 Solution

No. On $\Omega ={\mathbb{R}}^2$ letIt is a bijection onto ${\mathbb{R}}^2$, with inverse $(u,v)\mapsto (\sqrt[3]{u},v)$. Nevertheless $D{F}_{(s,t)}=\mathrm{diag}(3s^2,1)$ is singular whenever $s=0$. Set-theoretic invertibility, even together with a continuous inverse, does not force invertibility of the total derivative. If the inverse were assumed differentiable, the chain rule applied to $F^{-1}\circ F$ would force both derivatives to be invertible; that stronger assumption is absent here.