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

Since A U is faithful, this implies that A (Cα) = ˜C A α.Lemma 5.5. Let A = (A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be a comonadon the category A. Let Q : B → A be a functor such that ( Q, C ρ Q)is a left C-comodule functor. Assume that Φ : AC → CA is a mixed distributive law. Then(AQ, C ρ AQ)=(AQ, (ΦQ) ◦(A C ρ Q))is a left C-comodule functor.Proof. First of all we prove that C ρ AQ = (ΦQ) ◦ ( A C ρ Q)is coassociative. In fact wehave(C C ρ AQ)◦ C ρ AQdef C ρ AQ= (CΦQ) ◦(CA C ρ Q)◦ (ΦQ) ◦(A C ρ Q)Φ= (CΦQ) ◦ (ΦCQ) ◦ ( ) ( )AC C ρ Q ◦ A C ρ QC ρ Q coass= (CΦQ) ◦ (ΦCQ) ◦ ( A∆ C Q ) ◦ ( A C ρ Q)Φmdl= ( ∆ C AQ ) ◦ (ΦQ) ◦ ( A C ρ Q) def C ρ AQ=(∆ C AQ ) ◦ C ρ AQ .Now we prove the counitality condition. We have(ε C AQ ) ◦ C ρ AQdef C ρ AQ=(ε C AQ ) ◦ (ΦQ) ◦ ( A C ρ Q)Φmdl= ( Aε C Q ) ◦ ( A C ρ Q) C ρ Q couni= AQ.Proposition 5.6. 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 C-comodule functors and α : F → G be a functorialmorphism between them satisfyingC ρ G ◦ α = (Cα) ◦ (C ρ F),i.e. there exists C α : C F → C G such that C U C α = α. Then also Aα is a morphismbetween left C-comodule functors satisfying(CAα) ◦ C ρ AF = C ρ AG ◦ (Aα)i.e. there exists a functorial morphism C (Aα) : C (AF ) → C (AG) such thatC U C (Aα) = Aα. Moreover we haveC (Aα) = ÃC αwhere à is the lifted monad on the category C A, i.e. C Uà = AC U.Proof. Since F, G are left C-comodule functors, by Lemma 5.5 we know that(AF, C ρ AF)=(AF, (ΦF ) ◦(A C ρ F))and(AG, C ρ AG)=(AG, (ΦG) ◦(A C ρ G))areleft C-comodule functors. Then we have(CAα) ◦ C defρ C ρ AF( )AF = (CAα) ◦ (ΦF ) ◦ A C ρ FΦ= (ΦG) ◦ (ACα) ◦ ( )A C ρ FαmorpCcom= (ΦG) ◦ ( A C ρ G)◦ (Aα)def C ρ AG= C ρ AG ◦ (Aα)83□□