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

Since x A U is epi we get that̂k ◦ Ξ = k.Let now ̂k ′ : Q BB ̂QA → Y be another functorial morphism such that ̂k ′ ◦ Ξ = k.Then we have( ) )̂k ◦ p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)(191)= ̂k ◦ Ξ ◦ (x A U) = k ◦ (x A U) = ̂k ′ ◦ Ξ ◦ (x A U)(191)= ̂k) )′ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)) )Since we already observed from (196) that(p QB ̂QA ◦(Qpb Q◦ (Ql A U) ◦ (QP u AA U)is an epimorphism, we have ̂k ′ = ̂k.( )Hence we have proved that Q BB ̂QA , Ξ =Coequ Fun (m AA U, A A U A λ) and, in view of (202) , we get that (197)holds.Theorem 8.9. Within the assumptions and notations of Theorem 6.29, we have afunctorial isomorphism A A ∼ = A Q BB ̂QA .Proof. In Proposition 8.8 we ( have proved)the existence of a functorial morphismΞ : AU A → Q BB ̂QA such that Q BB ̂QA , Ξ = Coequ Fun (m AA U, A A Uλ A ). By Proposition3.13 and Proposition 3.14 also ( A U, A Uλ A ) = Coequ Fun ((m AA U, A A Uλ A )) , inview of uniqueness of a coequalizer up to isomorphisms, there exists a functorialisomorphismρ : A U A Q BB ̂QA = Q BB ̂QA → A U such that ρ ◦ Ξ = A Uλ A .Now since( A Uλ A ) ◦ (m AA U) = ( A Uλ A ) ◦ (A A Uλ A )and since ρ ◦ Ξ = A Uλ A , by Proposition 8.8, we deduce that) )(203) ρ ◦(p QB ̂QA ◦(Qpb Q◦ (Ql A U) = ( A Uλ A ) ◦ (m AA U) ◦ (xA A U)Now we want to prove that ρ lifts to a functorial morphism A Q BB ̂QA → A A of A-leftmodule functors. First we observe that ( A U, A Uλ A ) is an A-left module functor inview of Proposition 3.13. Also(Q BB ̂QA , A µ ̂Q)QB B A is an A-left module functor(see proof of Proposition 3.30) where A µ QB = A Uλ AA Q B : AQ B → Q B . To showthat ρ is morphism of A-left module functors we have to prove( A Uλ A ) ◦ (Aρ) = ρ ◦ ( A µ QB B ̂Q A ).We have(A ρ ◦ µ ̂Q))) )QB B A ◦(Ap QB ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U))(174)= ρ ◦(p (A )) )QB ̂QA ◦ µ Q BU B ̂QA ◦(xQ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U)◦ (QP QP x A U))) )(101)= ρ ◦(p QB ̂QA ◦(χ B U B ̂QA ◦(QP Qpb Q◦ (QP Ql A U) ◦ (QP QP x A U)169□