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

186so that h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) .2) Now, let h 1 ⊗ h 2 ⊗ a ∈ A ⊗ (A ⊗ A) co(H) , i.e. h 1 ⊗ (h 2 ) 0⊗ a 0 ⊗ (h 2 ) 1a 1 =h 1 ⊗ h 2 ⊗ a ⊗ 1 H or equivalently (h 1 ) 1 ⊗ (h 1 ) 2 ⊗ a 0 ⊗ h 2 a 1 = h 1 ⊗ h 2 ⊗ a ⊗ 1 H . Byapplying to this equality the map can ⊗ A we obtain1 A ⊗ h 1 ⊗ a 0 ⊗ h 2 a 1 = (can ⊗ A) ( (h 1 ) 1 ⊗ (h 1 ) 2 )⊗ a 0 ⊗ h 2 a 1= (can ⊗ A) ( h 1 ⊗ h 2 ⊗ a ⊗ 1 H)= 1A ⊗ h ⊗ a ⊗ 1 Hand henceh 1 ⊗ 1 A a 0 ⊗ h 2 a 1 = h ⊗ 1 A a ⊗ 1 Hso that h 1 ⊗ h 2 ⊗ a ∈ (H ⊗ A) co(H) . Therefore we proved that[(ϕ ⊗ A) (H ⊗ A) ∩ A ⊗ (A ⊗ A) co(H)] [= (ϕ ⊗ A) (H ⊗ A) co(H)] .We can take( ) ()Q, q = (H ⊗ A) co(H) , (ϕ ⊗ A) |(H⊗A)co(H) .By Theorem 7.5, using i ◦ ϕ = γ, we have thatχ := m A ◦ (A ⊗ m A ) ◦ (m A ⊗ A ⊗ A) ◦ (A ⊗ A ⊗ A ⊗ m A )()◦ (A ⊗ i ⊗ A ⊗ A) ◦ A ⊗ (ϕ ⊗ A) |(H⊗A)co(H) ⊗ Aχ : A ⊗ (H ⊗ A) co(H) ⊗ A → Aa ⊗ h ⊗ b ⊗ c ↦→ a ⊗ h 1 ⊗ h 2 ⊗ b ⊗ c ↦→ ( ah 1) ( h 2 (bc) ) = a ( h 1 h 2) (bc) (209)= abcε H (h)is a coherd in X = (H, (A ⊗ A) co(H) , (H ⊗ A) co(H) , A, δ H , δ D ) where δ H(H ⊗ A) co(H) ⊗ A is uniquely determined by[](ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ H = (H ⊗ i) ◦ ∆ H .: H →Since (C, i) = Equ Fun(ω l , ω r) , by the universal property of the equalizer, there existsa unique functorial morphism ϕ : H → C such that i◦ϕ = γ. In our case this meansthat Imγ ⊆ C and henceϕ : H → Ch ↦→ h 1 ⊗ h 2 .For every h ∈ H, we have([] )(ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ C = [ (C ⊗ i) ◦ ∆ C] = (C ⊗ i) ◦ (ϕ ⊗ ϕ) ◦ ∆ H ◦ ϕ −1= (ϕ ⊗ i ◦ ϕ) ◦ ∆ H ◦ ϕ −1 = (ϕ ⊗ γ) ◦ ∆ H ◦ ϕ −1and hence[](ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ = (ϕ ⊗ γ) ◦ ∆ H = (ϕ ⊗ A ⊗ A) ◦ (H ⊗ γ) ◦ ∆ Hso that(ϕ −1 ⊗ A ⊗ A ) []◦ (ϕ ⊗ A) |(H⊗A)co(H) ⊗ A ◦ δ C ◦ ϕ= ( ϕ −1 ⊗ A ⊗ A ) ◦ (ϕ ⊗ A ⊗ A) ◦ (H ⊗ γ) ◦ ∆ H