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

and hence there exists a ζ : C UZ → LY = C UK ϕ Y such that((57)CUh ) = (LηY ) ◦ ζ = (C UK ϕ ηY ) ◦ ζ.Let us prove that ζ gives rise to a morphism in C A. Since h is a morphism in C Awe have that(58) (ϕLRLY ) ◦ (LηRLY ) ◦ (C Uh ) = ( C C Uh ) ◦ (C Uγ C Z ) .Let us computeso that(CLηY ) ◦ (Cζ) ◦ (C Uγ C Z ) (57)= ( C C Uh ) ◦ (C Uγ C Z )(58)= (ϕLRLY ) ◦ (LηRLY ) ◦ (C Uh )(56)= (ϕLRLY ) ◦ (LRLηY ) ◦ (C Uh )(57)= (ϕLRLY ) ◦ (LRLηY ) ◦ (LηY ) ◦ ζϕ= (CLηY ) ◦ (ϕLY ) ◦ (LηY ) ◦ ζ(CLηY ) ◦ (Cζ) ◦ (C Uγ C Z ) = (CLηY ) ◦ (ϕLY ) ◦ (LηY ) ◦ ζ.Since (CɛLY ) ◦ (CLηY ) = CLRLY , CLηY is mono and hence we get(Cζ) ◦ (C Uγ C Z ) = (ϕLY ) ◦ (LηY ) ◦ ζi.e. ζ : C UZ → LY = C UK ϕ Y is a morphism of C-comodules.Proposition 4.66. Let (L, R) be an adjunction where L : B → A and R : A → B,let C = ( C, ∆ C , ε C) be a comonad on a category A and let ϕ : LR = (LR, LηR, ɛ) →C = ( C, ∆ C , ε C) be a comonad morphism. Let K ϕ = Υ (ϕ) = (L, (ϕL) ◦ (Lη)) andC UK ϕ (f) = L (f) for every morphism f in B. If ϕX is a monomorphism forevery X ∈ A, the assignment K Y,RLY ′ : Hom B (Y, RLY ′ ) → HomC A (K ϕ Y, K ϕ RLY ′ )defined by settingK Y,RLY ′ (f) = K ϕ (f)is an isomorphism whose inverse is defined byK −1Y,RLY ′ (h) = (RɛLY ′ ) ◦ ( R C Uh ) ◦ (ηY ) .Proof. Let f ∈ Hom B (Y, RLY ′ ). We compute( )˜K −1Y,RLY˜KY,RLY ′ ′ (f) = (RɛLY ′ ) ◦ ( R C UK ϕ f ) ◦ (ηY ) = (RɛLY ′ ) ◦ (RLf) ◦ (ηY )η= (RɛLY ′ ) ◦ (ηRLY ′ ) ◦ f = f.Let h ∈ HomC A (K ϕ Y, K ϕ RLY ′ ). This means that(ϕLRLY ′ ) ◦ (LηRLY ′ ) ◦ (C Uh ) = ( C C Uh ) ◦ (ϕLY ) ◦ (LηY )ϕ= (ϕLRLY ′ ) ◦ ( LR C Uh ) ◦ (LηY ) .Since ϕX is a monomorphism for every X ∈ A, we deduce that(59) (LηRLY ′ ) ◦ (C Uh ) = ( LR C Uh ) ◦ (LηY ) .79□