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

152Thus we obtainσ B ◦ ( ) (P µ ) B Q ◦ (jB) ◦ DσB= σ B ◦ ( ε D P Q )and since (Z, π Z ) = Coequ Fun((P µBQ)◦ (jB) ◦(DσB ) , ε D P Q ) , by the universalproperty of coequalizers, there exists a unique functorial morphism λ : Z → B suchthat(171) λ ◦ π Z = σ B .Since we already proved in 1) that σ B is a regular epimorphism, in particular wecan write ( B, σ B) = Coequ Fun (ξ, ζ). Let us computeso thatπ Z ◦ ξ ◦ ( ε D P Q ) ε D = π Z ◦ ( ε D P Q ) ◦ (Dξ)defπ Z= πZ ◦ ( P µ B Q)◦ (jB) ◦(DσB ) ◦ (Dξ)σ B coequ= π Z ◦ ( (P µ Q) ) B ◦ (jB) ◦ DσB◦ (Dζ)defπ=ZπZ ◦ ( ε D P Q ) ◦ (Dζ) εD = π Z ◦ ζ ◦ ( ε D P Q )π Z ◦ ξ ◦ ( ε D P Q ) = π Z ◦ ζ ◦ ( ε D P Q ) .Since σ B is a regular epimorphism, by Theorem 6.6, so is Bε D . Since by assumptionB reflects coequalizers, also ε D is an epimorphism so that we get(172) π Z ◦ ξ = π Z ◦ ζ.Since ( B, σ B) = Coequ Fun (ξ, ζ) , by the universal property of coequalizers thereexists a unique functorial morphism λ ′ : B → Z such that(173) λ ′ ◦ σ B = π Z .We prove that λ ′ is the two-sided inverse of λ. In factλ ′ ◦ λ ◦ π Z(171)= λ ′ ◦ σ B (173)= π Z .Since π Z is an epimorphism we deduce thatSimilarlyλ ′ ◦ λ = Id Z .λ ◦ λ ′ ◦ σ B (173)= λ ◦ π Z(171)= σ Band, since also σ B is an epimorphism, we deduce thatWe now want to prove thati.e.We computeλ ◦ λ ′ = Id B .ν B = λ ◦ ν Zλ ′ ◦ ν B = ν Z .λ ′ ◦ ν B ◦ y ′ (109)= λ ′ ◦ m B ◦ ( σ B σ B) ◦ (P iQ) ◦ (qQ)