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

19and fromµ A P ◦ ( µ A P A ) = µ A P ◦ (P m A ) and P = µ A P ◦ (P A)we deduce thatGµ A P ◦ ( Gµ A P A ) = Gµ A P ◦ (GP m A ) and GP = Gµ A P ◦ (GP A) .□Proposition 3.18. Let A = (A, m A , u A ) be a monad on a category A and let( A F, A U) be the adjunction associated. Then A U reflects isomorphisms.Proof. Let f : ( X, A µ X)→(Y, A µ Y)be a morphism in A A such that A Uf has atwo-sided inverse f −1 in A. SinceA µ X ′ ◦ (Af) = f ◦ A µ Xwe get thatf −1 ◦ A µ X ′ = A µ X ◦ ( Af −1) .□Lemma 3.19 ([BMV, Lemma 4.1]). Let A = (A, m A , u A ) be a monad on a categoryA, let ( P, µ A P)be a right A-module functor and let(Q, A µ Q)be a left A-modulefunctor where P : A → B, Q : B → A. Then any coequalizer preserved by P A isalso preserved by P and any coequalizer preserved by AQ is also preserved by Q.Proof. Consider the following coequalizerXfg Yz Zin the category A and assume that P A preserves it. By applying to it the functorsP A and P we get the following diagram in BP u A XP AXP Xµ A P XP AfP AgP fP gP u A YP AYP Yµ A P YP AzP zP u A ZP AZP Z.µ A P ZBy assumption, the first row is a coequalizer. Assume that there exists a morphismh : P Y → H such thath ◦ (P f) = h ◦ (P g) .Then, by composing with µ A P X we geth ◦ (P f) ◦ ( µ A P X ) = h ◦ (P g) ◦ ( µ A P X )and since µ A Pis a functorial morphism we obtainh ◦ ( µ A P Y ) ◦ (P Af) = h ◦ ( µ A P Y ) ◦ (P Ag) .Since (P AZ, P Az) = Coequ B (P Af, P Ag), there exists a unique morphism k :P AZ → H such that(3) k ◦ (P Az) = h ◦ ( µ A P Y ) .