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

Proposition 4.13. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let(X, C ρ X)be a comodule for C. Then we have(X, C ρ X)= EquA(C C ρ X , ∆ C X ) .In particular if ( Q, C ρ Q)is a left C-comodule functor, then(Q, C ρ Q)= EquFun(C C ρ Q , ∆ C Q ) .47Corollary 4.14. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let( CU, C F ) be the associated adjunction. Then (C U, (C Uγ C)) is a left C-comodulefunctor and( CU, (C Uγ C)) = Equ Fun(C C Uγ C , ∆ C C U ) .Proof. By Proposition 4.12 (C U, (C Uγ C)) is a left C-comodule functor. By Proposition4.13 we get that (C U, (C Uγ C)) (= Equ Fun C C Uγ C , ∆ C C U ) .□Proposition 4.15. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let(P, ρCP)where P : A → B a right C-comodule functor. Then we have(P, ρCP)= EquFun(ρCP C, P ∆ C) .Proof. By definition we have that(ρCP C ) ◦ ρ C P = ( P ∆ C) ◦ ρ C P .Now, let ζ : Z → P C be a functorial morphism such that ( ρ C P C) ◦ ζ = ( P ∆ C) ◦ ζand consider ζ := ( P ε C) ◦ ζ : Z → P. Then we haveρ C P ◦ ζ = ρ C P ◦ ( P ε C) ◦ ζ ρC P= ( P Cε C) ◦ ( ρ C P C ) ◦ ζ == ( P Cε C) ◦ ( P ∆ C) ◦ ζ Ccomonad= ζ.Moreover, let ζ ′ : Z → P C be another functorial morphism such that ( ρ C P C) ◦ζ ′ = ζ.Thenζ ′ = ( P ε C) ◦ ρ C P ◦ ζ ′ = ( P ε C) ◦ ζ = ζso that ζ is the unique functorial morphism such that ( ρ C P C) ◦ ζ = ζ.Lemma 4.16. Let C = ( C, ∆ C , ε C) be a comonad on a category A and let ( Q, C ρ Q)bea left and ( P, ρ C P)be a right C-comodule functors where Q : Q → A and P : A → P.Let F : X → Q and G : P → B be functors. Then(1) ( QF, C ρ Q F ) is a left C-comodule functor and(2) ( GP, Gρ C P)is a right C-comodule functor.Proposition 4.17. Let C = ( C, ∆ C , ε C) be a comonad on A and let (C U, C F ) bethe adjunction associated. Then C U reflects isomorphisms.Proof. Let f : ( X, C ρ X)→(Y, C ρ Y)be a morphism in C A such that C Uf has atwo-sided inverse f −1 in A. SinceC ρ Y ◦ f = (Cf) ◦ C ρ X□