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

z r= h 1 ◦ (P xQ B ) ◦ (z r P Q B ) ◦ (DP QP p Q ) .Since DP QP p Q is an epimorphism, we deduce thath 1 ◦ (P xQ B ) ◦ ( )z l P Q B = h1 ◦ (P xQ B ) ◦ (z r P Q B ) .( )Since ̂QQB lQ B = Coequ Fun ((P xQ B )◦ ( )z l P Q B , (P xQB )◦(z r P Q B )), there existsa functorial morphism h 2 : ̂QQ B → X such that(186) h 2 ◦ (lQ B ) = h 1 .We computeso that we gety ◦ ( P A µ Q)◦ (P xQ) ◦ (P χP Q)(101)= y ◦ (P χ) ◦ (P χP Q)(98)= y ◦ (P χ) ◦ (P QP χ)(101)= y ◦ ( P A µ Q)◦ (P xQ) ◦ (P QP χ)x= y ◦ ( P A µ Q)◦ (P Aχ) ◦ (P xQP Q)(187) y ◦ ( P A µ Q)◦ (P xQ) ◦ (P χP Q) = y ◦(P A µ Q)◦ (P Aχ) ◦ (P xQP Q) .We also have(lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)(101)= (lQ B ) ◦ (P Ap Q ) ◦ ( P A A µ Q BU ) ◦ (P AxQ B U) ◦ (P xQP Q B U)(174)= (lQ B ) ◦ ( )P A A µ QB ◦ (P AApQ ) ◦ (P AxQ B U) ◦ (P xQP Q B U)( )l= ̂QA µ QB ◦ (lAQ B ) ◦ (P AAp Q ) ◦ (P AxQ B U) ◦ (P xQP Q B U)( )x= ̂QA µ QB ◦ (lAQ B ) ◦ (P AxQ B ) ◦ (P AQP p Q ) ◦ (P xQP Q B U)( )̂QA µ QB ◦ (lAQ B ) ◦ (P AxQ B ) ◦ (P xQP QQ B ) ◦ (P QP QP p Q )( )= ̂QA µ QB ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )x=so that we get(188)=(lQ B ) ◦ (P Ap Q ) ◦ (P Aχ B U) ◦ (P xQP Q B U)( )̂QA µ QB ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )and hence we obtain)h 2 ◦(µ A Q bQ B ◦ (lAQ B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )(150)= h 2 ◦ (lQ B ) ◦ (P m A Q B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )(186)= h 1 ◦ (P m A Q B ) ◦ (P xxQ B ) ◦ (P QP QP p Q )(102)= h 1 ◦ (P xQ B ) ◦ (P χP Q B ) ◦ (P QP QP p Q )159