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

Remark 3.10. Let A = (A, m A , u A ) be a monad on a category A and let ( X, A µ X)∈AA. From the unitality property of A µ X we deduce that A µ X is epi for every(X, A µ X)∈ A A and that u A X is mono for every ( X, A µ X)∈ A A, i.e. u A is amonomorphism.Definition 3.1<strong>1.</strong> Corresponding to a monad A = (A, m A , u A ) on A, there is anadjunction ( A F, A U) where A U is the forgetful functor and A F is the free functorAU : AA → A AF : A → AA(X, A µ X)→ X X → (AX, mA X)f → f f → Af.Note that A U A F = A. The unit of this adjunction is given by the unit u A of themonad A:u A : A → A U A F = A.The counit λ A : A F A U → A A of this adjunction is defined by settingTherefore we haveAU ( λ A(X, A µ X))= A µ X for every ( X, A µ X)∈ A A.(λ AA F ) ◦ ( A F u A ) = A F and ( A Uλ A ) ◦ (u AA U) = A U.Proposition 3.1<strong>2.</strong> Let A = (A, m A , u A ) be a monad on a category A. Then A U isa faithful functor. Moreover, given Z, W ∈ A A we have thatZ = W if and only if A U (Z) = A U (W ) and A U (λ A Z) = A U (λ A W ) .In particular, if F, G : X → A A are functors, we haveF = G if and only if A UF = A UG and A U (λ A F ) = A U (λ A G)Proposition 3.13. Let A = (A, m A , u A ) be a monad on a category A. Then( A U, ( A Uλ A )) is a left A-module functor.Proof. We have to prove that( A U A λ) ◦ (A A U A λ) = ( A U A λ) ◦ (m AA U) and( A U A λ) ◦ (u AA U) = A U.Let us consider ( X, A µ X)∈ A A. We have to show that(AU A λ ( X, A µ X))◦(AA U A λ ( X, A µ X))=(AU A λ ( X, A µ X))◦(mAA U ( X, A µ X))17and that(AU A λ ( X, A µ X))◦(uAA U ( X, A µ X))= A U ( X, A µ X)i.e. thatA µ X ◦ ( A A µ X)= A µ X ◦ (m A X)andA µ X ◦ (u A X) = Xwhich hold since ( X, A µ X)is an A-module.□