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

Then, we can consider the morphism(136) cocan C := (A µ Q P C) ◦ ( Aδ C C): A C U = C Uà → QP C = C U C QP C .Then we have( ) QιP◦ cocan C = ( Qι ) P ◦ (A µ Q P C) ◦ ( )AδC C A µ Q(=Aµ Q P C U ) ◦ ( AQι ) P ◦ ( )AδCC(135)= (A µ Q P C U ) ◦ ( Aδ C C U ) ◦ ( A C Uγ C) = ( cocan C 1 U ) ◦ ( A C Uγ C)i.e.(137)Now, by assumption we have(QιP ) ◦ cocan C = ( cocan 1 C U ) ◦ ( A C Uγ C) .C U eA µC Q = A µ Q125so thatC U eA µC QP C = A µ Q P C .Moreover, by Lemma 6.39, there exists a morphism C δ C : C A → C QP C such thatC U C δ C C = δ C C .Since à is a lifting of the monad A, by Theorem 5.7 we have a mixed distributivelaw Φ : AC → CA so that we can apply Proposition 5.6 and we get thatAδ C C = A C U C δ C C = C UÃC δ C Cwhere ÃC δ C : à → ÃC QP C is a functorial morphism. Then we can consider themorphism( ) (ÃC )C cocan C eA:= µC QP C ◦ δCC : à → C QP Cand we get that(C ) ( )C U C cocan C = U eA µC QP C ◦C UÃC δCC= (A µ Q P C) ◦ ( A C U C δ C C)=( Aµ Q P C) ◦ ( Aδ C C)= cocan C .We compute(cocan1 C C U ) ◦ ( A∆ C C U ) = (A µ Q P C C U ) ◦ ( Aδ C C C U ) ◦ ( A∆ C C U )(125)= (A µ Q P C C U ) ◦ ( AQρ C P C U ) ◦ ( Aδ C C U )A µ Q=(QρCP C U ) ◦ (A µ Q P C U ) ◦ ( Aδ C C U ) = ( Qρ C P C U ) ◦ ( cocan 1 C U )so that we get((138) cocan1 C C U ) ◦ ( A∆ C C U ) = ( Qρ C P C U ) ◦ ( cocan C 1 U ) .Let us consider the following commutative diagram0 A C U AC Uγ C AC C Ucocan C cocan 1 C U0 QP C QιP QP C UA∆ C C UAC C Uγ CQρ C P C UQP C Uγ CACC C U QP C C Ucocan 1 C C U