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

(Q BB F, p QB F ) = Coequ Fun(µBQ B, Qm B)=(Q, µBQ).□(Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B)preserves them. Hence, using that(Q B , Q B λ B ) = Coequ Fun (Q B λ BB F B U, Q BB F B Uλ B ), in view of Lemmas <strong>2.</strong>10 and3.29, we have(Q BB P , Q B λ BB P ) = Coequ Fun (Q B λ BB F B U B P,Q BB F B Uλ BB P )= Coequ Fun(µBQ BU B P,Q B µ P)so that we get Q B λ BB P = p QB P .= (Q BB P , p QB P )Proposition 3.34. Let A = (A, m A , u A ) be a monad over a category A and letB = (B, m B , u B ) be a monad over a category B. Assume that both A and Bhave coequalizers and let Q : B → A be an A-B-bimodule functor. Then, withnotations(of Proposition)3.30, we can consider the functor Q B where (Q B , p Q ) =Coequ Fun µBQ BU, Q B Uλ B . Then(15) Q BB F = Q and p QB F = µ B Q.Proof. By construction we have that (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B). By applyingit to the functor B F we get that(Q BB F, p QB F ) = Coequ Fun(µBQ BU B F, Q B Uλ BB F )= Coequ Fun(µBQ B, Qm B).Since Q is a right B-module functor, by Proposition 3.16 we have that(Q, µBQ)= CoequFun(µBQ B, Qm B)so that we get31□Proposition 3.35. Let B = (B, m B , u B ) be a monad on a category B with coequalizerssuch that B preserves coequalizers. Let G : B B → A be a functor preservingcoequalizers. SetQ = G ◦ B F and let µ B Q = Gλ BB FThen ( Q, µ B Q)is a right B-module functor and(16) Q B = (G ◦ B F ) B= G.Proof. We computeandµ B Q ◦ ( µ B QB ) = (Gλ BB F ) ◦ (Gλ BB F B) λ B= (Gλ BB F ) ◦ (G B F B Uλ BB F )= (Gλ BB F ) ◦ (G ◦ B F m B ) = µ B Q ◦ (Qm B )µ B Q ◦ (Qu B ) = (Gλ BB F ) ◦ (G B F u B ) adj= G ◦ B F = Q.Thus ( Q, µ B Q)is a right B-module functor. Recall that (see Proposition 3.30)(Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B)