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

For every Y ∈ B, X ∈ A and for every ξ ∈ Hom A (LY, X) , let us computeHom B (Y, αX) ◦ a X,Y (ξ) = αX ◦ a X,Y (ξ) = (RϕX) ◦ (ηRX) ◦ a X,Y (ξ)defa= (RϕX) ◦ (ηRX) ◦ (Rξ) ◦ (ηY )η= (RϕX) ◦ (RLRξ) ◦ (RLηY ) ◦ (ηY ) defa= a CX,Y [(ϕX) ◦ (LRξ) ◦ (LηY )] ϕ == a CX,Y [(Cξ) ◦ (ϕLY ) ◦ (LηY )]Since Γ (ϕ) = (ϕL) ◦ (Lη) we have obtained thatLet us calculateHom B (Y, αX) ◦ a X,Y = a CX,Y ◦ [(C−) ◦ (Γ (ϕ) Y )] .( )Hom B Y, R C ρ X ◦ aX,Y (ξ) = ( )R C ρ X ◦ aX,Y (ξ) defa= ( )R C ρ X ◦ (Rξ) ◦ (ηY )defa (= a CCX,Y ρ X ◦ ξ )Therefore we get thatHom B(Y, R C ρ X)◦ aX,Y = a CX,Y ◦ (C ρ X ◦ − )Since K ϕ (Y ) = Υ (ϕ) (Y ) = (LY, (ϕLY ) ◦ (LηY )), for every χ ∈ Hom A (LY , X) wehave65and[C (−) ◦ Γ (ϕ) Y ] (χ) = Γ (ϕ) Y = (Cχ) ◦ (ϕLY ) ◦ (LηY ) = (Cχ) ◦ C ρ LY[ Cρ X ◦ − ] (χ) = C ρ X ◦ χso thatThus we get[C (−) ◦ Γ (ϕ) Y ] (χ) = [C ρ X ◦ − ] (χ) if and only ifχ ∈ HomC A(((LY ) , (ϕLY ) ◦ (LηY )) ,(X, C ρ X)).Equ HomA (LY ,X)( Cρ X ◦ −, C (−) ◦ Γ (ϕ) Y )= { f ∈ Hom A (LY , X) | C ρ X ◦ f = (Cf) ◦ (Γ (ϕ) Y ) }= { f ∈ Hom A (LY , X) | C ρ X ◦ f = (Cf) ◦ (ϕLY ) ◦ (LηY ) }= { (f ∈ Hom CA U (K ϕ Y ) , C U ( )) }X, C ρ X | C ρ X ◦ f = (Cf) ◦ C ρC U(K ϕY )(= HomC A Kϕ Y, ( ))X, C ρ Xso that Equ Fun( Cρ X ◦ −, C (−) ◦ Γ (ϕ) − ) = HomC A(Kϕ −, ( X, C ρ X)). □Part of the following Proposition is already in [GT], Proposition <strong>2.</strong>3.Proposition 4.47. 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 α = Θ (ϕ) = (Rϕ) ◦ (ηR). Thenthe functor K ϕ = Υ (ϕ) : B → C A has a right adjoint D ϕ : C A → B if and only if,