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

5and since g ◦ f is an epimorphism, we deduce that χ ′ = χ so that(H, h) = Coequ Fun(f −1 ◦ α, f −1 ◦ β ) .Conversely, let us assume that h is a regular epimorphism, i.e.(H, h) = Coequ Fun (α, b). Then we haveg ◦ f ◦ a = h ◦ a = h ◦ b = g ◦ f ◦ b.Now, let ξ : G → X be a functorial morphism such that ξ ◦ f ◦ a = ξ ◦ f ◦ b. Bythe universal property of (H, h) = Coequ Fun (α, b), there exists a unique functorialmorphism ξ : H → X such that ξ ◦ h = ξ ◦ f, i.e. ξ ◦ g ◦ f = ξ ◦ f and since f isan isomorphism we deduce that ξ ◦ g = ξ. Let us assume that there exists anotherfunctorial morphism ξ ′ : H → X such that ξ ′ ◦ g = ξ. Since we also have ξ ◦ g = ξwe get thatξ ′ ◦ h = ξ ′ ◦ g ◦ f = ξ ◦ f = ξ ◦ g ◦ f = ξ ◦ hand since h is an epimorphism, we deduce that ξ ′ = ξ. Therefore, (H, g) =Coequ Fun (f ◦ a, f ◦ b).□Lemma <strong>2.</strong>6. Let F, G, H be functors and let f : F → G, g : G → H and h : F → Hbe functorial morphisms such that h = g ◦ f. Assume that g is a functorial isomorphism.Then h is a regular epimorphism if and only if f is a regular epimorphism.Proof. Assume first that f is a regular epimorphism, i.e. (G, f) = Coequ Fun (α, β).Then we haveh ◦ α = g ◦ f ◦ α = g ◦ f ◦ β = h ◦ β.Let ξ : F → X be a functorial morphism such that ξ ◦ α = ξ ◦ β. By the universalproperty of the coequalizer (G, f) = Coequ Fun (α, β), there exists a unique functorialmorphism ξ such that ξ ◦ f = ξ. Then we haveξ ◦ g −1 ◦ h = ξ ◦ g −1 ◦ g ◦ f = ξ ◦ f = ξso that ξ factorizes through h via ξ ◦g −1 . Moreover, if there exists another functorialmorphism ξ ′ : F → X such that ξ ′ ◦ h = ξ, since we also have ξ = ξ ◦ g −1 ◦ h wehaveξ ′ ◦ g ◦ f = ξ ′ ◦ h = ξ = ξ ◦ g −1 ◦ h = ξ ◦ g −1 ◦ g ◦ f = ξ ◦ fand since f is epi we getfrom which we deduce thatTherefore we obtainedξ ′ ◦ g = ξξ ′ = ξ ◦ g −1 .(H, h) = Coequ Fun (α, β) .Conversely, let now assume that h is a regular epimorphism, i.e.(H, h) = Coequ Fun (a, b). Then we haveand since g is mono we get thatg ◦ f ◦ a = h ◦ a = h ◦ b = g ◦ f ◦ bf ◦ a = f ◦ b.