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

32and by Proposition 3.34 we have Q BB F = Q and p QB F = µ B Q . In particular we getQ BB F = Q = G B F.In order to prove that Q B = G it suffices to prove that(G, Gλ B ) = Coequ Fun(µBQ BU, Q B Uλ B). In fact, by Corollary 3.15, (B U, B Uλ B ) =Coequ Fun (B B Uλ B , m BB U) and, since by Lemma 3.20 B U reflects coequalizers, wehave(Id B B, λ B ) = Coequ Fun ( B F B Uλ B , λ BB F B U) .Since G preserves coequalizers, we get that(G, Gλ B ) = Coequ Fun (G B F B Uλ B , Gλ BB F B U)= Coequ Fun(QB Uλ B , µ B QBU ) = (Q B , p Q ) .Proposition 3.36. Let A = (A, m A , u A ) be a monad on a category A with coequalizerssuch that A preserves coequalizers. Let H : B → A A be a functor preservingcoequalizers. SetQ = A U ◦ H and let A µ Q = A Uλ A HThen ( Q, A µ Q)is a left A-module functor and(17) AQ = A ( A U ◦ H) = H.Proof. First we want to prove that A µ Q = A Uλ A H is associative. We haveA µ Q ◦ ( A A µ Q)= (A Uλ A H) ◦ (A A Uλ A H) λ A= ( A Uλ A H) ◦ ( A Uλ AA F A UH)□so that we get= ( A Uλ A H) ◦ (m AA UH) = A µ Q ◦ (m A Q)A µ Q ◦ ( A A µ Q)= A µ Q ◦ (m A Q) .Now we prove that A µ Q = A Uλ A H is unital. We computeso that we getA µ Q ◦ (u A Q) = ( A Uλ A H) ◦ (u AA UH) adj= A UH = QA µ Q ◦ (u A Q) = Q.Thus ( Q, A µ Q)is a left A-module functor. Recall that (see Lemma 3.29) there existsa unique functor A Q : B → A A such thatAU ◦ A Q = Q and A Uλ AA Q = A µ Q .Thus we haveAU ◦ A Q = Q = A U ◦ HandAUλ AA Q = A µ Q = A Uλ A Hso that, by Proposition 3.12, we obtain thatAQ = H.□