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

84i.e. Aα is a functorial morphism between left C-comodule functors. Then thereexists a functorial morphism C (Aα) : C (AF ) → C (AG) such that C U C (Aα) = Aα.Since we also haveC UÃC α = A C U C α = Aαwe deduce thatC U C (Aα) = C UÃC α.Since C U is faithful, this implies that C (Aα) = ÃC α.□5.<strong>2.</strong> Liftings of monads and comonads.Theorem 5.7 ([Be, Proposition p. 122] and [Mesa, Theorem <strong>2.</strong>1]). Let A =(A, m A , u A ) be a monad and let C = ( C, ∆ C , ε C) be a comonad on a category A.There is a bijection between the following collections of data:C liftings ( of C to a comonad ˜C on the category A A, that is comonads˜C = ˜C, ∆C e, ε e )Con A A such thatAU ˜C = C A U, AU∆ eC = ∆ C AU and AUε eC = ε C AUD mixed distributive laws Φ : AC → CAM liftings of A to a)monad à on the category C A, that is monadsà =(Ã, mA e, u A e on C A such thatC Uà = AC U,C Um e A= m A C Ugiven by( )a : C → D where a ˜C =andC Uu e A= u A C U( )BUλ B ˜CA F ◦ ( B U B F Cu A )b : D → C where A Ub (Φ) = C A U and A Uλ A b (Φ) = (C A Uλ A ) ◦ Φ i.e.b (Φ) (( )) ( ( ) )X, A µ X = CX, C A µ X ◦ (ΦX) and b (Φ) (f) = C (f))d : M → D where d(à = (C U C F Aε C) ( )◦C Uγ C à C Fm : D → M where C Um (Φ) = A C U and C Uγ C m (Φ) = Φ ◦ ( A C Uγ C) i.e.m (Φ) (( X, C ρ X))=(AX, (ΦX) ◦(A C ρ X))and m (Φ) (f) = A (f) .Proof. In order to prove the bijection between C and D, we apply Proposition 3.24,to the case (A, m A , u A ) = (B, m B , u B ) monad on A and Q = C. In particular wewill prove that the bijection a : F → M, b : M → F of Proposition 3.24 induces abijection between C and D.Let ˜C) ( )∈ C. We have to prove that Φ = a(˜C = AUλ A ˜CA F ◦ ( A U A F Cu A ) ∈ D.We have(CΦ) ◦ (ΦC) ◦ ( A∆ C) =()()C A Uλ A ˜CA F ◦ (C A U A F Cu A ) ◦ AUλ A ˜CA F C ◦ ( A U A F Cu A C) ◦ ( A∆ C)(= AU ˜Cλ)A ˜CA F ◦(AU ˜C) ()A F Cu A ◦ AUλ A ˜CA F C ◦ ( A U A F Cu A CX) ◦ ( A∆ C)