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

86Then we haveA µ CCX ◦ ( A∆ C X ) = ( CC A µ X)◦ (CΦX) ◦ (ΦCX) ◦(A∆ C X )andΦm.d.l.= ( CC A µ X)◦(∆ C AX ) ◦ (ΦX) ∆C= ( ∆ C X ) ◦ ( C A µ X)◦ (ΦX)(ε C X ) ◦ (A µ CX)=(ε C X ) ◦ ( C A µ X)◦ (ΦX)ε C = A µ X ◦ ( ε C AX ) ◦ (ΦX) Φm.d.l.= A µ X ◦ ( Aε C X ) .Thus ∆ C and ε C lift to functorial morphisms ∆ eC and ε eC uniquely defined byWe compute(AU ˜C∆ e ) (C◦ AU∆ e ) (C=AU∆ eC = ∆ C AU and AUε eC = ε C AU.C A U∆ e )C◦ ( ∆ C AU ) = ( C∆ C AU ) ◦ ( ∆ C AU )= [( C∆ C) ◦ ∆ C] AU Ccomonad [(= ∆ C C ) ◦ ∆ C] AU = ( ∆ C C A U ) ◦ ( ∆ C AU )(= ∆ C AU ˜C) (◦ AU∆ e ) (C= AU∆ e ) (C ˜C ◦ AU∆ e )Cand since A U is faithful , we deduce( ) (˜C∆C e◦ ∆ eC = ∆ e )C ˜C ◦ ∆ eC .We compute(AU ˜Cε e ) (C◦ AU∆ e ) (C= C A Uε e )C◦ ( ∆ C AU ) = ( Cε C AU ) ◦ ( ∆ C AU )= [( Cε C) ◦ ∆ C] AU Ccomonad= C A U = A U ˜Cand since A U is faithful, we obtain(˜Cεe C)◦ ∆ eC = ˜C.Similarly we compute(AUε e ) (C ˜C ◦ AU∆ e )Cand since A U is faithful, we obtain(ε e C ˜C)=(ε C AU ˜C)◦ ( ∆ C AU ) = ( ε C C A U ) ◦ ( ∆ C AU )= [( ε C C ) ◦ ∆ C] AU Ccomonad= C A U = A U ˜C◦ ∆ eC = ˜C.Therefore ˜C =(˜C, ∆e C, ε e C)is a comonad on A A.Similarly, in order to prove the bijection between D and M, we apply Proposition4.23, taking both ( C, ∆ C , ε C) , ( D, ∆ D , ε D) = ( C, ∆ C , ε C) comonad on A and T =A. In particular we will prove that the bijection a : F → M, b : M → F ofProposition 4.23 induces a bijection between M and D.