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

Since Q is a left A-module functor, by Lemma 3.17, also Q B U is a left A-modulefunctor. Now p Q is an epimorphism and hence, since A preserves coequalizers, alsoAAp Q is an epimorphism. Therefore we can apply Lemma 3.23 to ”ϕ” = A µ QB ,”h” = p Q and ”µ” = A µ Q BU and hence we obtain that ( )Q B , A µ QB is a left A-module functor that is A µ QB is associative and unital. By Lemma 3.29 appliedto ( )Q B , A µ QB there exists a functor A (Q B ) : B B → A A such that A U A (Q B ) =Q B and A Uλ AA (Q B ) = A µ QB . Moreover A (Q B ) is unique with respect to theseproperties.□Proposition 3.3<strong>1.</strong> 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 B havecoequalizers and A preserves them. Let Q : B → A be an A-B-bimodule functorwith functorial morphisms A µ Q : AQ → Q and µ B Q : QB → Q. Then the functorAQ : B → A A is a right B-module functor via µ B AQ : AQB → A Q where µ B AQ isuniquely determined by(11) AUµ B AQ = µ B Q.Let (( A Q) B, p A Q) = Coequ Fun(µBAQBU, A Q B Uλ B). Then we have( A Q) B= A (Q B ) : B B → A A.Proof. Since Q is endowed with a left A-module structure, by Lemma 3.29 thereexists a unique functor A Q : B → A A such that A U A Q = Q and A Uλ AA Q =A µ Q . Note that, since Q is an A-B-bimodule functor, in particular the compatibilityconditionA µ Q ◦ ( )Aµ B Q = µBQ ◦ (A µ Q B ) .holds. Thus, by Lemma 3.29, there exists a functorial morphism µ B AQ : AQB → A Qsuch thatAUµ B = µAQ B Q.By the associativity and unitality properties of µ B Q and since AU is faithful, weget that also µ B AQ is associative and unital, so that ( AQ, µ AQ) B is a right B-modulefunctor. Thus we can consider the coequalizer(12) AQB B Uµ B A Q BUAQ B Uλ B AQ B Up A Q ( A Q) Bso that we get a functor ( A Q) B: B B → A A. Since A preserves coequalizers, byLemma 3.21, also A U preserves coequalizers. Then, by applying the functor A U to12 we still get a coequalizer29that can be written as AU A QB B U A Uµ B A Q BUAU A Q B UAU A Q B Uλ BAUp A QAU ( A Q) BQB B Uµ B QBUQ B Uλ B Q B UAUp A QAU ( A Q) B