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

181On the other side,(θ r P ) ◦ ( )P can −11 ◦ (P CuA ) : − ⊗ R C ⊗ R Σ ∗ T −→ − ⊗ R Σ ∗ T ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T B− ⊗ R c ⊗ R f ↦→ − ⊗ R 1 A ⊗ R c ⊗ R f ↦→ − ⊗ R c 1 ⊗ T c 2 ⊗ R f↦→ − ⊗ R c 1 ⊗ T c 2 ⊗ R f ⊗ T 1 B .MaybewhereQ = − ⊗ R XX = Ker (ϕ)ϕ : C ⊗ R Σ ∗ T → Σ R ⊗ R Σ ∗ T ⊗ T Bc ⊗ R f ↦→ c 1 ⊗ T(c2 ) 0 ⊗ R β 1 ⊗ T β 2 · f = c 1 ⊗ T c 2 ⊗ R f ⊗ T 1 BIn case all the computations above make sense, we could write a coherd associatedto the pretorsor. By Theorem 7.5, the coherd is defined byχ : = µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ ( QP [ can −11 ◦ (Cu A ) ] Q ) ◦ (QqQ)i.e.= µ B Q ◦ (A µ Q B ) ◦ ( AQσ B) ◦ ( σ A QP Q ) ◦ ( QP can −11 Q ) ◦ (QP Cu A Q) ◦ (QqQ)χ : QQQ ⊆ QP CQ = − ⊗ T Σ R ⊗ R C ⊗ R Σ ∗ T ⊗ T Σ R → Q = − ⊗ T Σ R− ⊗ T x ⊗ R c ⊗ R f ⊗ T y ↦→ − ⊗ T x ⊗ R 1 A ⊗ R c ⊗ R f ⊗ T y↦→ − ⊗ T x ⊗ R c 1 ⊗ T c 2 ⊗ R f ⊗ T y ↦→ − ⊗ T x ⊗ R c 1 ⊗ T c 2 ⊗ R f (y)↦→ − ⊗ T x · c 1 ⊗ T c 2 ⊗ R f (y) ↦→ − ⊗ T x · c 1 ⊗ T c 2 f (y)↦→ − ⊗ T(x · c1 ) ( c 2 f (y) ) = − ⊗ T xc 1 ( c 2 f (y) ) = − ⊗ T xc 1 ( c 2) f (y)where for every x ∈ Σ A and h ∈ Σ ∗In particular, we have= − ⊗ T x ( 1 A ε C (c) ) f (y)x · h ∈ B = Hom A (Σ A , Σ A ) is defined by setting(x · h) (t) = xh (t) for every t ∈ Σ.x · c 1 : Σ A → Σ At ↦→ xc 1 (t) .9.<strong>1.</strong> H-Galois extension. In order to understand better the situation of the previousexample, we decide to consider a very particular case, the Schauenburg setting.Let H be a Hopf algebra and let A/k be a right H-Galois extension. Let us recallsome useful equalities related to the translation mapγ := can −1 (1 A ⊗ −) : H → A ⊗ A : h → can −1 (1 A ⊗ h) =: h 1 ⊗ h 2 .For every h, l ∈ H, a ∈ A, we have(206)(207)h 1 ( h 2) 0 ⊗ ( h 2) 1= 1 A ⊗ hh 1 ⊗ ( h 2) 0 ⊗ ( h 2) 1= (h 1 ) 1 ⊗ (h 1 ) 2 ⊗ h 2