= Solution
{c}
\b[No.] On $\Omega=\mathbb R^2$ let
$$
\boxed{F(s,t)=(s^3,t).}
$$
It is a <bijection> onto $\mathbb R^2$, with inverse $(u,v)\mapsto(\sqrt[3]u,v)$. Nevertheless $DF_{(s,t)}=\operatorname{diag}(3s^2,1)$ 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.
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