$\newcommand{\GLm}{\text{GL}_n^-}$Let $A$ be a real $n \times n$ matrix with non-positive determinant. Suppose that the smallest singular value of $A$ is **strictly** smaller than all the others (it has multiplicity $1$).

Question:Do there exist an open neighbourhood $O$ of $A$, and smooth maps $U:O \to \operatorname{SO}_n$, $V:O \to \operatorname{O}_n^-$ such that $$ X=U(X)\Sigma(X)V(X)^T$$ holds for every $X \in O$, where $\Sigma(X) = \operatorname{diag}\left( \sigma_1(X),\dots\sigma_n(X) \right)$, and $\sigma_1(X)$ is thesmallestsingular value of $X$?

Note that I don't care about the ordering of $\sigma_2,\dotsc,\sigma_n$, but I specifically want the minimal singular value to be in a *fixed* position.

**Comment 1:** If we replace the requirement that $\sigma_1 $ has multiplicity $1$ by the requirement that all the singular values are distinct, then the answer is positive: In that case the map
\begin{align*}
\mu: \operatorname{SO}_n\times \mathcal D\times \operatorname{O}_n^-\to Y \\
(U,\Sigma,V)\mapsto U\Sigma V^T
\end{align*}
is a local diffeomorphism, so is locally invertible (here $\mathcal D$ is the space of $n\times n$ diagonal matrices with strictly increasing positive entries, and $Y=\{A \,|\, \text{$\det A<0$ and all the singular values of $A$ are distinct}\}$). This works when $\det A <0$. When $\det A=0$ (and all its singular values are distinct) a slight adaptation of this argument works.

**Comment 2:** If such maps $U$, $V$ exist, then the map $A \to \Sigma(A)$ is also smooth. In general the *ordered* singular values cannot be chosen smoothly when they "cross", but here I don't require to keep the ordering of $\sigma_2,\dotsc,\sigma_n$ *fixed*, so I think there is no obstruction, but I may be wrong.