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

69and thus[D((X, LR ρ X)), d(X, LR ρ X)]= EquB(ηRX, R( LRρ X)).Proof. We can apply Proposition 4.47 with ”ϕ” = Id LR .Remark 4.49 ([GT]). In the setting of Proposition 4.47, for every Y ∈ B, we notethat the unit of the adjunction (K ϕ , D ϕ ) is given by( )̂ηY = â KϕY,Y IdKϕY : Y → Dϕ K ϕ (Y ) .We will consider the diagram (43) in the particular case of ( X, C ρ X)= Kϕ Y. Notethat since K ϕ Y = (LY, (ϕLY ) ◦ (LηY )) = (LY, βY ) we havei.e.(D ϕ K ϕ (Y ) , d ϕ K ϕ (Y )) = (D ϕ ((LY, βY )) , d ϕ K ϕ (Y )) = Equ B (αLY, RβY )(45) (D ϕ K ϕ (Y ) , d ϕ K ϕ (Y )) = Equ B (αLY, RβY )where β = Γ (ϕ) = (ϕL) ◦ (Lη). We compute(d ϕ K ϕ Y ) ◦ (̂ηY ) = Hom B (Y, d ϕ K ϕ Y ) (̂ηY ) = Hom B (Y, d ϕ K ϕ Y ) ( ( ))â KϕY,Y IdKϕY= [ ] ( ) (43)Hom B (Y, d ϕ K ϕ Y ) ◦ â KϕY,Y IdKϕY = a C KϕY,Y U ( )Id KϕY( )= a KϕY,Y IdC UK ϕY = aKϕY,Y (Id RY ) = ηYso that(46) (d ϕ K ϕ Y ) ◦ (̂ηY ) = ηY.On the other hand, for every ( X, C ρ X)∈ C A, the counit of the adjunction (K ϕ , D ϕ )is given bŷɛ ( X, C ρ X)= â−1X,D ϕ(X, C ρ X )(IdDϕ(X, C ρ X )): Kϕ D ϕ((X, C ρ X))→(X, C ρ X).Then we have thatâ X,Dϕ(X, C ρ X )(̂ɛ(X, C ρ X))= IdDϕ(X, C ρ X ).By commutativity of the diagram (43), we deduce that( ) ( ) ( )))d ϕ X, C ρ X = dϕ X, C ρ X ◦(âX,Dϕ(X, C ρ X )(̂ɛ X, C ρ XThus we obtain that(47)= a X,Dϕ(X, C ρ X )( CÛɛ ( X, C ρ X)).C Ûɛ ( X, C ρ X)= a−1X,D ϕ(X, C ρ X )(dϕ(X, C ρ X))= (ɛX) ◦(Ldϕ(X, C ρ X)).Observe that, for every X ∈ A, we have that C F (X) = ( CX, ∆ C X ) ∈ C A. Moreover(Dϕ C F (X) , d ϕ C F (X) ) = ( D ϕ(CX, ∆ C X ) , d ϕ(CX, ∆ C X )) =□so that we get(48)= Equ B(αCX, R∆ C X ) (4.15)= (RX, αX)(Dϕ C F , d ϕ C F ) = (R, α) .