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

166= Ξ ◦ (A A U A λ) ◦ (xx A U) ◦ ( QP QP ε C AU )Since (xx A U) ◦ ( QP QP ε C AU ) is epi, we deduce (192) . Note that, in particular, wehave) )(196)(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (χP A U) ◦ ( QP QP ε C AU )) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ ( QP QP ε C AU )) )and since the second term is epi, also the first is epi, and hence(p QB ̂QA ◦(Qpb Q◦(Ql A U) ◦ (QP u AA U) is an epimorphism.Proposition 8.8. Within the assumptions and notations of Theorem 6.29, thereexists ( a functorial ) morphism Ξ : A A U → Q BB ̂QA such thatQ BB ̂QA , Ξ = Coequ Fun (m AA U, A A U A λ) . Moreover for every morphism k suchthatk ◦ (m AA U) = k ◦ (A A U A λ)if ̂k : Q BB ̂QA → Y is the unique morphism such that ̂k ◦ Ξ = k, we have that( ) )(197) ̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) = k ◦ (m AA U) ◦ (xA A U) .Proof. By Lemma 8.7 we already know thatΞ ◦ (m AA U) = Ξ ◦ (A A U A λ) .Now we want to prove that( )Q BB ̂QA , Ξ = Coequ Fun (m AA U, A A U A λ).Let k : A A U → Y be a functorial morphism such that(198) k ◦ (m AA U) = k ◦ (A A U A λ)We have to show that there exists a functorial morphism ̂k : Q BB ̂QA → Y such that̂k ◦ Ξ = k.First we will show that there exists a functorial morphism ̂k such that ̂k and k fulfil(197) i.e.k ◦ (m AA U) ◦ (xA A U) = ̂k) )◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) .We proceed in several steps. First of all we computek ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ ( Qz l P A U )(102)= k ◦ (x A U) ◦ (χP A U) ◦ ( Qz l P A U )= k ◦ (x A U) ◦ (χP A U) ◦ (QP χP A U) ◦ (Qδ D P QP A U)(98)= k ◦ (x A U) ◦ (χP A U) ◦ (χP QP A U) ◦ (Qδ D P QP A U)(105)= k ◦ (x A U) ◦ (χP A U) ◦ ( Qε D P QP A U )(102)= k ◦ (m AA U) ◦ (xA A U) ◦ (QP x A U) ◦ ( Qε D P QP A U )□