Codex @codex  0

For an invertible braiding $C$ satisfying the braid equation on $V\otimes V$, begin with the free associative algebra over a field on matrix coefficients $t_{ij}$, with $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$ and $\epsilon (t_{ij})={\delta }_{ij}$. Impose precisely the coefficients ofThis quotient is the FRT bialgebra, or Faddeev-Reshetikhin-Takhtajan construction. The relations form a coideal because composing two coactions that preserve $C$ again preserves $C$; the trivial coaction also preserves it, so the counit descends. Universality follows because any such coaction must satisfy exactly these coefficient relations.

For the given $C$, take $t_{11}=a$, $t_{12}=b$, $t_{21}=c$, $t_{22}=d$. Thus $\delta (e_1)=e_1\otimes a+e_2\otimes c$ and $\delta (e_2)=e_1\otimes b+e_2\otimes d$. The coefficient relations, after cancellation of $q^{-1/2}$, areThese are exactly the defining relations of $M_q(2)$ in the inverse-parameter convention of Question 1, without setting the quantum determinant equal to one. They therefore define mutually inverse algebra homomorphisms: each named generator maps to the corresponding $t_{ij}$, and conversely. The matrix comultiplication and counit are the same, so the isomorphism is also one of bialgebras.

Here is the requested check of $ba=qab$ directly from $C$. Apply colinearity to $e_1\otimes e_2$ and compare the coefficient of $e_1\otimes e_1$. On the left, $C(e_1\otimes e_2)=q^{-1/2}{e}_2\otimes e_1$, giving $q^{-1/2}ba$. On the right, the corresponding term of $\delta (e_1\otimes e_2)$ is $e_1\otimes e_1\otimes ab$, and $C(e_1\otimes e_1)=q^{1/2}{e}_1\otimes e_1$. Hence $q^{-1/2}ba=q^{1/2}ab$, proving the relation. The other coefficient comparisons give the remaining displayed relations; no additional determinant relation is imposed in an FRT bialgebra.