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

136and hence we get(160) x ◦ (χP ) ◦ ( QP w l) = x ◦ (χP ) ◦ (QP w r ) .We observe thatν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ ( BP w l) ◦ (yP QP C)= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (yP QP ) ◦ ( P QP w l)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ ( P QP w l)(109)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ ( P QP w l)(149)= l ◦ (P x) ◦ (P χP ) ◦ ( P QP w l)(160)= l ◦ (P x) ◦ (P χP ) ◦ (P QP w r )(149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P QP w r )(109)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P QP w r )= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (yP QP ) ◦ (P QP w r )= ν ′ 0 ◦ (m B P ) ◦ (ByP ) ◦ (BP w r ) ◦ (yP QP C)since yP QP C is an epimorphism, we obtain thatν 0 ′ ◦ (m B P ) ◦ (ByP ) ◦ ( BP w l) = ν 0 ′ ◦ (m B P ) ◦ (ByP ) ◦ (BP w r ) .(Since B preserves coequalizers, we have that B ̂Q,)Bν 0′ = Coequ Fun ((ByP ) ◦( ) BP wl, (ByP ) ◦ (BP w r )) so that there exists a unique functorial morphism B µb Q:( )̂Q, B µb QB ̂Q → ̂Q which satisfies (151).Now we want to show thatB-module functor. First let us prove that B µb Qis associative that is) ( )(B B µb Qm B ̂Q .B µb Q◦= B µb Q◦is a leftWe have)B µb Q◦(B B µb Q◦ (BBν ′ 0) (151)= B µb Q◦ (Bν ′ 0) ◦ (Bm B P )(151)= ν ′ 0 ◦ (m B P ) ◦ (Bm B P ) m Bass(151)= B µb Q◦ (Bν ′ 0) ◦ (m B BP ) m B= B µb Q◦= ν 0 ′ ◦ (m B P ) ◦ (m B BP )( )m B ̂Q ◦ (BBν 0) ′ .Since BBν 0 ′ is an epimorphism, we get that B µb Qis associative. Let us prove thatB µb Qis unital that is( )B µb Q◦ u B ̂Q = ̂Q.We calculate( )B µb Q◦ u B ̂Q◦ ν ′ 0 = B µb Q◦ (Bν ′ 0) ◦ (u B BP ) (151)= ν ′ 0 ◦ (m B P ) ◦ (u B BP ) = ν ′ 0.