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

Proof. Apply Proposition 3.24 to the case A = X , B = A, A = Id X and B = A.Then ˜Q = A F is the lifting of F and Φ = A µ F satisfies A µ F ◦(m A F ) = A µ F ◦ ( )A A µ Fand A µ F ◦ (u A F ) = F that is ( )F, A µ F is a left A-module functor.□Corollary 3.26. Let (L, R) be an adjunction with L : B → A and R : A → Band let A = (A, m A , u A ) be a monad on B. Then there is a bijective correspondencebetween the following collections of dataK Functors K : A → A B such that A U ◦ K = R,L functorial morphism α : AR → R such that (R, α) is a left module functorfor the monad Agiven byΦ : K → L where Φ (K) = A Uλ A K : AR → RΩ : L → K where Ω (α) (X) = (RX, αX) and A UΩ (α) (f) = R (f) .Proof. Apply Corollary 3.25 to the case ”F ” = R : A → B where (L, R) is anadjunction with L : B → A and R : A → B and A = (A, m A , u A ) a monad onB. □In the following Proposition we give a more precise version of Lemma 3 in [J].Proposition 3.27. Let A = (A, m A , u A ) be a monad on a category A and letB = (B, m B , u B ) be a monad on a category B. Let Q : A → B be a functor andlet ˜Q : A A → B B be a lifting of Q (i.e. BU ˜Q = Q A U) and Φ : BQ(→ QA)asin Proposition 3.24. Then Φ is an isomorphism if and only if φ = λ B ˜QA F ◦( B F Qu A ) : B F Q → ˜Q A F is an isomorphism.Proof. By construction in Proposition 3.24 we have that Φ = B Uφ. Assume thatΦ is an isomorphism. Since, by Proposition 3.18, B U reflects isomorphisms, φ :BF Q → ˜Q A F is an isomorphism. Conversely, assume that φ : B F Q → ˜Q A F is anisomorphism. Then Φ = B Uφ is also an isomorphism.□Corollary 3.28. Let (L, R) be an adjunction where L : B → A and R : A → Band let B = (B, m B , u B ) be a monad on B. Let K : A → B B be a functor such thatBU ◦ K = R and let (R, α) be a left B-module functor as in Corollary 3.26. Then αis an isomorphism if and only if λ B K : B F R → K is an isomorphism.Proof. Apply Proposition 3.27 with Q = R, A = Id A . Then ˜Q = K is the lifting ofR and Φ = α : BR → R, given by α = B Uφ = B Uλ B K.□<strong>Some</strong> <strong>results</strong> in the following part of this section can be found in the literature(see e.g. [BM] and [BMV]). To introduce our main tools of investigation, for thereader’s sake, we give here a full description.Lemma 3.29. Let A = (A, m A , u A ) be a monad over a category A with coequalizers.Let Q : B → A be a left A-module functor with functorial morphisms A µ Q : AQ → Q.Then there exists a unique functor A Q : B → A A such thatAU ◦ A Q = Q and A Uλ AA Q = A µ Q .27