64and[(Ω ◦ Γ) (ϕ)] (Y ) = (LY, (ϕLY ) ◦ (LηY )) = Υ (ϕ) (Y ) and [(Ω ◦ Γ) (ϕ)] (f) = Lf.Remark 4.44. When C = LR = (LR, LηR, ɛ) and ϕ = Id LR the functor K =Υ (ϕ) : B → LR A such that LR U ◦ K = L is called the Eilenberg-Moore comparis<strong>on</strong>functor.Corollary 4.45. Let C = ( C, ∆ C , ε C) and D = ( D, ∆ D , ε D) be com<strong>on</strong>ads <strong>on</strong> acategory A. There exists a bijective corresp<strong>on</strong>dence between the following collecti<strong>on</strong>sof data:K Functors K : C A → D A such that D U ◦ K = C U,M com<strong>on</strong>ad morphisms ϕ : C → Dgiven byΨ : K → M where Ψ (K) = (Cɛ) ◦ ([D U ( γ D K )] C F )Υ : M → K where Υ (ϕ) (Y ) = (C UY, ( ϕ C UY ) ◦ (C Uγ C Y )) and Υ (ϕ) (f) = C U (f) .Proof. Apply Theorem 4.43 to the case L = C U : C A → A and R = C F : A → C Aand note that (LR, LηR, ɛ) = (C U C F , C Uγ C C F , ε C) = ( C, ∆ C , ε C) .□Propositi<strong>on</strong> 4.46. Let (L, R) be an adjuncti<strong>on</strong> where L : B → A and R : A →B , let C = ( C, ∆ C , ε C) be a com<strong>on</strong>ad <strong>on</strong> the category A and let ϕ : LR =(LR, LηR, ɛ) → C = ( C, ∆ C , ε C) be a com<strong>on</strong>ad morphism. Let α = Θ (ϕ) =(Rϕ) ◦ (ηR). Then the isomorphism a X,Y : Hom A (LY, X) → Hom B (Y, RX) of theadjuncti<strong>on</strong> (L, R) induces an isomorphismâ X,Y: HomC A(Kϕ Y, ( X, C ρ X))→ EquSets(HomB (Y, αX) , Hom B(Y, R C ρ X)).Proof. Leta X,Y : Hom A (LY, X) → Hom B (Y, RX)be the isomorphism of the adjuncti<strong>on</strong> (L, R) for every Y ∈ B and for every X ∈ A.Recall that a X,Y (ξ) = (Rξ) ◦ (ηY ) and a −1X,Y(ζ) = (ɛX) ◦ (Lζ) . Let us check thatwe can apply Lemma <str<strong>on</strong>g>2.</str<strong>on</strong>g>15 to the case Z = Hom A (L−, X) , Z ′ = Hom B (−, RX) ,W = Hom A (L−, CX) , W ′ = Hom( B (−, RCX), ) a = C ρ X ◦ −, b = C − ◦Γ (ϕ) Y,a ′ = Hom( B (−, αX), b ′ = Hom B −, R C ρ X and ϕ = aX,− , ψ = a CX,− , E =Equ CFun ρ X ◦ −, C (−) ◦ Γ (ϕ) Y ) (and( ))E ′ = Equ Fun HomB (−, αX) , Hom B −, R C ρ X(Equ CFun ρ X ◦ −, C (−) ◦ Γ (ϕ) − ) â ( ( ))X,− Equ Fun HomB (−, αX) , Hom B −, R C ρ X□iZ = Hom A (L−, X)a= C ρ X ◦−b=C−◦Γ(ϕ)−W = Hom A (L−, CX)a X,−a CX,−Z ′ = Hom B (−, RX)a ′ =Hom B (−,αX)i ′b ′ =Hom B(−,R C ρ X) W ′ = Hom B (−, RCX)
For every Y ∈ B, X ∈ A and for every ξ ∈ Hom A (LY, X) , let us computeHom B (Y, αX) ◦ a X,Y (ξ) = αX ◦ a X,Y (ξ) = (RϕX) ◦ (ηRX) ◦ a X,Y (ξ)defa= (RϕX) ◦ (ηRX) ◦ (Rξ) ◦ (ηY )η= (RϕX) ◦ (RLRξ) ◦ (RLηY ) ◦ (ηY ) defa= a CX,Y [(ϕX) ◦ (LRξ) ◦ (LηY )] ϕ == a CX,Y [(Cξ) ◦ (ϕLY ) ◦ (LηY )]Since Γ (ϕ) = (ϕL) ◦ (Lη) we have obtained thatLet us calculateHom B (Y, αX) ◦ a X,Y = a CX,Y ◦ [(C−) ◦ (Γ (ϕ) Y )] .( )Hom B Y, R C ρ X ◦ aX,Y (ξ) = ( )R C ρ X ◦ aX,Y (ξ) defa= ( )R C ρ X ◦ (Rξ) ◦ (ηY )defa (= a CCX,Y ρ X ◦ ξ )Therefore we get thatHom B(Y, R C ρ X)◦ aX,Y = a CX,Y ◦ (C ρ X ◦ − )Since K ϕ (Y ) = Υ (ϕ) (Y ) = (LY, (ϕLY ) ◦ (LηY )), for every χ ∈ Hom A (LY , X) wehave65and[C (−) ◦ Γ (ϕ) Y ] (χ) = Γ (ϕ) Y = (Cχ) ◦ (ϕLY ) ◦ (LηY ) = (Cχ) ◦ C ρ LY[ Cρ X ◦ − ] (χ) = C ρ X ◦ χso thatThus we get[C (−) ◦ Γ (ϕ) Y ] (χ) = [C ρ X ◦ − ] (χ) if and <strong>on</strong>ly ifχ ∈ HomC A(((LY ) , (ϕLY ) ◦ (LηY )) ,(X, C ρ X)).Equ HomA (LY ,X)( Cρ X ◦ −, C (−) ◦ Γ (ϕ) Y )= { f ∈ Hom A (LY , X) | C ρ X ◦ f = (Cf) ◦ (Γ (ϕ) Y ) }= { f ∈ Hom A (LY , X) | C ρ X ◦ f = (Cf) ◦ (ϕLY ) ◦ (LηY ) }= { (f ∈ Hom CA U (K ϕ Y ) , C U ( )) }X, C ρ X | C ρ X ◦ f = (Cf) ◦ C ρC U(K ϕY )(= HomC A Kϕ Y, ( ))X, C ρ Xso that Equ Fun( Cρ X ◦ −, C (−) ◦ Γ (ϕ) − ) = HomC A(Kϕ −, ( X, C ρ X)). □Part of the following Propositi<strong>on</strong> is already in [GT], Propositi<strong>on</strong> <str<strong>on</strong>g>2.</str<strong>on</strong>g>3.Propositi<strong>on</strong> 4.47. Let (L, R) be an adjuncti<strong>on</strong> where L : B → A and R : A → B,let C = ( C, ∆ C , ε C) be a com<strong>on</strong>ad <strong>on</strong> a category A and let ϕ : LR = (LR, LηR, ɛ) →C = ( C, ∆ C , ε C) be a com<strong>on</strong>ad morphism. Let α = Θ (ϕ) = (Rϕ) ◦ (ηR). Thenthe functor K ϕ = Υ (ϕ) : B → C A has a right adjoint D ϕ : C A → B if and <strong>on</strong>ly if,
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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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- Page 15 and 16: f ↦→ Rfis bijective for every X
- Page 17 and 18: Remark 3.10. Let A = (A, m A , u A
- Page 19 and 20: 19and fromµ A P ◦ ( µ A P A ) =
- Page 21 and 22: and thusk ◦ (u A QZ) ◦ (Qz) = h
- Page 23 and 24: and since A preserves equalizers, A
- Page 25 and 26: Conversely, let Φ be a functorial
- Page 27 and 28: Proof. Apply Proposition 3.24 to th
- Page 29 and 30: Since Q is a left A-module functor,
- Page 31 and 32: (Q BB F, p QB F ) = Coequ Fun(µBQ
- Page 33 and 34: Theorem 3.37. Let B = (B, m B , u B
- Page 35 and 36: where A UG B F : B → A is such th
