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

and since d is mono we get that(ε C Z ′′) ◦ C ρ Z ′′ = Z ′′76Let us prove that ( Z ′′ , C ρ Z ′′)∈ C A and thus formula (54) will say that d is amorphism in C A. Since ∆ C is a functorial morphism and by definition of C ρ Z ′′, thelower left square serially commutes. We have(CCd) ◦ ( C C ρ Z ′′)◦ C ρ Z ′′(54)= ( C C ρ Z ′)◦ (Cd) ◦ C ρ Z ′′(54)= ( )C C ρ Z ′ ◦ C ρ Z ′ ◦ d C ρ Z ′coass= ( ∆ C Z ′) ◦ C ρ Z ′ ◦ d(54)= ( ∆ C Z ′) ◦ (Cd) ◦ C ∆ρ CZ ′′ = (CCd) ◦ ( ∆ C Z ′′) ◦ C ρ Z ′′and since CCd is a monomorphism we get( )C C ρ Z ′′ ◦ C ρ Z ′′ = ( ∆ C Z ′′) ◦ C ρ Z ′′that is that C ρ Z ′′is coassociative. Moreover we haved ◦ ( ε C Z ′′) ◦ C ρ Z ′′ε C = ( ε C Z ′) ◦ (Cd) ◦ C ρ Z ′′(54)= ( ε C Z ′) ◦ C ρ Z ′ ◦ d C ρ Z ′counit= dso that C ρ Z ′′ is also counital. Therefore ( Z ′′ , C ρ Z ′′)∈ C A and d is a morphism inC A. Now we want to prove that it is an equalizer in C A. Let ( E, C ρ E)∈ C A andf : ( E, C ρ E)→(Z ′ , C ρ Z ′)be a morphism in C A such that (K ϕ d 0 ) ◦ f = (K ϕ d 1 ) ◦ f.Then, by regarding f as a morphism in A we also have that(Ld 0 ) ◦ f = (Ld 1 ) ◦ f.Since (Z ′′ , d) = Equ A (Ld 0 , Ld 1 ) , there exists a unique morphism h : E → Z ′′ suchthatd ◦ h = f.Now we want to prove that h is a morphism in C A. In fact, let us consider thefollowing diagramEC ρ EhC ρ Z ′′Z ′′ d Z ′C ρ Z ′CE Ch CZ ′′ Cd CZ ′ .Since d ∈ C A, the right square commutes. Since f ∈ C A we haveso that we have(Cd) ◦ (Ch) ◦ C ρ E = (Cf) ◦ C ρ E = C ρ Z ′ ◦ f = C ρ Z ′ ◦ d ◦ h(Cd) ◦ C ρ Z ′′ ◦ h (54)= C ρ Z ′ ◦ d ◦ h = (Cd) ◦ (Ch) ◦ C ρ Eand since Cd is a monomorphism, we deduce thatC ρ Z ′′ ◦ h = (Ch) ◦ C ρ E