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

82Proof. First of all we prove that A µ CQ = ( C A µ Q)◦ (ΦQ) is associative. In fact wehaveA µ CQ ◦ ( )A A defµ A µ CQ( ) ( )CQ = C A µ Q ◦ (ΦQ) ◦ AC A µ Q ◦ (AΦQ)Φ= ( C A µ Q)◦(CA A µ Q)◦ (ΦAQ) ◦ (AΦQ)A µ Q ass= ( C A µ Q)◦ (CmA Q) ◦ (ΦAQ) ◦ (AΦQ)Φmdl= ( C A µ Q)◦ (ΦQ) ◦ (mA CQ) defA µ CQ= A µ CQ ◦ (m A CQ) .Now we prove the unitality condition. We haveA µ CQ ◦ (u A CQ) defA µ CQ( )= C A µ Q ◦ (ΦQ) ◦ (uA CQ)Φmdl= ( C A µ Q)◦ (CuA Q) A µ CQ uni= CQ.Proposition 5.4. Let A = (A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be acomonad on the category A. Assume that Φ : AC → CA is a mixed distributive lawbetween them. Let F, G be left A-module functors and α : F → G be a functorialmorphism between them satisfyingA µ G ◦ (Aα) = α ◦ (A µ F),i.e. there exists a functorial morphism A α : A F → A G such that A U A α = α. Thenalso Cα is a functorial morphism between left A-module functors satisfyingA µ CG ◦ (ACα) = (Cα) ◦ A µ CFi.e. there exists a functorial morphism A (Cα) : A (CF ) → A (CG) such thatAU A (Cα) = Cα. Moreover we haveA (Cα) = ˜C A αwhere ˜C is the lifted comonad on the category A A, i.e. A U ˜C = C A U.Proof. By Lemma 3.29 there exists A α : A F → A G such that A U A α = α. Moreover,by Lemma 5.3, we know that ( CF, A µ CF)=(CF,(C A µ F)◦ (ΦF ))and(CG, A µ CG)=(CG,(C A µ G)◦ (ΦG))are left A-module functors. Then we have□A µ CG ◦ (ACα) defA µ=CG( )C A µ G ◦ (ΦG) ◦ (ACα)Φ= ( C A µ G)◦ (CAα) ◦ (ΦG)αmorpAmod= (Cα) ◦ ( C A µ F)◦ (ΦG)def A µ CF= (Cα) ◦ A µ CFi.e. Cα is a functorial morphism between left A-module functors. Then there existsa functorial morphism A (Cα) : A (CF ) → A (CG) such that A U A (Cα) = Cα. Sincewe also haveAU ˜C A α = C A U A α = Cαwe deduce thatAU A (Cα) = A U ˜C A α.