- Page 37 and 38: Proposition 3.44. Let A = (A, m A ,
- Page 39 and 40: Note that, since f and g are A-bili
- Page 41 and 42: Proposition 3.54. Let (L, R) be an
- Page 43 and 44: Corollary 3.58. Let (L, R) be an ad
- Page 45 and 46: Definition 4.2. A
- Page 47 and 48: Proposition 4.13. Let C = ( C, ∆
- Page 49 and 50: Then we have(P Cx) ◦ ( ρ C P X )
- Page 51 and 52: and since C preserves coequalizers,
- 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 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
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- 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
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- 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
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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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and[µBQ ◦ ( Qσ B)] (− ⊗ T x
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178so that− ⊗ R 1 A ⊗ R c = (
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180− ⊗ T x ⊗ R 1 A ⊗ A f
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182(208)(209)(210)(211)(h1 ) 0 ⊗
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184= abd 0 ⊗ d 1 1 ⊗ d 2 1b⊗d
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186so that h 1 ⊗ h 2 ⊗ a ∈ A
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188= 〈( h (1) y (1))εH ( h (2) y
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190H C is faithfully coflat. Assume
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192=(Qε C H C) ( ∑ )−□ C k i
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194Following Theorem 6.29, we now c
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196)û E(ε C H C (h) = û E (π (h
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198Letandα l = (ϕ ⊗ H) ( (x ⊗
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200This map is well-defined, in fac
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202We now have to prove that this m
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2041) − ⊗ B Σ A preserves the
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206functorial isomorphism. In parti
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208coaction ρ C Σ : Σ → Σ ⊗
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210Now, we consider a particular ca
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212Definition 9.27. Let k be a comm
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214∆coass= a ⊗ c (1) ⊗ A 1 A
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216Definition 9.32.</strong
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218Let us compute, for every d ∈
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220• 2-cells: monad functor trans
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222We now want to prove that ρ Q·
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224Proof. Let us consider the follo
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226and since p Q•B Q ′ ,Q ′
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228(241)= (1 Q • B l Q ′) ζ Q,
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230On the other hand, we can first
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232so that we define the map φ F (
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234Since we have(B • B (Q · A) ,
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2362-cells. This means that a comon
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238defined by settingu Q·A = ( u (
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240the unique A-bimodule morphism s
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242Let F be a finite subset of Hom
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244Lemma A.4. Let A be an abelian c
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246We haveT (ζ) ◦ ξ ◦ T H (p)
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248where k : Ker (Coker (f ◦ p))
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250be the codiagonal map of the ρ
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252Proposition A.12 ([ELGO2, Propos
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254(⇒) Let {A i } i∈Ibe a famil
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256We will prove that h : ∐ B i
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258Proposition A.19. Let (T, H) be
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260Since P is finite Hom A (P, P )
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262andP (J ′ )e f ′−→ P (I
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264hence there exists a unique morp
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266[RW] R. Rosebrugh, R.J. Wood, Di