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

158In fact we haveTherefore we deduce that(yy) ◦ (P χP Q) = (yB) ◦ (P Qy) ◦ (P χP Q)χ= (yB) ◦ (P χB) ◦ (P QP Qy)(101)= (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .y ◦ ( ) ( )(183) P A µ Q ◦ P AµBQ ◦ (P AQy) ◦ (P xQP Q)= m B ◦ (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .Now we computeh ◦ (y B U) ◦ ( P A µ QB U ) ◦ ( P Aµ B QBU ) ◦ (P AQy B U) ◦ (P xQP Q B U)(183)= h ◦ (m BB U) ◦ (yB B U) ◦ ( P A µ Q B B U ) ◦ (P xQB B U) ◦ (P QP Qy B U)assumpth= h ◦ (B B Uλ B ) ◦ (yB B U) ◦ ( P A µ Q B B U ) ◦ (P xQB B U) ◦ (P QP Qy B U)y= h ◦ (y B U) ◦ (P Q B Uλ B ) ◦ ( P A µ Q B B U ) ◦ (P xQB B U) ◦ (P QP Qy B U)A µ Q= h ◦ (yB U) ◦ ( P A µ QB U ) ◦ (P AQ B Uλ B ) ◦ (P xQB B U) ◦ (P QP Qy B U)x= h ◦ (y B U) ◦ ( P A µ QB U ) ◦ (P AQ B Uλ B ) ◦ (P AQy B U) ◦ (P xQP Q B U)Since (P AQy B U) ◦ (P xQP Q B U) is an epimorphism, we obtainh ◦ (y B U) ◦ ( P A µ QB U ) ◦ ( P Aµ B QBU ) = h ◦ (y B U) ◦ ( P A µ QB U ) ◦ (P AQ B Uλ B )Since (P AQ B , P Ap Q ) = Coequ Fun(P AµBQ BU, P AQ B Uλ B), there exists a functorialmorphism h 1 : P AQ B → X such that(184) h 1 ◦ (P Ap Q ) = h ◦ (y B U) ◦ ( P A µ Q BU ) .Now we prove that(185) y ◦ ( P A µ Q)◦ (P xQ) ◦(z l P Q ) = y ◦ ( P A µ Q)◦ (P xQ) ◦ (z r P Q)In fact, we havey ◦ ( P A µ Q)◦ (P xQ) ◦(z l P Q ) (101)= y ◦ (P χ) ◦ ( z l P Q )(156)= y ◦ (P χ) ◦ (z r P Q) (101)= y ◦ ( P A µ Q)◦ (P xQ) ◦ (z r P Q) .Using the previous equalities, we obtainh 1 ◦ (P xQ B ) ◦ ( )z l P Q B ◦ (DP QP pQ ) = zlh 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ ( z l P Q B U )x= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ ( z l P Q B U )(184)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ ( z l P Q B U )(185)= h ◦ (y B U) ◦ ( P A µ Q BU ) ◦ (P xQ B U) ◦ (z r P Q B U)(184)= h 1 ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (z r P Q B U)x= h 1 ◦ (P xQ B ) ◦ (P QP p Q ) ◦ (z r P Q B U)