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

200This map is well-defined, in fact, let us consider the map θ : H ⊗ H → H ⊗ LL⊗H(L⊗H)L + defined by settingθ (x ⊗ y) = x ⊗ L [1 L ⊗ y]b Q.For every (h ⊗ g) · (t− ε H (t) ) ∈ (H ⊗ H) L + , we have (h ⊗ g) · (t− ε H (t) ) =ht (1) ⊗ gt (2) − h ⊗ gε H (t) and using that ∆ (L) ⊆ L ⊗ H, we computeht (1) ⊗ L(1L ⊗ gt (2))− h ⊗L(1L ⊗ gε H (t) )= h ⊗ L t (1) · (1L ⊗ gt (2))− h ⊗L(1L ⊗ gε H (t) )= h ⊗ L(t(1) ⊗ gt (2))− h ⊗L(1L ⊗ gε H (t) )= h ⊗ L(t(1) ⊗ gt (2) − 1 L ⊗ gε H (t) )= h ⊗ L((1L ⊗ g) · t − (1 L ⊗ g) ε H (t) )= h ⊗ L((1L ⊗ g) · (t− ε H (t) )) ∈ H ⊗ L (L ⊗ H) L +so that θ factors throughH⊗H(H⊗H)L + → H ⊗ LL⊗H(L⊗H)L +giving rise to the map ̂θ. Wecompute, using definition of ̂Q L = ⊗ L ̂Q and µ A b Q(̂θ ◦ ̂β) (h ⊗ L [x ⊗ h ′ ]b Q)= ̂θ ([hx ⊗ h ′ ] E) = hx ⊗ L [1 L ⊗ h ′ ]b Q= h ⊗ L x · [1 L ⊗ h ′ ]b Q= h ⊗ L [x1 L ⊗ h ′ ]b Q= h ⊗ L [x ⊗ h ′ ]b Qand (̂β ◦ ̂θ)([x ⊗ y] E) = ̂β)(x ⊗ L [1 L ⊗ y]b Q= [x ⊗ y] E.Let us show that ̂β is an isomorphism of left E-modules. Using definition of ̂µ E Q ,(215) i.e. ∑ ∑i j ki,j ⊗ S (h i,j ) g i ∈ H ⊗ L, definition of ̂m E we compute(∑ ∑ [̂βk i,j ⊗ h i,j] )·i jE gi ⊗ L [x ⊗ h ′ ]b Q= ̂β(∑i∑j ki,j S ( h i,j) )g i ⊗ L [x ⊗ h ′ ]b Q= ̂β(∑i∑j ki,j ⊗ L S ( h i,j) )g i · [x ⊗ h ′ ]b Q= ̂β(∑[ (i∑j ki,j ⊗ ) L S hi,jg i x ⊗ h ′] )Q b[∑=i∑j ki,j S ( h i,j) ]g i x ⊗ h ′ = ∑ ∑ [k i,j S ( h i,j) g i x ⊗ h ′]E i jE= ∑ ∑ [k i,j ⊗ h i,j] · [g i x ⊗ h ′] = ∑ ∑ [i jE E k i,j ⊗ h i,j] · ̂β)(g i ⊗i jE L [x ⊗ h ′ ]b Q.Similarly we want to understand the other isomorphism. Given a right L-modulefunctor G we will denote simply by − ⊗ L G the functor defined byCoequ Fun(µLG LU, G L Uλ L).Let us consider the functorL ⊗ HLQ EE ̂QL = − ⊗ L(L ⊗ H) L ⊗ + E H : L A = Mod-L → L A = Mod-L.We want to prove that L Q EE ̂QL is functorially isomorphic to Id L A. Now, for any(X, L µ X)∈ L A we have(X, L µ X)⊗L L = Coequ Fun(µLL LU ( X, L µ X), LL Uλ L(X, L µ X))