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

Now we will give details of the isomorphisms associated to the equivalence of categories.Given a right E-module functor F we will denote simply by − ⊗ E F thefunctor defined byLet us consider the functorE ̂Q LL Q E = − ⊗ E H L ⊗ LCoequ Fun(µEF EU, F E Uλ E).L ⊗ H(L ⊗ H) L + : EB = E (Comod-C) → E B = E (Comod-C) .We want to prove that E ̂QLL Q E is functorially isomorphic to Id E B. Now, for any(X, E µ X)∈ E B we have(X, E µ X)⊗E E = Coequ Fun(µEE EU ( X, E µ X), EE Uλ E(X, E µ X))= Coequ Fun(mE X, E E µ X) 3.14= ( X, E µ X).Thus to this aim it is enough to construct an isomorphism of left E-modules ̂β :L⊗HH L ⊗ L → H⊗H . This will imply that β = −□(L⊗H)L + (H⊗H)L + C ̂β : ̂QLL Q = −□ C H L ⊗ LL⊗HH⊗H→ E = −□(L⊗H)L + C gives rise to a functorial isomorphism(H⊗H)L +E ̂QLL Q E ≃Id E B. We want to show that ̂β is the following morphismL ⊗ H(L ⊗ H) L → H ⊗ H+ (H ⊗ H) L +h ⊗ L [x ⊗ h ′ ]b Q↦→ [hx ⊗ h ′ ] E.̂β : H ⊗ LFirst we have to prove that it is a well-defined map. Let us consider the mapβ : H ⊗ L ⊗ H →H ⊗ H(H ⊗ H) L +h ⊗ x ⊗ h ′ ↦→ [hx ⊗ h ′ ] E.For every (x ⊗ h ′ ) · (t− ε H (t) ) ∈ (L ⊗ H) L + we haveβ ( h ⊗ [ (x ⊗ h ′ ) · (t− ε H (t) )]) = β ( h ⊗ xt (1) ⊗ h ′ t (2) − h ⊗ x ⊗ h ′ ε H (t) )so that β factors through β : H ⊗we have= hxt (1) ⊗ h ′ t (2) − hx ⊗ h ′ ε H (t) ∈ (H ⊗ H) L +199L⊗H → H⊗H . Moreover, for every l ∈ L,(L⊗H)L + (H⊗H)L +β (hl ⊗ x ⊗ h ′ ) = [(hl) x ⊗ h ′ ] E= [h (lx) ⊗ h ′ ] E= β (h ⊗ lx ⊗ h ′ )so that β is also L-balanced and gives rise to the map ̂β : H ⊗ LThe inverse of ̂β is given bŷθ :H ⊗ H(H ⊗ H) L → H ⊗ L ⊗ H+ L(L ⊗ H) L +[x ⊗ y] E↦→ x ⊗ L [1 L ⊗ y]b Q.L⊗H→ H⊗H .(L⊗H)L + (H⊗H)L +