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

Note that, since f and g are A-bilinear morphism, g ◦ f is still an A-bilinear morphism,so that we can also consider (g ◦ f) Asuch that(23) (g ◦ f) A◦ p P = p T ◦ [(g ◦ f) A U] = p T ◦ (g A U) ◦ (f A U) .First we prove that (g ◦ f) A= g A ◦ f A . In fact we have(g ◦ f) A◦ p P(23)= p T ◦ (g A U) ◦ (f A U)(22)= g A ◦ p Q ◦ (f A U) (21)= g A ◦ f A ◦ p Pand since p P is an epimorphism we obtain(g ◦ f) A= g A ◦ f A .The, we can both consider A (g ◦ f) A = A ((g ◦ f) A ) such thatAU A (g ◦ f) A = A U A ((g ◦ f) A ) = (g ◦ f) Aand the composite of the liftings A g A ◦ A f A such thatWe haveAU [ A g A ◦ A f A ] = ( A U A g A ) ◦ ( A U A f A ) = g A ◦ f A .AU A (g ◦ f) A ◦ p P = A U A ((g ◦ f) A ) ◦ p P = (g ◦ f) A ◦ p P(23)= p T ◦ (g A U) ◦ (f A U) = g A ◦ p Q ◦ (f A U) = g A ◦ f A ◦ p Pand since p P is an epimorphism we deduce thatSince A U reflects we conclude thatAU A (g ◦ f) A = g A ◦ f A = A U A g A ◦ A U A f A .A (g ◦ f) A = A g A ◦ A f Awhere A U A (g ◦ f) A = (g ◦ f) A and (g ◦ f) A◦ p P = p T ◦ [(g ◦ f) A U] .Proposition 3.48. Let A = (A, m A , u A ) be a monad on a category A such that theunderlying functor A preserves coequalizers, let A P A , A Q A , A T A ∈ Ob (( A A ← A A))and let A f A : A P A → A Q A , A g A : A Q A → A T A , A h A : A T A → A W A be morphisms in( A A ← A A). ThenAh A ◦ ( A g A ◦ A f A ) = ( A h A ◦ A g A ) ◦ A f A .Proof. By Proposition 3.47 we have that, for every morphisms A f A : A P A → A Q A ,Ag A : A Q A → A T A in ( A A ← A A) , also the morphism A g A ◦ A f A is in ( A A ← A A)and A (g ◦ f) A = A g A ◦ A f A . Hence we have that39□Ah A ◦ ( A g A ◦ A f A ) (20)= A h A ◦ ( A (g ◦ f) A ) (20)= ( A (h ◦ (g ◦ f)) A )Astrictly monoidal= ( A ((h ◦ g) ◦ f) A ) (20)= A (h ◦ g) A ◦ A f A(20)= ( A h A ◦ A g A ) ◦ A f A .□