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

46Definition 4.9. A comodule for a comonad C = ( C, ∆ C , ε C) on a category A is apair ( X, C ρ X)where X ∈ A and C ρ X : X → CX is a morphism in A such that(C C ρ X)◦ C ρ X = ( ∆ C X ) ◦ C ρ X and X = ( ε C X ) ◦ C ρ X .A morphism between two C-comodules ( X, C ρ X)and(X ′ , C ρ X ′)is a morphism f :X → X ′ in A such thatC ρ X ◦ f = (Cf) ◦ C ρ X .We denote by C A the category of C-comodule and their morphisms.Definition 4.10. Corresponding to a comonad C = ( C, ∆ C , ε C) on A, there is anadjunction (C U, C F ) where C U is the forgetful functor and C F is the free functorC U : C A → A C F : A → C A(X, C ρ X)→ X X →(CX, ∆ C X )f → f f → CfNote that C U C F = C. The counit of the adjunction is given by the counit ε C of thecomonad Cε C : C = C U C F → A.The unit γ C : C A → C F C U of this adjunction is defined by settingC U ( γ C ( X, C ρ X))= C ρ X for every ( X, C ρ X)∈ C A.Therefore we have(εC C U ) ◦ (C Uγ C) = C U and( CF ε C) ◦ ( γ C C F ) = C F .Proposition 4.1<strong>1.</strong> Let C = ( C, ∆ C , ε C) be a comonad on a category A and letZ, W ∈ C A. Then Z = W if and only if C U (Z) = C U (W ) and C U ( γ C Z ) =C U ( γ C W ) . In particular, if F, G : X → C A are functors, we haveF = G if and only if C UF = C UG and C U ( γ C F ) = C U ( γ C G ) .Proposition 4.1<strong>2.</strong> Let C = ( (C, ∆ C , ε C) be a comonad on a category A.CU, (C Uγ C)) is a left C-comodule functor.ThenProof. We have to prove these two equalities(C C Uγ C) ◦ (C Uγ C) = ( ∆ C C U ) ◦ (C Uγ C)(εC C U ) ◦ (C Uγ C) = C ULet us consider ( X, C ρ X)∈ C A, we have to show that(C C Uγ C) ( X, C ρ X)◦( CUγ C) ( X, C ρ X)=(∆C C U ) ( X, C ρ X)◦( CUγ C) ( X, C ρ X)and thati.e.(εC C U ) ( X, C ρ X)◦( CUγ C) ( X, C ρ X)= C U ( X, C ρ X)(C C ρ X)◦ C ρ X = ( ∆ C X ) ◦ C ρ X and (ε C X) ◦ C ρ X = Xwhich both hold in view of the definition of C-comodule.□