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

157Now we want to prove that(̂QAA Q B , α)= Coequ Fun (m BB U, B B Uλ B ).Let us show the fork property for α, that is(180) α ◦ (m BB U) = α ◦ (B B Uλ B ) .We have(179)=α ◦ (B B Uλ B ) ◦ (yy B U) = α ◦ (B B Uλ B ) ◦ (yB B U) ◦ (P Qy B U)y= α ◦ (y B U) ◦ (P Q B Uλ B ) ◦ (P Qy B U)) (pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P Q B Uλ B ) ◦ (P Qy B U))=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (P Q B Uλ B ) ◦ (P Qy B U)( )defp Q= pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ ( P µ B QBU ) ◦ (P Qy B U))(107)=(pb Q AQ B ◦ (lQ B ) ◦ (P u A Q B ) ◦ (P p Q ) ◦ (P χ B U))u=A(pb Q AQ B ◦ (lQ B ) ◦ (P Ap Q ) ◦ (P u A Q B U) ◦ (P χ B U)(179)= α ◦ (y B U) ◦ (P χ B U) (109)= α ◦ (m BB U) ◦ (yy B U)and, since yy B U is an epimorphism, we conclude. Now, let us consider a functorialmorphism h : B B U → X such that h ◦ (m BB U) = h ◦ (B B Uλ B ) . We have to showthat there exists a unique functorial morphism ĥ : ̂Q AA Q B → X such thatĥ ◦ α = h.First we will show that there exists a functorial morphism ĥ such that ĥ and h fulfill(177) i.e.)ĥ ◦(pb Q AQ B ◦ (l A U A Q B ) ◦ (P Ap Q ) = h ◦ (y B U) ◦ ( P A µ QB U ) .To do this, we need a series of equalities. First of all, let us show that(181) y ◦ ( P A µ Q)◦(P AµBQ)◦ (P AQy) ◦ (P xQP Q) = mB ◦ (yy) ◦ (P χP Q) .In fact, we haveNow let us prove thaty ◦ ( P A µ Q)◦(P AµBQ)◦ (P AQy) ◦ (P xQP Q)(107)= y ◦ ( P A µ Q)◦ (P Aχ) ◦ (P xQP Q)x= y ◦ ( )P A (101)µ Q ◦ (P xQ) ◦ (P QP χ) = y ◦ (P χ) ◦ (P QP χ)(98)= y ◦ (P χ) ◦ (P χP Q) (109)= m B ◦ (yy) ◦ (P χP Q) .(182) (yy) ◦ (P χP Q) = (yB) ◦ ( P A µ Q B ) ◦ (P xQB) ◦ (P QP Qy) .