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

164(194)=) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP χP A U) ◦ (QP δ C QP A U)(99)=◦ (QP Cδ C A U)) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ ( QP ε C QP A U )◦ (QP Cδ C A U)) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP x A U) ◦ (QP δ C A U) ◦ ( QP Cε C AU )) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ ( QP ε C AU ) ◦ ( QP Cε C AU )) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (w r AU) ◦ ( QP Cε C AU )ε C =(103)=since QP Cε C AU is epi we deduce that) ) (p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ ( w l AU )) )=(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (w r AU) .(Since (A A U, x A U) = Coequ Fun wl A U, w r AU ) , there exists a functorial morphismΞ : A A U → Q BB ̂QA such that) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) = Ξ ◦ (x A U) .Let us prove thatBy the definition of pb Qwe have that( )µ A Q bAUso thatand hencepb Q◦Ξ ◦ (m AA U) = Ξ ◦ (A A Uλ A ) .◦ (lA A U) = pb Q◦(̂QA U A λpb Q◦ (l A U) ◦ (P m AA U) (150)= pb Q◦p bQ coequ= pb Q◦(̂QA U A λ)◦ (lA A U)( )µ A Q bAU◦ (lA A U))◦ (lA A U)(195) pb Q◦ (l A U) ◦ (P m AA U) = pb Q◦(̂QA U A λWe calculate(191)=Ξ ◦ (m AA U) ◦ (xx A U) ◦ ( QP QP ε C AU ))◦ (lA A U) .(102)= Ξ ◦ (x A U) ◦ (χP A U) ◦ ( QP QP ε C AU )) )(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) ◦ (χP A U)◦ ( QP QP ε C AU )(χ ̂Q)A U ◦ (QP Ql A U) ◦ (QP QP u AA U)) )χ=(p QB ̂QA ◦(Qpb Q◦