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

68i.e.(44) (d ϕ K ϕ Y ) ◦ (̂ηY ) = ηY.Now, we have to prove the universal property of the equalizer. Let Z ∈ B and letζ : Z → RX be a morphism such that (αX) ◦ ζ = ( R C ρ X)◦ ζ, i.e.(RϕX) ◦ (ηRX) ◦ ζ = ( )R C ρ X ◦ ζ.( ( )) (This means ζ ∈ Equ Sets HomB (Y, αX) , Hom B Y, R C ρ X ≃ HomC A Kϕ Y, ( ))X, C ρ Xby Proposition 4.46. Then,a −1X,Z (ζ) = (ɛX) ◦ (Lζ) ∈ ( ( ))HomC A (LZ, (ϕLZ) ◦ (LηZ)) , X, C ρ X= HomC A(Kϕ Z, ( X, C ρ X)).We want to prove that there exists ζ ′ : Z → D ϕ(X, C ρ X)such that dϕ(X, C ρ X)◦ζ ′ =ζ. By hypothesis the map( HomC A Kϕ Y, ( ))X, C ba (X, C ρX ( ( ))ρ ),Y X Hom B Y, Dϕ X, C ρ Xis bijective. Hence, given (ɛX) ◦ (Lζ) ∈ HomC A(Kϕ Z, ( X, C ρ X)),â (X, C ρ X ),Z ((ɛX) ◦ (Lζ)) = (D ϕ ɛX) ◦ (D ϕ Lζ) ◦ (̂ηZ) ∈ Hom B(Z, Dϕ(X, C ρ X)). Wewant to prove that(dϕ (X, C ρ X))◦ (Dϕ ɛX) ◦ (D ϕ Lζ) ◦ (̂ηZ) = ζ.We compute(dϕ(X, C ρ X))◦ (Dϕ ɛ) ◦ (D ϕ Lζ) ◦ (̂ηZ) dϕ = (RɛX) ◦ (RLζ) ◦ (d ϕ K ϕ Z) ◦ (̂ηZ)(44)= (RɛX) ◦ (RLζ) ◦ (ηZ) η = (RɛX) ◦ (ηRX) ◦ ζ (L,R)= ζ.Let us denote by ζ ′ = (D ϕ ɛX)◦(D ϕ Lζ)◦(̂ηZ) the morphism such that ( d ϕ(X, C ρ X))◦ζ ′ = ζ. We have to prove that ζ ′ is unique with respect to this property. Letζ ′′ : Z → D ϕ(X, C ρ X)be another morphism in B such that(dϕ(X, C ρ X))◦ ζ ′′ = ζ.Then we haveHom B(Z, dϕ(X, C ρ X))(ζ ′′ ) = ( d ϕ(X, C ρ X))◦ ζ ′′ = ζ= ( d ϕ(X, C ρ X))◦ ζ ′ = Hom B(Z, dϕ(X, C ρ X))(ζ ′ )and since Hom B(Z, dϕ(X, C ρ X))is mono we deduce thatζ ′′ = ζ ′ .Corollary 4.48. Let (L, R) be an adjunction where L : B → A and R : A → B.Let α = Θ (Id LR ) = ηR. Then the functor K = Υ (Id LR ) : B → LR A has a rightadjoint D : LR A → B if and only if, for every ( X, LR ρ X)∈ LR A, there existsEqu B(ηRX, R LR ρ X). In this case there exists a functorial morphism d : D → R LR Usuch that(D, d) = Equ Fun(ηR LR U, R LR Uγ LR) .□