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

217so that we have( ) ( )ρDA ◦ µ A A (a ⊗ b) = ρDA ◦ m A (a ⊗ b) = ρDA (ab)= ∑ ∑(ab) 0⊗ µ H D (g ⊗ (ab) 1) AisHcomalg= a0 b 0 ⊗ g · (a 1 b 1 )∑a0 b 0 ⊗ (g · a 1 ) · b 1 = ∑ ( )a 0 b 0 ⊗ µ H D µHD (g ⊗ a 1 ) ⊗ b 1= ∑ ( )m A (a 0 ⊗ b 0 ) ⊗ µ H D µHD (g ⊗ a 1 ) ⊗ b 1DisHmodcoalg== ∑ µ A A (a 0 ⊗ b 0 ) ⊗ µ H D (g · a 1 ⊗ b 1 ) .Then we deduce that A ∈ M D A (H). Note that this last computation says that m Ais a morphism of right D-comodules. It remains to prove that u A is also a morphismof right D-comodules. Let us compute(ρDA ◦ u A)(1k ) = ρ D A (1 A ) = (1 A ) 0⊗ g · (1 A ) 1AisHcomalgDisHmodcoalg= 1 A ⊗ g · 1 H = 1 A ⊗ g= (u A ⊗ D) (1 k ⊗ g) = [ ](u A ⊗ D) ◦ ρ D k (1k )so that we conclude that ( (A, m A , u A ) , ρA) D is a right D-comodule algebra. □Theorem 9.34 ([MeZu, Theorem 3.29 (a) ⇔ (f)]). Let H = ( )H, ∆ H , ε H , m H , u Hbe a k-bialgebra, let A = ( )(((A, m A , u A ) , ρ H A be a right H-comodule algebra, let D =D, ∆ D , ε D) ), µ H D be a right H-module coalgebra and let g ∈ D be a grouplikeelement. Then ( (A, m A , u A ) , ρA) D is a right D-comodule algebra and D = A ⊗ D isan A-coring. Assume that A D is flat (i.e. D is k-flat). Let B = A coD . Then thefollowing statements are equivalent:(a) the functor − ⊗ B A : Mod-B → M D A (H) is an equivalence of categories;(b) the canonical map can : A ⊗ B A → A ⊗ D is an isomorphism (i.e. B ⊆ A isa D-Galois extension) and B A is faithfully flat.Proof. We set µ H D (d ⊗ h) = d · h. By Lemma 9.33, we know that ( )(A, m A , u A ) , ρ D Ais a right D-comodule algebra. First of all we want to prove that D = A ⊗ D is anA-coring. Let us consider ψ : D ⊗ A → A ⊗ D defined by setting, for every a ∈ Aand d ∈ D,ψ (d ⊗ a) = a 0 ⊗ d · a 1where we denote ρ H A (a) = a 0 ⊗ a 1 . Let us prove that (A, D, ψ) is then an entwiningstructure over k. We have to prove (222). Let us compute, for every a, b ∈ A, d ∈ D,[(m A ⊗ D) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)] (d ⊗ a ⊗ b)= [(m A ⊗ D) ◦ (A ⊗ ψ)] (a 0 ⊗ d · a 1 ⊗ b)= (m A ⊗ D) (a 0 ⊗ b 0 ⊗ (d · a 1 ) · b 1 ) = a 0 b 0 ⊗ (d · a 1 ) · b 1so that(D,µ H D)Hmodcoalg= a 0 b 0 ⊗ d · (a 1 b 1 ) (A,ρH A)H-com alg= (ab) 0⊗ d · (ab) 1= ψ (d ⊗ ab) = [ψ ◦ (D ⊗ m A )] (d ⊗ a ⊗ b)(m A ⊗ D) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A) = ψ ◦ (D ⊗ m A ) .