Let $F:\mathbb{R}^n \to \mathbb{R}^m$ ($m < n$) be a Lipschitz function whose Jacobian determinant $J F$ does not vanish on a compact set $A \subseteq \mathbb{R^n}$. Assume $J F$ exists everywhere.
Does this imply $J F(x)$ is bounded away from zero for all $x \in A$?
I know this would follow from $x \mapsto J F (x)$ being a continuous map or even a closed map, but I don't see why either of those should be true.
The Jacobian determinant of $F$ is defined as $$J F(x) = \sqrt{ \text{det}( DF(x) DF(x)^T ) },$$ where $D F$ is the $m \times n$ matrix of partial derivatives of $F$.
Take a counterexample with $n=m=1,$ and compose with a projection to get a map $\mathbb R^n\to\mathbb R$ satisfying the requriements.