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

42[= Hom A (d ψ K ψ X, X) ◦ ã −1 (IdKψ )X,K ψ X]X(24)= a −1X,K ψ X AU ( ) (Id Kψ X = a−1X,K ψ X IdA UK ψ X)= a −1X,K ψ X (Id RX) = ɛXso that(˜ɛX) ◦ (d ψ K ψ X) = ɛX.(Since ˜ɛX = ã −1X,K ψ X IdKψ X)and ã−1X,K ψ Xis an isomorphism, we deduce that ˜ɛX :D ψ K ψ (X) → X is defined as the unique morphism such that(26) (˜ɛX) ◦ (d ψ K ψ X) = ɛX.On the other hand, for every ( )Y, A µ Y ∈ A B, the unit of the adjunction (D ψ , K ψ ) ,˜η : A B → K ψ D ψ , is given by˜η ( ) ( ) ( ) (( ))Y, A µ Y = ãDψ (Y, A µ Y ),Y IdDψ (Y, A µ Y ) : Y, A µ Y → Kψ D ψ Y, A µ Y .Then by commutativity of the diagram (24), we deduce thatAU ˜η ( ) (Y, A µ Y = A Uã Dψ (Y, A µ Y ),Y IdDψ (Y, A µ Y ))( (( )) ( )) ( )= a Dψ (Y, A µ Y ),Y ◦ Hom A dψ Y, A µ Y , Dψ Y, A µ Y IdDψ (Y, A µ Y )( (( ))) ( ( ))= a Dψ (Y, A µ Y ),Y dψ Y, A µ Y = Rdψ Y, A µ Y ◦ (ηY ) .Thus we obtain that(27) AU ˜η ( Y, A µ Y)=(Rdψ(Y, A µ Y))◦ (ηY ) .Observe that, for every Y ∈ B we have that A F (Y ) = (AY, m A Y ) . Moreoverso that we get(D ψA F (Y ) , d ψA F (Y )) = (D ψ (AY, m A Y ) , d ψ (AY, m A Y ))= Coequ A (rAY, Lm A Y ) (2)= (LY, rY )(28) (D ψA F , d ψA F ) = (L, r) .In particular(29) d ψ (AY, m A Y ) = rY.Theorem 3.57. Let (L, R) be an adjunction where L : B → A and R : A → B, letA = (A, m A , u A ) be a monad on the category B and let ψ : A = (A, m A , u A ) → RL =(RL, RɛL, η) be a monad morphism. Let r = Θ (ψ) = (ɛL) ◦ (Lψ). Assume that, forevery ( Y, A µ Y)∈ A B, there exists Coequ A(rY, L A µ Y). Then we can consider thefunctor K ψ = Υ (ψ) : A → A B. Its left adjoint D ψ : A B → A is full and faithful ifand only if1) R preserves the coequalizer(D ψ , d ψ ) = Coequ Fun (r A U, L A Uλ A )2) ψ : A → RL is a monad isomorphism.