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

<strong>2.</strong>15 and hence we get that C Ûɛ is a functorial isomorphism. Since,by Proposition4.17, C U reflects isomorphism we deduce that ̂ɛ is a functorial isomorphism. □Corollary 4.54. Let (L, R) be an adjunction where L : B → A and R : A → B.Let α = Θ (Id LR ) = ηR and assume that, for every ( X, LR ρ X)∈ LR A, there existsEqu B(ηRX, R LR ρ X). Then we can consider the functor K = Υ (IdLR ) : B → LR Aand its right adjoint D : LR A → B. D is full and faithful if and only if L preservesthe equalizer(D, d) = Equ Fun(ηR LR U, R LR Uγ LR) .Proof. We can apply Theorem 4.53 with ”ϕ” = Id LR .Theorem 4.55 ([GT, Theorem <strong>2.</strong>7]). Let (L, R) be an adjunction where L : B → Aand 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 α = Θ (ϕ) = (Rϕ) ◦(ηR) and assume that, for every ( ) ( )X, C ρ X ∈ C A, there exists Equ B αX, R C ρ X .Then we can consider the functor K ϕ = Υ (ϕ) : B → C A and its right adjointD ϕ : C A → B. The functor K ϕ is an equivalence of categories if and only if1) L preserves the equalizer((D ϕ , d ϕ ) = Equ Fun α C U, R C Uγ C)2) L reflects isomorphisms and3) ϕ : LR → C is a comonad isomorphism.Proof. If K ϕ is an equivalence then, by Lemma <strong>2.</strong>33, D ϕ is an equivalence of categoriesso that, by Theorem 4.53, 1) and 3) hold. By Proposition 4.17, the functorC U reflects isomorphisms. Since L = C UK ϕ we get that 2) holds.Conversely assume that 1), 2) and 3) hold. By Theorem 4.53 , D ϕ is full andfaithful and hence by Corollary 4.51 ̂ɛ is a functorial isomorphism. Let us provethat ̂η is an isomorphism as well. Since L reflects isomorphisms, it is enough toprove that L̂η is an isomorphism. As observed in Remark 4.49, by (44), ̂ηY is theunique morphism such thatHence we getso that(d ϕ K ϕ Y ) ◦ (̂ηY ) = ηY.(Ld ϕ K ϕ Y ) ◦ (L̂ηY ) = LηY(ɛLY ) ◦ (Ld ϕ K ϕ Y ) ◦ (L̂ηY ) = (ɛLY ) ◦ (LηY ) = LY.We now want to prove that (ɛLY ) ◦ (Ld ϕ K ϕ Y ) is also a right inverse for L̂ηY . Wecompute(Ld ϕ K ϕ Y ) ◦ (L̂ηY ) ◦ (ɛLY ) ◦ (Ld ϕ K ϕ Y ) (44)= (LηY ) ◦ (ɛLY ) ◦ (Ld ϕ K ϕ Y )Since L preserves the equalizer(L,R)adj= (Ld ϕ K ϕ Y ) .(D ϕ , d ϕ ) = Equ Fun(α C U, R C Uγ C)73□