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

since (C, u A C) = Equ Fun (υ, ϖ) , there exists a unique functorial morphism ξ : X →C such that(u A C) ◦ ξ = (xC) ◦ ( Qρ C P)◦ ξ.Then, by composing to the left with the isomorphism cocan 1 := (A µ Q P ) ◦ (Aδ C ) weget ( Aµ Q P ) ◦ (Aδ C ) ◦ (u A C) ◦ ξ = (A µ Q P ) ◦ (Aδ C ) ◦ (xC) ◦ ( )Qρ C P ◦ ξi.e. by (121) we have(122) δ C ◦ ξ = ξ.Then ξ : X → C is the unique morphism satisfying condition (122) so that(C, δ C ) = Equ Fun(υ ◦ (xC) ◦(QρCP), ϖ ◦ (xC) ◦(QρCP))i.e. also δ C is a regular monomorphism.6) We have thatso thatδ D = ( P µ B Q)◦ (P QuB ) ◦ δ Du B= ( P µ B Q)◦ (δD B) ◦ (Du B )(123) δ D = ( P µ B Q)◦ (δD B) ◦ (Du B ) .Since ( P µ Q) B ◦ (δD B) = cocan 1 is an isomorphism, we will prove that if Du B isa regular monomorphism, so is δ D . In fact, let (D, Du B ) = Equ Fun (ζ, θ) whereζ, θ : DB → L. We know that ( P µ Q) B ◦ (δD B) = cocan 1 is an isomorphism withinverse (Dy) ◦ (D ρ P Q ) so that we haveζ ◦ (cocan 1 ) −1 ◦ δ D(123)= ζ ◦ (cocan 1 ) −1 ◦ ( P µ B Q)◦ (δD B) ◦ (Du B )cocan 1 iso= ζ ◦ (Du B ) Du Bequ= θ ◦ (Du B )cocan 1 iso= θ ◦ (cocan 1 ) −1 ◦ ( P µ B Q)◦ (δD B) ◦ (Du B )(123)= θ ◦ (cocan 1 ) −1 ◦ δ Dso thatζ ◦ (cocan 1 ) −1 ◦ δ D = θ ◦ (cocan 1 ) −1 ◦ δ Di.e.ζ ◦ (Dy) ◦ (D ρ P Q ) ◦ δ D = θ ◦ (Dy) ◦ (D ρ P Q ) ◦ δ D .Moreover, for every ν : Y → P Q such thatζ ◦ (Dy) ◦ (D ρ P Q ) ◦ ν = θ ◦ (Dy) ◦ (D ρ P Q ) ◦ ν,since (D, Du B ) = Equ Fun (ζ, θ) , there exists a unique functorial morphism ν : Y →D such that(Du B ) ◦ ν = (Dy) ◦ (D ρ P Q ) ◦ ν.Then, by composing to the left with the isomorphism cocan 1 := we get( )P µBQ ◦ (δD B) ◦ (Du B ) ◦ ν = ( )P µ B Q ◦ (δD B) ◦ (Dy) ◦ (D ρ P Q ) ◦ νi.e. by (123) we have(124) δ D ◦ ν = ν.111