A decomposable color with nonscalar twist has no single writhe correction

Let $\mathcal{C}$ have objects finite-dimensional rational vector spaces $U$ equipped with a linear involution $S_U$ ($S_U^2=1$), and morphisms commuting with the involutions. Use $S_{U\otimes W}=S_U\otimes S_W$, the ordinary flip braiding, standard vector-space duality, and define ${\theta }_U=S_U$. This is a ribbon category: the twist is natural because every morphism commutes with $S$; its tensor rule is $S_{U\otimes W}=S_U\otimes S_W$ and the double flip is the identity; and on the dual, $S_{U^{\ast }}=(S_U^{-1}{)}^T=S_U^T=({\theta }_U{)}^{\ast }$, giving ribbon dual compatibility. The unit is $\mathbb{Q}$ with trivial involution, so $\mathrm{End}(\mathbf{1})=\mathbb{Q}$. By [F7], the Drinfeld morphism in this symmetric vector-space category is the canonical double-dual identification: on $v\in U$ and $f\in U^{\ast }$ its evaluation is $u_U(v)(f)=f(v)$. Thus the ribbon-induced pivotal map $j_U=u_U{\theta }_U$ is the canonical double-dual identification composed with $S_U$.

Take $X=\mathbb{Q}{e}_{+}\oplus \mathbb{Q}{e}_{-}$ with $S_X=\mathrm{diag}(1,-1)$. It is decomposable and its twist is invertible but nonscalar. For the identity $e\in B_1$, [F1] gives $t_1(e)=\mathrm{Tr}(S_X)=1+(-1)=0$. For ${\sigma }_1\in B_2$, the braid action is the flip $P$. By [F8], under the canonical vector-space double-dual comparison, $j_{X\otimes X}$ has matrix $S_X\otimes S_X$, so the trace of $(S_X\otimes S_X)P$ is $2$: the basis tensors $e_{+}\otimes e_{+}$ and $e_{-}\otimes e_{-}$ contribute diagonal entries $1$, while the two mixed basis tensors are exchanged and contribute no diagonal entries.

By [A1] and [F9], countable choice holds and the traces in step 2.1 are the actual blackboard-framed closure values; the comparison applies to this nonsimple color. The braid ${\sigma }_1$ is the positive Markov stabilization of the identity braid in $B_1$, so [F3] says their oriented unframed closures are equivalent. Their exponent sums are $w(e)=0$ and $w({\sigma }_1)=1$. For any proposed $\lambda \in {\mathbb{Q}}^{\times }$, the corrected values are $0$ and $2{\lambda }^{-1}$, which are unequal because $2\mathrm{\ne }0$ and $\lambda$ is invertible. Thus these framed evaluations admit no scalar writhe correction giving an unframed invariant. The counterexample has precisely the missing hypothesis identified by [F2]: its invertible twist is not scalar on the decomposable color.