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

Proof. Clearly (qP )◦(αP ) = (qP )◦(βP ). Let y : F ′ P → Y be a functorial morphismsuch that y ◦ (αP ) = y ◦ (βP ). Then,(yD) ◦ (αP D) = (yD) ◦ (βP D)and hence, since by Lemma <strong>2.</strong>7 (QP D, qP D) = Coequ B (αP D, βP D), there existsa unique d D : QP D → Y D such thatd D ◦ (qP D) = yD.Let us prove that the assignment D ↦→ d D yields a functorial morphism d : QP → Y .Let h : D → D ′ be a morphism in D. We compute(Y h) ◦ d D ◦ (qP D) = (Y h) ◦ (yD) y = (yD ′ ) ◦ (F ′ P h)Since qP D is an epimorphism, we conclude.= d D ′ ◦ (qP D ′ ) ◦ (F ′ P h) q = d D ′ ◦ (QP h) ◦ (qP D) .Lemma <strong>2.</strong>10 ([BM, Lemma <strong>2.</strong>2]). Let G, G ′ : C → K be functors, and let γ, θ :G → G ′ be functorial morphisms. Assume that every pair of parallel morphisms inK has an equalizer and let (E, i) = Equ Fun (γ, θ). Under these assumptions, for anyfunctor P : D → C, Equ Fun (γP, θP ) = (EP, iP ).Lemma <strong>2.</strong>1<strong>1.</strong> Consider the following serially commutative diagram in an arbitrarycategory KAfgBn mn ′m ′n ′′ m ′′A ′ f ′ B ′ i ′ C ′g ′ee ′ e ′′A ′′ f ′′ g ′′ B ′′ i ′′ C ′′Assume that all columns are coequalizers and also the first and second rows arecoequalizers. Then also the third row is a coequalizer.Proof. In order to see that the third row is a fork, note that, by commutativity ofthe diagram and fork property of the second row,i ′′ ◦ f ′′ ◦ e = i ′′ ◦ e ′ ◦ f ′ = e ′′ ◦ i ′ ◦ f ′ = e ′′ ◦ i ′ ◦ g ′ = i ′′ ◦ e ′ ◦ g ′= i ′′ ◦ g ′′ ◦ e.Since e is an epimorphism, this proves that the third row is a fork that is i ′′ ◦ f ′′ =i ′′ ◦ g ′′ .To conclude we want to prove the universality of i ′′ . To do so, let us take anymorphism x : B ′′ → X such that x ◦ f ′′ = x ◦ g ′′ . Then we want to prove that thereexist a unique functorial morphism z : C ′′ → X such that z ◦ i ′′ = x.We observex ◦ e ′ ◦ f ′ = x ◦ f ′′ ◦ e = x ◦ g ′′ ◦ e = x ◦ e ′ ◦ g ′ .so we get thatx ◦ e ′ ◦ f ′ = x ◦ e ′ ◦ g ′ .iC7□