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

Conversely, let Φ be a functorial morphism satisfying Φ ◦ (m B Q) = (Qm A ) ◦ (ΦA) ◦(BΦ) ( and ) Φ ◦ (u B Q) = Qu A . We define ˜Q : AA → B B by setting, for everyX, A µ X ∈ A A,˜Q (( X, A µ X))=(QX,(Q A µ X)◦ (ΦX)).We have to check that ( Q (X) , ( Q A µ X)◦ (ΦX))∈ B B, that is)B µe QX◦(B B µe QX= B µe QX◦ (m B QX) and B µe QX◦ (u B QX) = QX.We compute)B µe QX◦(B B µe QXpropertyofΦ=Moreover we have= ( Q A µ X)◦ (ΦX) ◦(BQ A µ X)◦ (BΦX)Φ= ( Q A µ X)◦(QA A µ X)◦ (ΦAX) ◦ (BΦX)Xmodule= ( Q A µ X)◦ (QmA X) ◦ (ΦAX) ◦ (BΦX)(Q A µ X)◦ (ΦX) ◦ (mB QX) = B µe QX◦ (m B QX) .B µe QX◦ (u B QX) = ( Q A µ X)◦ (ΦX) ◦ (uB QX)propertyofΦ=(Q A µ X)◦ (QuA X) Xmodule= QX.Now, let f : ( X, A µ X)→(Y, A µ Y)a morphism of left A-modules, that is a morphismf : X → Y in A such thatA µ Y ◦ (Af) = f ◦ A µ X .We have to prove that ˜Q( ) ( )(f) : ˜QX = QX, B µ QX → ˜QY = QX, B µ QY is amorphism of left B-modules. We set ˜Q (f) = Q (f) and we compute(B µe ˜Qf) ( )?QY◦ B = ˜Qfi.e. by definition of the functor ˜Qin fact◦ B µe QXB µ QY ◦ (BQf) ? = (Qf) ◦ B µ QXB µ QY ◦ (BQf) = ( Q A µ Y)◦ (ΦY ) ◦ (BQf) Φ = ( Q A µ Y)◦ (QAf) ◦ (ΦX)fmorphA-mod= (Qf) ◦ ( Q A µ X)◦ (ΦX) = (Qf) ◦ B µ QX .Let now check that ˜Q is a lifting of Q. Let ( X, A µ X)∈ A A and computeBU ˜Q (( X, A µ X))= B U ( QX, B µ QX)= QX = QA U (( X, A µ X))and thus on the objectsBU ˜Q = Q A U.Let f : ( ) ( )X, A µ X → Y, A µ Y ∈ A A be a morphism, we haveBU ˜Q (f) : QX → QY = Q A U (f) : QX → QY.Therefore ˜Q is a lifting of the functor Q.25