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

189[(ε C ⊗ H ) ◦ (π ⊗ H) ◦ ∆ H] (h) = ( ε C π ) ( h (1))⊗ h(2) = ε H ( h (1))⊗ h(2) ≃ h[(H ⊗ εC ) ◦ (H ⊗ π) ◦ ∆ H] (h) = h (1) ⊗ ( ε C π ) ( h (2))= h(1) ⊗ ε H ( h (2))≃ hso that H is a C-coring. Moreover, H has a right L-module structurewhich is left C-colinear i.e.µ L H : H ⊗ L → Hh ⊗ b ↦→ m H (h ⊗ b) = hb(212) π ( h (1) b (1))⊗ h(2) b (2) = π (h 1 ) ⊗ h (2) b(see [BrHaj, Lemma 3.3]), so that [( H ⊗ µ L H)◦(∆ H ⊗ L )] (H ⊗ L) ⊆ H□ C H.Assume that H is a right L-Galois coextension over C, that iscocan = ( H ⊗ µ L H)◦(∆ H ⊗ L ) : H ⊗ L → H□ C Hh ⊗ b ↦→ h (1) ⊗ h (2) bis an isomorphism and assume also that H is flat over k. In particular, if H L isfaithfully flat, we know that co(C) H = L (see [Schn2, Lemma <strong>1.</strong>3 (2)] and [BrWi,34.2 p. 343]) where we denoteco(C) H = { h ∈ H | C ρ H (h) = π (1 H ) ⊗ h } .In this case, we can also define the inverse of the cocanonical map, i.e.cocan −1 : H□ C H → H ⊗ L∑h i ⊗ g i ↦→ ∑ h i (1) ⊗ S ( )h i (2) g i .For every ∑ h i ⊗g i ∈ H□ C H, we have ∑ )h i (1)(h ⊗π i (2)⊗g i = ∑ ( )h i ⊗π g(1)i ⊗g(2) i .By means of the left H-linearity of π and of this equality we have∑hi(1) ⊗ π ( S ( ) ) ( )h i (3) gi(1) ⊗ S hi(2) gi(2) = ∑ h i (1) ⊗ S ( ) ( ) ( )h i (3) π gi(1) ⊗ S hi(2) gi(2)= ∑ h i (1) ⊗ S ( ( ) ( )h(3)) i π hi(4) ⊗ S hi(2) g i = ∑ h i (1) ⊗ π ( S ( ) ) ( )h i (3) hi(4) ⊗ S hi(2) gi= ∑ h i (1) ⊗ π (1 H ) ⊗ S ( )h i (2) giso that∑hi(1) ⊗ S ( )h i (2) g i ∈ Ker ( H ⊗ [C ρ H − π (1 H ) ⊗ (−) ]) = H ⊗ co(C) H = H ⊗ Lwhere in the first equality we have used that H is flat over k. Therefore cocan −1 isa well-defined map. Note that, by applying ε H ⊗ L to this element, we also deducethat, for every ∑ h i ⊗ g i ∈ H□ C H, we have(213)∑S(hi ) g i ∈ L.Now, let k be a commutative ring, let H be a k-Hopf algebra and letL ⊆ H be a right coideal subalgebra. Assume that H is a right L-Galoiscoextension over the coalgebra C = H/HL + , assume that H L is faithfullyflat, so that co(C) H = L, assume that H k is faithfully flat and assume that