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

f ↦→ Rfis bijective for every X, X ′ ∈ A. Since this φ (f) = R (f) = R X,X ′ (f) , φ is anisomorphism if and only if R X,X ′ is an isomorphism for every X, X ′ ∈ A and, byLemma <strong>2.</strong>30, if and only if ɛX is an isomorphism for every X ∈ A.□Lemma <strong>2.</strong>33. Let (L, R) be an adjunction where L : B → A and R : A → B suchthat L is an equivalence of categories. Then R is also an equivalence of categories.Proof. By assumption L : B → A is an equivalence of categories with R ′ : A → B.Then it is well-known that (L, R ′ ) is an adjunction. By the uniqueness of theadjoint, we have that R ≃ R ′ which is an equivalence. Thus R is also an equivalenceof categories.□3. MonadsDefinition 3.<strong>1.</strong> A monad on a category A is a triple A = (A, m A , u A ) , whereA : A → A is a functor, m A : AA → A and u A : A → A are functorial morphismssatisfying the associativity and the unitality conditions:m A ◦ (m A A) = m A ◦ (Am A ) and m A ◦ (Au A ) = A = m A ◦ (u A A) .Definition 3.<strong>2.</strong> A morphism between two monads A = (A, m A , u A ) and B =(B, m B , u B ) on a category A is a functorial morphism ϕ : A → B such thatϕ ◦ m A = m B ◦ (ϕϕ) and ϕ ◦ u A = u B .Example 3.3. Let (A, m A , u A ) be an R-ring where R is an algebra. Then• A is an R-R-bimodule• m A : A ⊗ R A → A is a morphism of R-R-bimodules• u A : R → A is a morphism of R-R-bimodules satisfying the followingm A ◦(m A ⊗ R A) = m A ◦(A ⊗ R m A ) , m A ◦(A ⊗ R u A ) = r A and m A ◦(u A ⊗ R A) = l Awhere r A : A ⊗ R R → A and l A : R ⊗ R A → A are the right and leftconstraints. LetA = − ⊗ R A : Mod-R → Mod-Rm A = − ⊗ R m A : − ⊗ R A ⊗ R A → − ⊗ R Au A = (− ⊗ R u A ) ◦ r−−1 : − → − ⊗ R R → − ⊗ R AWe prove that A = (A, m A , u A ) is a monad on the category Mod-R. Forevery M ∈ Mod-R we compute[m A ◦ (m A A)] (M) = (M ⊗ R m A ) ◦ (M ⊗ R A ⊗ R m A ) = M ⊗ R [m A ◦ (A ⊗ R m A )]= M ⊗ R [m A ◦ (m A ⊗ R A)] = (M ⊗ R m A ) ◦ (M ⊗ R m A ⊗ R A)= [m A ◦ (Am A )] (M)[m A ◦ (Au A )] (M) = (M ⊗ R m A ) ◦ [ ](M ⊗ R u A ) ◦ r −1M ⊗R A= (M ⊗ R m A ) ◦ (M ⊗ R u A ⊗ R A) ◦ ( r −1M ⊗ R A )= (M ⊗ R [m A ◦ (u A ⊗ R A)]) ◦ ( r −1M ⊗ R A )= (M ⊗ R l A ) ◦ ( r −1M⊗ R A ) = M ⊗ R A = AM15