= Solution
{c}
Take $\Omega=\mathbb R^2$ and
$$
\boxed{F(s,t)=(e^s\cos t,e^s\sin t).}
$$
Its <Jacobian matrix> and <determinant> are
$$
DF_{(s,t)}=e^s\begin{pmatrix}\cos t&-\sin t\\\sin t&\cos t\end{pmatrix},\qquad \det DF_{(s,t)}=e^{2s}>0.
$$
Thus its <total derivative> is invertible everywhere, and the <inverse function theorem> makes it locally invertible. But $F(s,t+2\pi)=F(s,t)$, so it is not <injective> and has no global inverse on its image. The domain is open and <connected> as required.
Back to article page
