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

884.23) is a functor à : C A → C A that is a lifting of A (i.e. C Uà = AC U). We have toprove that such a à gives rise to a monad on the category C A. Let us prove that m Aand u A are C-comodule morphisms. Indeed, for every ( X, C ρ X)∈ C A, by Lemma5.5 we haveC ρ AX = (ΦX) ◦ ( A C ρ X)and alsoC ρ AAX = (ΦAX) ◦ ( A C ρ AX)= (ΦAX) ◦ (AΦX) ◦(AA C ρ X).Then we haveand(Cm A X) ◦ C ρ AAX = (Cm A X) ◦ (ΦAX) ◦ (AΦX) ◦ ( AA C ρ X)Φm.d.l.= (ΦX) ◦ (m A CX) ◦ ( AA C ρ X)m A= (ΦX) ◦ ( A C ρ X)◦ (mA X) = C ρ AX ◦ (m A X)C ρ AX ◦ (u A X) = (ΦX) ◦ ( A C ρ X)◦ (uA X)u A= (ΦX) ◦ (u A CX) ◦ C ρ XΦm.d.l.= (Cu A X) ◦ C ρ XThus m A and u A lift to functorial morphisms m e Aand u e Auniquely defined byC Um e A= m A C UandC Uu e A= u A C U.We compute( C) (Um A e ◦ C AliftUm A eÃ) = ( m C A U ) ( )◦ m C AliftA Uà = ( m C A U ) ◦ ( m A A C U )Amonad= ( m C A U ) ◦ ( Am C A U ) Alift= (C ) ( )Um A e ◦ A C AliftUm A e = (C ) ( )Um A e ◦ C UÃm A eand since C U is faithful , we deduce( )m A e ◦ m A eÃ= m A e ◦We computeand since C U is faithful, we obtain(Ãm e A).( C) (Um A e ◦ C AliftUu A eÃ) = ( m C A U ) ( )◦ u C A UÃAlift= ( m A C U ) ◦ ( u A A C U ) Amonad= A C U Alift= C UÃm e A◦( )u A eÃ= Ã.Similarly we compute( CUm e A)◦ (C UÃu e A)Alift= ( m A C U ) ◦ ( A C Uu e A) Alift= ( m A C U ) ◦ ( Au A C U )Amonad= A C U Alift= C UÃand since C U is faithful, we obtain)m A e ◦(ÃuA e = Ã.