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

26We have to prove that it is a bijection. Let us start with ˜Q : A A → B B a liftingof the functor Q : A → B. Then we construct Φ : BQ → QA given by( )Φ = BUλ B ˜QA F ◦ ( B U B F Qu A )and using this functorial morphism we define a functor Q : A A → B B as follows: forevery ( X, A µ X)∈ A AQ (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX)).Since both ˜Q and Q are lifting of Q, we have that B U ˜Q = Q A U = B UQ. We haveto prove that B U ( λ B Q ) ( )= B U λ B ˜Q . Let Z ∈ A A. We computeBU ( λ B QZ ) ()= (Q A Uλ A Z) ◦ BUλ B ˜QA F A UZ ◦ ( B U B F Qu AA UZ)(eQliftingQ= BU ˜Qλ) ()A Z ◦ BUλ B ˜QA F A UZ ◦ ( B U B F Qu AA UZ)) (λ=B(BUλ B ˜QZ ◦ BU B F B U ˜Qλ)A Z ◦ ( B U B F Qu AA UZ))=(BUλ B ˜QZ ◦ ( B U B F [Q A Uλ A Z ◦ Qu AA UZ]) (u A,λ A )adj= BUλ B ˜QZ.Conversely, let us start with a functorial morphism Φ : BQ → QA satisfying Φ ◦(m B Q) = (Qm A ) ◦ (ΦA) ◦ (BΦ) and Φ ◦ (u B Q) = Qu A . Then we construct a functor˜Q : A A → B B by setting, for every ( X, A µ X)∈ A A,˜Q (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX))which lifts Q : A → B. Now, we define a functorial morphism Ψ : BQ → QA givenby( )Ψ = BUλ B ˜QA F ◦ ( B U B F Qu A ) .Then we have( )Ψ = BUλ B ˜QA F◦ ( B U B F Qu A ) def e Q= (Q A Uλ AA F ) ◦ (Φ A F ) ◦ ( B U B F Qu A )= (Qm A ) ◦ (ΦA) ◦ (BQu A ) Φ = (Qm A ) ◦ (QAu A ) ◦ Φ Amonad= Φ.Corollary 3.25. Let X , A be categories, let A = (A, m A , u A ) be a monad on a categoryA and let F : X → A be a functor. Then there exists a bijective correspondencebetween the following collections of data:H Left A-module actions A µ F : AF → FG Functors A F : X → A A such that A U A F = F ,given byã : H → G where A Uã (A µ F)= F and A Uλ A ã (A µ F)= A µ F i.e.ã (A µ F)(X) =(F X, A µ F X ) and ã (A µ F)(f) = F (f)˜b : G → H where ˜b (A F ) = A Uλ AA F : AF → F.□