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

72We have to prove that (LD ϕ , Ld ϕ ) = Equ Fun(Lα C U, LR C Uγ C) . Since L is a functor,we clearly have ( Lα C U ) ◦(Ld ϕ ) = ( LR C Uγ C) ◦(Ld ϕ ). Let Z : Z → C A be a functorand let ξ : Z → LR C U be a functorial morphism such that(Lα C U ) ◦ ξ = ( LR C Uγ C) ◦ ξ.Recall that ( ε C C U ) ◦ (C Uγ C) = C U and (C F ε C) ◦ ( γ C C F ) = C F . We compute(ϕ C U ) ◦ ξ = Id C C U ◦ ϕ C U ◦ ξ Ccomonad (= ε C C C U ) ◦ ( ∆ CC U ) ◦ ( ϕ C U ) ◦ ξ =ϕcomonadmorp (= ε C C C U ) ◦ ( ϕϕ C U ) ◦ ( LηR C U ) ◦ ξ == ( ε C C C U ) ◦ ( ϕC C U ) ◦ ( LRϕ C U ) ◦ ( LηR C U ) ◦ ξ == ( ε C C C U ) ◦ ( ϕC C U ) ◦ ( Lα C U ) ◦ ξ = ( ε C C C U ) ◦ ( ϕC C U ) ◦ ( LR C Uγ C) ◦ ξ =ϕ= ( ε C C C U ) ◦ ( C C Uγ C) ◦ ( ϕ C U ) ◦ ξ εC = (C Uγ C) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ =Since ϕ is iso, we get(52)= ( ϕ C U ) ◦ (Ld ϕ ) ◦ (C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ.ξ = (Ld ϕ ) ◦ [(C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ ] .Let now w : Z → LD ϕ be a functorial morphism such thatWe computeξ = (Ld ϕ ) ◦ w.( CUγ C) ◦ (C Ûɛ ) ◦ [(C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ ] =(51)= ( ϕ C U ) ◦ (Ld ϕ ) ◦ [(C Ûɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ ] == ( ϕ C U ) ◦ ξ = ( ϕ C U ) ◦ (Ld ϕ ) ◦ w (51)= (C Uγ C) ◦ (C Ûɛ ) ◦ wand since C Uγ C is a monomorphism (since it is an equalizer) and ̂ɛ is an isomorphismwe obtain that( CÛɛ −1) ◦ ( ε C C U ) ◦ ( ϕ C U ) ◦ ξ = w.Conversely, assume that 1) and 2) hold. Then ϕ is a functorial isomorphism. Considerthe diagramLD ϕ(X, C ρ X) C Ubɛ(X, C ρ X)C U ( X, C ρ X)Ld ϕ(X, C ρ X) LR C U ( )X, C ϕ C U(X, C ρ X)ρ XC U C γ(X, C ρ X)C C U ( )X, C ρ XLα C U(X, C ρ X) LR C U C γ(X, C ρ X) ∆ C C U(X, C ρ X) C C U C γ(X, C ρ X)LRC C U ( )X, C ϕC C U(X, C ρ X)ρ X CC C U ( )X, C ρ Xof Lemma 4.52 where the last row is always an equalizer (see Proposition 4.13) andthe first row is also an equalizer by the assumption 1). Then we can apply Lemma