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

162so that we obtain:(190)We computeB µb Q◦ (Bl) ◦ (yP A) ◦ (P QP x) = l ◦ (P x) ◦ (P χP ) .β ◦ ( B µb QA AQ B ) ◦(175)= β ◦ (pb Q AQ B ) ◦= β ◦ (pb Q AQ B ) ◦(B µb Q AU A Q B))(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (yP xp Q ) =(B µb Q AU A Q B)◦ (Bl A U A Q B ) ◦ (yP xp Q )◦ (Bl A U A Q B ) ◦ (yP AQ B ) ◦ (P QP xQ B )◦ (P QP QP p Q )(190)= β ◦ (pb Q AQ B ) ◦ (l A U A Q B ) ◦ (P x A U A Q B ) ◦ (P χP A U A Q B )◦ (P QP QP p Q )χ= β ◦ (pb Q AQ B ) ◦ (l A U A Q B ) ◦ (P x A U A Q B ) ◦ (P QP p Q ) ◦ (P χP Q B U)x= β ◦ (pb Q AQ B ) ◦ (l A U A Q B ) ◦ (P Ap Q ) ◦ (P xQ B U) ◦ (P χP Q B U)(189)= ( B Uλ B ) ◦ (y B U) ◦ (P χ B U) ◦ (P χP Q B U)(98)= ( B Uλ B ) ◦ (y B U) ◦ (P χ B U) ◦ (P QP χ B U)(109)= ( B Uλ B ) ◦ (m BB U) ◦ (yy B U) ◦ (P QP χ B U)BUλ B coequ= ( B Uλ B ) ◦ (B B Uλ B ) ◦ (By B U) ◦ (yP Q B U) ◦ (P QP χ B U)y= ( B Uλ B ) ◦ (B B Uλ B ) ◦ (By B U) ◦ (BP χ B U) ◦ (yP QP Q B U))(189)= ( B Uλ B ) ◦ (Bβ) ◦(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (BP Ap Q ) ◦ (BP xQ B U)◦ (yP QP Q B U))= ( B Uλ B ) ◦ (Bβ) ◦(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (yP xp Q ))Since(Bpb Q AQ B ◦ (Bl A U A Q B ) ◦ (yP xp Q ) is an epimorphism, we get that ( B Uλ B ) ◦(Bβ) = β ◦ ( B µb QA A Q B). Therefore β is a morphism of B-left module functorsand ̂Qhence, in view of Proposition 3.25, it gives rise to a functorial isomorphismB AA Q B∼ = B B.□Now, we prove the second isomorphism.Within the assumptions and notations of Theorem 6.29, we will construct a functorialisomorphism A Q BB ̂QA ∼ = A A.Lemma 8.7. Within the assumptions and notations of Theorem 6.29, there exists afunctorial morphism Ξ : A A U → Q BB ̂QA uniquely determined by) )(191)(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U) = Ξ ◦ (x A U)such that(192) Ξ ◦ (m AA U) = Ξ ◦ (A A U A λ) .