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

188= 〈( h (1) y (1))εH ( h (2) y (2))− h(1) ε H ( h (2))ε H (y) | h ∈ H, y ∈ L 〉= 〈 h (1) y (1) ε H ( h (2))εH ( y (2))− h(1) ε H ( h (2))ε H (y) | h ∈ H, y ∈ L 〉= 〈 hy − hε H (y) | h ∈ H, y ∈ L 〉 = 〈 hy ′ | h ∈ H, y ′ ∈ L +〉 = HL +where we denote L + = L ∩ Ker ( ε H) . Let us prove that such J is a coideal of H (seealso [BrHaj, Lemma 3.2]). In fact, since ∆ H (y) = y (1) ⊗ y (2) ∈ L ⊗ H we have∆ H ( hy − hε H (y) ) = h (1) y (1) ⊗ h (2) y (2) − h (1) ⊗ h (2) ε H (y)= h (1)(y(1) − ε H ( y (1)))⊗ h(2) y (2) + h (1) ε H ( y (1))⊗ h(2) y (2) − h (1) ⊗ h (2) ε H (y)= h (1)(y(1) − ε H ( y (1)))⊗ h(2) y (2) + h (1) ⊗ h (2) ε H ( y (1))y(2) − h (1) ⊗ h (2) ε H (y)= h (1)(y(1) − ε H ( y (1)))⊗ h(2) y (2) + h (1) ⊗ h (2)(y − ε H (y) ) ∈ HL + ⊗ H + H ⊗ HL +and obviouslyε H ( hy − hε H (y) ) = ε H (h) ε H (y) − ε H (h) ε H (y) = 0.Then we can construct the coalgebra C := H/J = H/HL + and we can considerthe canonical projection π : H → C = H/J which is a coalgebra map and a leftH-linear map. We can defineC ρ H := (π ⊗ H) ◦ ∆ H : H → C ⊗ H and ρ C H := (H ⊗ π) ◦ ∆H : H → H ⊗ Ch ↦→ π ( h (1))⊗ h(2) h ↦→ h (1) ⊗ π ( h (2))so that H is a C-bicomodule. Note that C ρ H : H → C□ C H in fact, using that π isa coalgebra map, the coassociativity and the naturality of ∆ H , we compute(∆ C ⊗ H ) ◦ C ρ H = ( ∆ C ⊗ H ) ◦ (π ⊗ H) ◦ ∆ H = (π ⊗ π ⊗ H) ◦ ( ∆ H ⊗ H ) ◦ ∆ H= (π ⊗ π ⊗ H) ◦ ( H ⊗ ∆ H) ◦ ∆ H = (C ⊗ π ⊗ H) ◦ (π ⊗ H ⊗ H) ◦ ( H ⊗ ∆ H) ◦ ∆ H= (C ⊗ π ⊗ H) ◦ ( C ⊗ ∆ H) ◦ (π ⊗ H) ◦ ∆ H = ( C ⊗ C ρ H)◦ C ρ H .Similarly, we also have that ρ C H : H → H□ CC in fact, using that π is a coalgebramap, the coassociativity and naturality of ∆ H , we have(H ⊗ ∆C ) ◦ ρ C H = ( H ⊗ ∆ C) ◦ (H ⊗ π) ◦ ∆ H = ( H ⊗ ∆ C ◦ π ) ◦ ∆ H= ( H ⊗ (π ⊗ π) ◦ ∆ H) ◦ ∆ H = (H ⊗ π ⊗ π) ◦ ( H ⊗ ∆ H) ◦ ∆ H= (H ⊗ π ⊗ π) ◦ ( ∆ H ⊗ H ) ◦ ∆ H = (H ⊗ π ⊗ C) ◦ (H ⊗ H ⊗ π) ◦ ( ∆ H ⊗ H ) ◦ ∆ HNow, the map= (H ⊗ π ⊗ C) ◦ ( ∆ H ⊗ C ) ◦ (H ⊗ π) ◦ ∆ H = ( ρ C H ⊗ C ) ◦ ρ C H.∆ H : H → H□ C Hh ↦→ h (1) ⊗ h (2)))is well defined. In fact h (1)(1) ⊗π(h (1)(2) ⊗h (2) = h (1) ⊗π(h (2)(1) ⊗h (2)(2) . Moreover,the map π : H → C is a counit for H, in fact[(π□C H) ◦ ∆ H] (h) = π ( )h (1) ⊗ h(2) = C ρ H (h) ≃ h[(H□C π) ◦ ∆ H] (h) = h (1) ⊗ π ( )h (2) = ρCH (h) ≃ h