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

80We compute(LRɛLY ′ ) ◦ ( LR C Uh ) ◦ (LηY ) (59)= (LRɛLY ′ ) ◦ (LηRLY ′ ) ◦ (C Uh )= C Uhand since L = C UK ϕ and C U reflects, we getThen we deduce thatK Y,RLY ′(K ϕ RɛLY ′ ) ◦ ( K ϕ R C Uh ) ◦ (K ϕ ηY ) = h.(K−1Y,RLY(h) ) (= K ′ Y,RLY ′ (RɛLY ′ ) ◦ ( R C Uh ) ◦ (ηY ) )= (K ϕ RɛLY ′ ) ◦ ( K ϕ R C Uh ) ◦ (K ϕ ηY ) = h.Proposition 4.67. 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 K ϕ is full and faithful then, forevery Y ∈ B, we haveProof. By Lemma 4.65 we have(Y, ηY ) = Equ B (RLηY, ηRLY ) .(K ϕ Y, K ϕ ηY ) = EquC A (K ϕ RLηY, K ϕ ηRLY ) .Then we can apply Lemma <strong>2.</strong>16 and deduce that (Y, ηY ) = Equ B (RLηY, ηRLY ) .□Theorem 4.68 (Generalized Beck’s Theorem for comonads). Let (L, R) be an adjunctionwhere L : B → A and R : A → B, let C = ( C, ∆ C , ε C) be a comonadon a category A and let ϕ : LR = (LR, LηR, ɛ) → C = ( C, ∆ C , ε C) be a comonadsmorphism such that ϕX is a monomorphism for every X ∈ A. Let K ϕ =Υ (ϕ) = (L, (ϕL) ◦ (Lη)) and C UK ϕ (f) = L (f) for every morphism f in B. ThenK ϕ : B → C A is full and faithful if and only if for every Y ∈ B we have that(Y, ηY ) = Equ B (ηRLY, RLηY ) .Proof. If K ϕ is full and faithful then we can apply Proposition 4.67 to get that forevery Y ∈ B we have that (Y, ηY ) = Equ B (RLηY, ηRLY ) .Conversely assume that for every Y ∈ B we have that (Y, ηY ) = Equ B (ηRLY, RLηY ) .We want to prove that K Y,Y ′ is bijective for every Y, Y ′ ∈ B. Let us consider the□