q ′) defp= p Q,Q ′ (1 Q · λ Q ′)(238)= p Q,Q ′ (1 Q · l Q ′) (1 Q · p B,Q ′)= (1 Q • B l Q ′) p Q,B•B Q ′ (1 Q · p B,Q ′)
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Contents1.
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linearity and compatibility conditi
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5and since g ◦ f is an epimorphis
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Proof. Clearly (qP )◦(αP ) = (qP
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Lemma 2.13 ([BM, L
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i.e. Hom B (Y, iX) equalizes Hom B
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13such thatd 0 ◦ v = Id Yd 1 ◦
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f ↦→ Rfis bijective for every X
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Remark 3.10. Let A = (A, m A , u A
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19and fromµ A P ◦ ( µ A P A ) =
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and thusk ◦ (u A QZ) ◦ (Qz) = h
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and since A preserves equalizers, A
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Conversely, let Φ be a functorial
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Proof. Apply Proposition 3.24 to th
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Since Q is a left A-module functor,
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(Q BB F, p QB F ) = Coequ Fun(µBQ
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Theorem 3.37. Let B = (B, m B , u B
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where A UG B F : B → A is such th
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Proposition 3.44. Let A = (A, m A ,
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Note that, since f and g are A-bili
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Proposition 3.54. Let (L, R) be an
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Corollary 3.58. Let (L, R) be an ad
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Definition 4.2. A
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Proposition 4.13. Let C = ( C, ∆
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Then we have(P Cx) ◦ ( ρ C P X )
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and since C preserves coequalizers,
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Proof. Apply Corollary 4.24 to the
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Let( (CQ ) ()D, ι Q) C = Equ Fun
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58F D right D-comodule functors Q :
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60prove that C ν D : C F D → (C
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624.2. The compari
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64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY,
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66for every ( X, C ρ X)∈ C A, th
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68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂η
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70In particular(49) d ϕ(CX, ∆ C
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72We have to prove that (LD ϕ , Ld
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74we have that Ld ϕ K ϕ Y is mono
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and since d is mono we get that(ε
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78Corollary 4.63 (Beck’s Precise
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80We compute(LRɛLY ′ ) ◦ ( LR
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82Proof. First of all we prove that
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84i.e. Aα is a functorial morphism
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86Then we haveA µ CCX ◦ ( A∆ C
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884.23) is a functor Ã : C A → C
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90Let θ l = ( σ B P Q ) ◦ (P τ
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925)σ A = ( ε C A ) ◦ ( Cσ A)
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94(ii) the functorial morphism can
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96defΦ= ( QP A µ Q)◦(QP σ A Q
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98AU A can AA F = can AA F = ( CσA
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100Similarly, one can prove the sta
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102(b) A comonad C = ( C, ∆ C ,
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104We calculateso that we getx ◦
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106There exist functorial morphisms
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108andsatisfying(B, y) = Coequ Fun(
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1104) With notations of Theorem 6.2
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112Then ν : Y → D is the unique
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114= A µ Q ◦ ( Aε C Q ) ◦ (AC
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116= ( Aε C Q ) ◦ ( cocan1 −1
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118so that we getχ= (Cx) ◦ (C ρ
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120We want to prove that Γ is an o
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122and since Dε D is an epimorphis
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124χ= (Cχ) ◦ (C ρ Q P Q ) ◦
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126Now, since cocan 1 : AC → QP i
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1287. Herds and Coherds7.1.
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130◦ ( σ A QQQ ) ◦ (A µ Q P Q
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132= µ B Q ◦ (A µ Q B ) ◦ ( A
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134Assume now that there is another
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136and hence we get(160) x ◦ (χP
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138Proposition 7.7. In the setting
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140We calculateA µ Q ◦ ( σ A Q
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142x=and=δ C=(l= QlQ ̂QQ)◦ (QP
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144(◦ ρ D ̂QQ)Q◦ (QlQ) ◦ (Q
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146given byWe computeσ B = m B ◦
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148andy= ′m B ◦ (ν B B) ◦ (y
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150Now we compute(hQ) ◦ ( Qχ )
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152Thus we obtainσ B ◦ ( ) (P µ
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154Thus hQ is an isomorphism with i
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156) ( )l=(pb Q AQ B ◦ ̂QA µ QB
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158In fact we haveTherefore we dedu
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160χ= h 1 ◦ (P xQ B ) ◦ (P QP
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162so that we obtain:(190)We comput
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164(194)=) )(p QB ̂QA ◦(Qpb Q◦
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166= Ξ ◦ (A A U A λ) ◦ (xx A
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168)(155)= k 2 ◦(Qpb Q◦ (Ql A U
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170) ) (χ= ρ ◦(p QB ̂QA ◦(Qp
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172Theorem 8.13. Let A and B be cat
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174l = eC ρ L : L = − ⊗ B A
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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
Page 198 and 199: 198Letandα l = (ϕ ⊗ H) ( (x ⊗
Page 200 and 201: 200This map is well-defined, in fac
Page 202 and 203: 202We now have to prove that this m
Page 204 and 205: 2041) − ⊗ B Σ A preserves the
Page 206 and 207: 206functorial isomorphism. In parti
Page 208 and 209: 208coaction ρ C Σ : Σ → Σ ⊗
Page 210 and 211: 210Now, we consider a particular ca
Page 212 and 213: 212Definition 9.27. Let k be a comm
Page 214 and 215: 214∆coass= a ⊗ c (1) ⊗ A 1 A
Page 216 and 217: 216Definition 9.32.</strong
Page 218 and 219: 218Let us compute, for every d ∈
Page 220 and 221: 220• 2-cells: monad functor trans
Page 222 and 223: 222We now want to prove that ρ Q·
Page 224 and 225: 224Proof. Let us consider the follo
Page 228 and 229: 228(241)= (1 Q • B l Q ′) ζ Q,
Page 230 and 231: 230On the other hand, we can first
Page 232 and 233: 232so that we define the map φ F (
Page 234 and 235: 234Since we have(B • B (Q · A) ,
Page 236 and 237: 2362-cells. This means that a comon
Page 238 and 239: 238defined by settingu Q·A = ( u (
Page 240 and 241: 240the unique A-bimodule morphism s
Page 242 and 243: 242Let F be a finite subset of Hom
Page 244 and 245: 244Lemma A.4. Let A be an abelian c
Page 246 and 247: 246We haveT (ζ) ◦ ξ ◦ T H (p)
Page 248 and 249: 248where k : Ker (Coker (f ◦ p))
Page 250 and 251: 250be the codiagonal map of the ρ
Page 252 and 253: 252Proposition A.12 ([ELGO2, Propos
Page 254 and 255: 254(⇒) Let {A i } i∈Ibe a famil
Page 256 and 257: 256We will prove that h : ∐ B i
Page 258 and 259: 258Proposition A.19. Let (T, H) be
Page 260 and 261: 260Since P is finite Hom A (P, P )
Page 262 and 263: 262andP (J ′ )e f ′−→ P (I
Page 264 and 265: 264hence there exists a unique morp
Page 266: 266[RW] R. Rosebrugh, R.J. Wood, Di
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Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

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