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

and thusk ◦ (u A QZ) ◦ (Qz) = h.Let l := k ◦ (u A QZ) : QZ → H. Then we havel ◦ (Qz) = k ◦ (u A QZ) ◦ (Qz) u A= k ◦ (AQz) ◦ (u A QY )(4)= h ◦ (A µ Q Y ) ◦ (u A QY ) = h.Let l ′ : QZ → H be another morphism such thatThen we havel ′ ◦ (Qz) = h.l ◦ (A µ Q Z ) ◦ (AQz) = l ◦ (Qz) ◦ (A µ Q Y ) = h ◦ (A µ Q Y )= l ′ ◦ (Qz) ◦ (A µ Q Y ) = l ′ ◦ (A µ Q Z ) ◦ (AQz) .Since AQ preserves coequalizers, we have that AQz is an epimorphism. Since A µ Q Zis also an epimorphism, we deduce that l = l ′ . Therefore we obtain that (QZ, Qz) =Coequ B (Qf, Qg) .□Lemma 3.20 ([BMV, Lemma 4.2]). Let A = (A, m A , u A ) be a monad on a categoryA and let f, g : ( X, A µ X)→(Y, A µ Y)be morphisms in A A. Assume that there exists(C, c) = Coequ A ( A Uf, A Ug) and assume that AA preserves coequalizers. Then thereexists (Γ, γ) = Coequ A A (f, g) and A U (Γ, γ) = (C, c) .Proof. Since AA preserves coequalizers and (A, m A ) is a right A-module functor, alsoA preserves coequalizers by Lemma 3.19, in particular, A preserves (C, c) . Sincec ◦ A µ Y ◦ (A A Uf) f∈ AA= c ◦ ( A Uf) ◦ A µ Xccoequ= c ◦ ( A Ug) ◦ A µ Xg∈ A A= c ◦ A µ Y ◦ (A A Ug)by the universal property of the coequalizer (AC, Ac) there exists a unique morphismA µ C : AC → C such thatc ◦ A µ Y = A µ C ◦ (Ac) .Moreover, by composing with u A Y this identity we getSince c is an epimorphism we obtainc = A µ C ◦ (Ac) ◦ (u A Y ) u A= A µ C ◦ (u A C) ◦ c.C = A µ C ◦ (u A C) .Now, consider the following serially commutative diagram21m A XAAX A A µ XAA A Uf AA A Ug m A YAAYAAcAACAA µ Ym A CAA µ CA A UfAXAY ACA A UgAcA µ XA µ YXAUfAUgYcA µ C C.