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

10Let us computea ◦ ϕ −1 ◦ i ′ = ψ −1 ◦ a ′ ◦ i ′ def i ′= ψ −1 ◦ b ′ ◦ i ′ = b ◦ ϕ −1 ◦ i ′and since (E, i) = Equ Fun (a, b) there exists a unique functorial morphism ̂ϕ ′ : E ′ →E such thati ◦ ̂ϕ ′ = ϕ −1 ◦ i ′ .Then we havei ◦ ̂ϕ ′ ◦ ̂ϕ = ϕ −1 ◦ i ′ ◦ ̂ϕ = ϕ −1 ◦ ϕ ◦ i = iand since i is a monomorphism we deduce thatSimilarlŷϕ ′ ◦ ̂ϕ = Id E .i ′ ◦ ̂ϕ ◦ ̂ϕ ′ = ϕ ◦ i ◦ ̂ϕ ′ = ϕ ◦ ϕ −1 ◦ i ′ = i ′and since i ′ is a monomorphism we obtain that̂ϕ ◦ ̂ϕ ′ = Id E ′.Lemma <strong>2.</strong>16. Let K : B → A be a full and faithful functor and let f, g : X → Y bemorphisms in B. If (KE, Ke) = Equ A (Kf, Kg) then (E, e) = Equ B (f, g).Proof. Since K is faithful, from (Kf) ◦ (Ke) = (Kg) ◦ (Ke) we get that f ◦ e = g ◦ e.Let h : Z → X be a morphism in B such that f ◦ h = g ◦ h. Then in A we get(Kf) ◦ (Kh) = (Kg) ◦ (Kh) and hence there exists a unique morphism ξ : KZ →KE such that (Ke) ◦ ξ = (Kh). Since ξ ∈ Hom A (KZ, KE) and K is full, thereexists a morphism ζ ∈ Hom B (Z, E) such that ξ = Kζ. Since K is faithful, from(Ke) ◦ (Kζ) = Kh we get e ◦ ζ = h. From the uniqueness of ξ, the one of ζ easilyfollows.□Lemma <strong>2.</strong>17. Let α, γ : F → G be functorial morphisms where F, G : A → Bare functors. Assume that, for every X ∈ A there exists Equ B (αX, γX). Let(E, i) = Equ Fun (α, γ) , where i : E → F . Then, for every X ∈ A and Y ∈ B wehave that(Hom B (Y, EX) , Hom B (Y, iX)) = Equ Sets (Hom B (Y, αX) , Hom B(Y, γX))which means thatwhere(Hom B (−, E) , Hom B (−, i)) = Equ Fun (Hom B (−, α) , Hom B(−, γ))Hom B (−, E) and Equ Fun (Hom B (−, α) , Hom B(−, γ)) : B op × A → Sets.Proof. We have thatHom B (Y, αX) ◦ Hom B (Y, iX) = Hom B (Y, (αX) ◦ (iX))= Hom B (Y, (γX) ◦ (iX)) = Hom B (Y, γX) ◦ Hom B (Y, iX)□