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

214∆coass= a ⊗ c (1) ⊗ A 1 A ⊗ c (2)(1) ⊗ A 1 A ⊗ c (2)(2)= ( C ⊗ ∆ C) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2)).We define the counit of C, ε C : C → A, as followsε C (a ⊗ c) = aε (c) .It is straightforward to check that ε C is left A-linear. Let us check it is also rightA-linear. Let us computeε C ((a ⊗ c) b ′ ) = ε C (aψ (c ⊗ b ′ )) = ε C (ab ′ α ⊗ c α ) = ab ′ αε (c α )(225)= aε (c) b ′ = ( ε C (a ⊗ c) ) b ′ .Let now check the counitality(rC ◦ ( C ⊗ ε C) ◦ ∆ C) (a ⊗ c) = ( r C ◦ ( C ⊗ ε C)) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= r C(a ⊗ c(1) ⊗ A ε ( c (2)))= a ⊗ cand similarly(lC ◦ ( ε C ⊗ C ) ◦ ∆ C) (a ⊗ c) = ( l C ◦ ( ε C ⊗ C )) ( a ⊗ c (1) ⊗ A 1 A ⊗ c (2))= l C(aε(c(1))⊗A 1 A ⊗ c (2))= a ⊗ cthe right counitality is proved.2) Assume that A ⊗ C is an A-coring with the coproduct and counit as above, i.e.∆ C (a ⊗ c) = a ⊗ c (1) ⊗ A 1 A ⊗ c (2) and ε C (a ⊗ c) = aε (c) .Let us setψ (c ⊗ a) = (1 A ⊗ c) · a.We want to prove that ψ is an entwining for A and C. Since A ⊗ C is an A-coring,it is in particular a right A-module, so thati.e.((a ⊗ c) · a ′ ) · b ′ = (a ⊗ c) · (a ′ b ′ )(227) aa ′ αb ′ β ⊗ c αβ = a (a ′ b ′ ) α⊗ c αLet us compute, for every a, b ∈ A and c ∈ C[(m ⊗ C) ◦ (A ⊗ ψ) ◦ (ψ ⊗ A)] (c ⊗ a ⊗ b)= [(m ⊗ C) ◦ (A ⊗ ψ)] (ψ (c ⊗ a) ⊗ b)= [(m ⊗ C) ◦ (A ⊗ ψ)] (a α ⊗ c α ⊗ b) = (m ⊗ C)(a α ⊗ b β ⊗ (c α ) β)= a α b β ⊗ (c α ) β (227)= (ab) α⊗ c α= ψ (c ⊗ ab) = [ψ ◦ (C ⊗ m)] (c ⊗ a ⊗ b)and[ ]ψ ◦ (C ⊗ u) ◦ r−1C (c) = [ψ ◦ (C ⊗ u)] (c ⊗ 1k ) = ψ (c ⊗ 1 A )A⊗CrightAmod= (1 A ⊗ c) · 1 A = 1 A ⊗ c = (u ⊗ C) (1 k ⊗ c) = [ ](u ⊗ C) ◦ l −1 (c) .C