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

140We calculateA µ Q ◦ ( σ A Q ) ◦ (QlQ) ◦ (QP xQ) (161)= A µ Q ◦ (m A Q) ◦ (xAQ) ◦ (QP xQ)= A µ Q ◦ (m A Q) ◦ (xxQ) (102)= A µ Q ◦ (xQ) ◦ (χP Q)(101)= χ ◦ (χP Q) (128)= χ ◦ (QP χ) (107)= µ B Q ◦ (Qy) ◦ (QP χ)(109)= µ B Q ◦ (Qm B ) ◦ (Qyy) = µ B Q ◦ (Qm B ) ◦ (QBy) ◦ (QyP Q)(162)= µ B Q ◦ ( Qσ B) ◦ (Qν ′ 0Q) ◦ (QyP Q)(149)= µ B Q ◦ ( Qσ B) ◦ (QlQ) ◦ (QP xQ) .Since (QlQ)◦(QP xQ) is an epimorphism, we deduce that A µ Q ◦ ( σ A Q ) = µ B Q ◦( Qσ B) .We compute( (µ A Q b ◦ ̂QσA)◦ ν 0Q ′ ̂Q) (◦ yP Q ̂Q)◦ (P QP Ql) ◦ (P QP QP x)( ((149)= µ A Q b ◦ ̂QσA)◦ lQ ̂Q) (◦ P xQ ̂Q)◦ (P QP Ql) ◦ (P QP QP x)l= µ A Q b ◦ (lA) ◦ ( P Aσ A) (◦ P xQ ̂Q)◦ (P QP Ql) ◦ (P QP QP x)(150)= l ◦ (P m A ) ◦ ( P Aσ A) (◦ P xQ ̂Q)◦ (P QP Ql) ◦ (P QP QP x)x= l ◦ (P m A ) ◦ (P xA) ◦ ( P QP σ A) ◦ (P QP Ql) ◦ (P QP QP x)(161)= l ◦ (P m A ) ◦ (P xA) ◦ (P QP m A ) ◦ (P QP xA) ◦ (P QP QP x)= l ◦ (P m A ) ◦ (P xA) ◦ (P QP m A ) ◦ (P QP xx)(102)= l ◦ (P m A ) ◦ (P xA) ◦ (P QP x) ◦ (P QP χP )(151)= l ◦ (P m A ) ◦ (P xx) ◦ (P QP χP )(102)= l ◦ (P x) ◦ (P χP ) ◦ (P QP χP )(128)= l ◦ (P x) ◦ (P χP ) ◦ (P χP QP )(149)= ν ′ 0 ◦ (yP ) ◦ (P χP ) ◦ (P χP QP )(109)= ν ′ 0 ◦ (m B P ) ◦ (yyP ) ◦ (P χP QP )= B µb Q◦ (Bν ′ 0) ◦ (yBP ) ◦ (P QyP ) ◦ (P χP QP )y= B µb Q◦(149)= B µb Q◦χ= B µb Q◦(109)= B µb Q◦( )y ̂Q ◦ (P Qν 0) ′ ◦ (P QyP ) ◦ (P χP QP )( )y ̂Q ◦ (P Ql) ◦ (P QP x) ◦ (P χP QP )( ) ( )y ̂Q ◦ P χ ̂Q ◦ (P QP Ql) ◦ (P QP QP x)( ) (m B ̂Q ◦ yy ̂Q)◦ (P QP Ql) ◦ (P QP QP x)