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

22Since we already observed that the columns are coequalizers and also the first andthe second row are coequalizers by Proposition 3.14, in view of Lemma <strong>2.</strong>11 also thethird row is a coequalizer, so that (C, c) has a left A-module structure, i.e. thereexists (Γ, γ) ∈ A A such that (Γ, γ) = Coequ A A (f, g) and A U (Γ, γ) = (C, c) . □Lemma 3.21 ([BMV, Lemma 4.3]). Let A = (A, m A , u A ) be a monad on a categoryA with coequalizers and let ( A F, A U) be the adjunction associated. The followingstatements are equivalent:(i) A : A → A preserves coequalizers(ii) AA : A → A preserves coequalizers(iii) A A has coequalizers and they are preserved by A U : A A → A(iv) A U : A A → A preserves coequalizers.Proof. (i) ⇒ (ii) and (iii) ⇒ (iv) are clear.(ii) ⇒ (iii) follows by Lemma 3.20.(iv) ⇒ (i) Note that A F is a left adjoint, so that in particular it preserves coequalizers.Then A U A F = A also preserves coequalizers.□Lemma 3.2<strong>2.</strong> Let A = (A, m A , u A ) be a monad over a category A and assume that Apreserves equalizers. Then A F preserves equalizers where ( A F, A U) is the adjunctionassociated to the monad.Proof. LetE e Xbe an equalizer in A. Let us consider the fork obtained by applying the functor A Fto the equalizeri.e.(AE, m A E)AF E A F e AF XAe (AX, m A X)fgAF fAF gAfAg Y AF Y (AY, m A Y )Now, let ( Z, A µ Z)∈ A A and z : ( Z, A µ Z)→ (AX, mA X) be a morphism in A A suchthat (Af) ◦ z = (Ag) ◦ z. Since A preserves equalizers, we know that (AE, Ae) =Equ A (Af, Ag) . By the universal property of the equalizer (AE, Ae) in A, thereexists a unique morphism z ′ : Z → AE in A such that (Ae) ◦ z ′ = z. We now wantto prove that z ′ is a morphism in A A, i.e. that (m A E) ◦ (Az ′ ) = z ′ ◦ A µ Z . Since z isa morphism in A A we have that(m A X) ◦ (Az) = z ◦ A µ Zand since also Ae is a morphism in A A we have thatThen we have(m A X) ◦ (AAe) = (Ae) ◦ (m A E) .(Ae) ◦ (m A E) ◦ (Az ′ ) Ae∈ AA= (m A X) ◦ (AAe) ◦ (Az ′ )propz= (m A X) ◦ (Az) z∈ AA= z ◦ A µ Zpropz= (Ae) ◦ z ′ ◦ A µ Z