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

1104) With notations of Theorem 6.29, we havex ◦ cocan 1 ◦ (xC) (119)= x ◦ (χP ) ◦ (QP δ C )= x ◦ w l xcoequ= x ◦ w r = x ◦ ( QP ε C) x = ( Aε C) ◦ (xC) .Since xC is an epimorphism, we deduce thatand henceSimilarly, we havex ◦ cocan 1 = Aε Cx = ( Aε C) ◦ (cocan 1 ) −1 .y ◦ cocan 1 ◦ (Dy) (120)= y ◦ (P χ) ◦ (δ D P Q)= y ◦ z l ycoequ= y ◦ z r = y ◦ ( ε D P Q ) ε D = ( ε D B ) ◦ (Dy) .Since Dy is an epimorphism, we deduce thatand hence5) We have thatso thaty ◦ cocan 1 = ε D By = ( ε D B ) ◦ (cocan 1 ) −1 .δ C = (A µ Q P ) ◦ (u A QP ) ◦ δ Cu A= (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)(121) δ C = (A µ Q P ) ◦ (Aδ C ) ◦ (u A C) .Since (A µ Q P ) ◦ (Aδ C ) = cocan 1 is an isomorphism, we will prove that if u A C isa regular monomorphism, so is δ C . In fact, let (C, u A C) = Equ Fun (υ, ϖ) whereυ, ϖ : AC → T. We know that (A µ Q P ) ◦ (Aδ C ) = cocan 1 is an isomorphism withinverse (xC) ◦ ( )Qρ C P so that we haveso thati.e.υ ◦ (cocan 1 ) −1 ◦ δ C(121)= υ ◦ (cocan 1 ) −1 ◦ (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)cocan 1 iso= υ ◦ (u A C) u ACequ= ϖ ◦ (u A C)cocan 1 iso= ϖ ◦ (cocan 1 ) −1 ◦ (A µ Q P ) ◦ (Aδ C ) ◦ (u A C)(121)= ϖ ◦ (cocan 1 ) −1 ◦ δ Cυ ◦ (cocan 1 ) −1 ◦ δ C = ϖ ◦ (cocan 1 ) −1 ◦ δ Cυ ◦ (xC) ◦ ( Qρ C P)◦ δC = ϖ ◦ (xC) ◦ ( Qρ C P)◦ δC .Moreover, for every ξ : X → QP such thatυ ◦ (xC) ◦ ( Qρ C P)◦ ξ = ϖ ◦ (xC) ◦(QρCP)◦ ξ,