Scalar twist correction removes the framing

By [F6], the assumed Axiom of Choice supplies countable choice, so [F1]'s closure comparison identifies every $t_n$ in the Statement with the algebraic trace. Fix $\gamma \in B_n$ and put $j=j_{X^{\otimes n}}$, $\rho ={\rho }_n$. Apply [F2] with $a=j\rho (\gamma )\rho (\beta )$ and $c=\rho (\gamma {)}^{-1}$ to obtain $t_n(\gamma \beta {\gamma }^{-1})={\mathrm{Tr}}_L((\rho (\gamma {)}^{-1}{)}^{\vee \vee }j\rho (\gamma )\rho (\beta ))$. Naturality of $j$ gives $(\rho (\gamma {)}^{-1}{)}^{\vee \vee }j=j\rho (\gamma {)}^{-1}$, so this is $t_n(\beta )$. The exponent sum is unchanged by [F4], hence $J_X$ is invariant under conjugation.

For a positive or negative stabilization, [F3] multiplies $t_n$ by ${\lambda }^{\pm 1}$ and [F4] changes $w$ by $\pm 1$. Hence ${\lambda }^{-w({\iota }_n\beta {\sigma }_n^{\pm 1})}{t}_{n+1}({\iota }_n\beta {\sigma }_n^{\pm 1})={\lambda }^{-w(\beta )\mp 1}{\lambda }^{\pm 1}{t}_n(\beta )={\lambda }^{-w(\beta )}{t}_n(\beta )$. The scalar is available because [A1] makes the invertible twist a unique $\lambda \in k^{\times }$; the positive sign is fixed by [F3]. Thus $J_X$ is invariant under both stabilizations and their inverses.

By [F5], any two braids with equivalent oriented unframed closures are connected by finitely many conjugations, stabilizations, and destabilizations. Steps 1.1 and 2.1 preserve $J_X$ under each such move, so the corrected value depends only on the oriented unframed link type. The Axiom of Choice [A2] is used in [F5] and supplies the countable-choice hypothesis of [F3] by [F6]; no choice is used in the finite trace or scalar calculations.