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

By Lemma 3.29, we have A Q = L = − ⊗ T Σ : Mod-T → Mod-A. We have thatP A is a right adjoint of A Q, so that, by the uniqueness of the adjoint, we haveP A = W = − ⊗ A Σ ∗ T : Mod-A → Mod-T.Note that, by Corollary 6.22,, A σ A A : AQP A → A A = Mod-A is the counit ɛ of theadjunction ( A Q,P A ) = (L, W ) , i.e.Now, we can considerAσ A AM = ɛM : A QP A M = M ⊗ A Σ ∗ ⊗ T Σ A → Mm ⊗ A f ⊗ T x ↦→ mf (x) .Acan A =(˜CA σ A A)( ) eC◦ ρ L W : LW → ˜C.For every M ∈ Mod-A, we have( ) ( )Acan A M = ˜CA σAA eC◦ ρ L W M : M ⊗ A Σ ∗ ⊗ T Σ A → M ⊗ R Cm ⊗ A f ⊗ T x ↦→ mf (x 0 ) ⊗ R x 1 .Therefore we deduce that A can A = can. Recall now that, by Lemma 6.17 A can A isan isomorphism if and only ifcan 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CAis an isomorphism. For every M ∈ Mod-R, we havecan 1 M : QP M = M ⊗ R Σ ∗ T ⊗ T Σ R −→ CAM = M ⊗ R A ⊗ R Cm ⊗ R f ⊗ T x ↦→ ( Cσ A M ) (m ⊗ R f ⊗ T x 0 ⊗ R x 1 ) = m ⊗ R f (x 0 ) ⊗ R x 1 .Assume now that (Σ A , ρ e CΣ ) is a right Galois comodule. Thus we deduce that(L, eC ρ L)is a left Galois functor and thus can 1 := ( Cσ A) ◦ (C ρ Q P ) : QP → CAis an isomorphism. Therefore, we can consider the compositeτ := ( (can 1 ) −1 Q ) ◦ (Cu A Q) ◦ C ρ Q : Q → QP Qand we can apply Theorem 6.24, that implies that the functorial morphism τ is aregular herd. It is defined bywhereτ : M ⊗ T Σ R −→ M ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T Σ Rm ⊗ T x ↦→ m ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1(( ) ) ( )m ⊗ T x 0 ⊗ R x 1 1 x21 ⊗T0 x2 = (can 1 1 1QM) ( )m ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1= [ (can 1 QM) ◦ ( (can 1 ) −1 QM ) ◦ (Cu A QM) ◦ (C ρ Q M )] (m ⊗ T x)For every c ∈ C, we denote by= [ (Cu A QM) ◦ (C ρ Q M )] (m ⊗ T x)= m ⊗ T x 0 ⊗ R 1 A ⊗ R x 1 .− ⊗ R c 1 ⊗ T c 2 = (can 1 ) −1 (− ⊗ R 1 A ⊗ R c)177