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

28Moreover if ϕ : Q → T is a functorial morphism between left A-module functors andϕ satisfiesA µ T ◦ (Aϕ) = ϕ ◦ (A µ Q)then there is a unique functorial morphism A ϕ : A Q → A T such thatAU A ϕ = ϕ.Proof. Corollary 3.25 applied to the case where F = Q gives us the first statement.Let B ∈ B. Then we have( Aµ T B ) ◦ (AϕB) = (ϕB) ◦ (A µ Q B )which means that ϕB yields a morphism A ϕB in A A.Proposition 3.30. 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 that A preserves coequalizers. Let Q : B → A be a functor and letA µ Q : AQ → Q and µ B Q : QB → Q be functorial morphisms. Assume that A µ Q isassociative and unital and that A µ Q ◦ ( Aµ B Q)= µBQ ◦ (A µ Q B ) . Set(6) (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B)Then Q B : B B → A is a left A-module functor where A µ QB : AQ B → Q B is uniquelydetermined by(7) p Q ◦ (A µ QB U ) = A µ QB ◦ (Ap Q ) .Moreover there exists a unique functor A (Q B ) : B B → A A such that(8) AU A (Q B ) = Q B and A Uλ AA (Q B ) = A µ QB .Proof. By Lemma <strong>2.</strong>7 we can consider (Q B , p Q ) = Coequ Fun(µBQ BU, Q B Uλ B). SinceA µ Q ◦ ( Aµ B Q)= µBQ ◦ (A µ Q B )we deduce that(9) ( A µ QB U) ◦ ( Aµ B QBU ) = ( µ B QBU ) ◦ (A µ Q B B U ) .Also, in view of the naturality of A µ Q , we have(10) ( A µ QB U) ◦ (AQ B Uλ B ) = (Q B Uλ B ) ◦ (A µ Q B B U ) .We computeand hence we obtainp Q ◦ ( A µ QB U) ◦ (AQ B Uλ B ) (10)= p Q ◦ (Q B Uλ B ) ◦ (A µ Q B B U )p Q coeq= p Q ◦ (µ B QBU) ◦ (A µ Q B B U ) (9)= p Q ◦ ( A µ QB U) ◦ ( Aµ B QBU )p Q ◦ ( A µ QB U) ◦ (AQ B Uλ B ) = p Q ◦ ( A µ Q BU) ◦ ( Aµ B QBU ) .Since A preserves coequalizers, we get(AQ B , Ap Q ) = Coequ Fun(AµBQ BU, AQ B Uλ B).Hence there exists a unique functorial morphism A µ QBp Q ◦ (A µ Q BU ) = A µ QB ◦ (Ap Q ) .: AQ B → Q B such that□