Solution
= Solution
Yes. For any $R>1$, the formula
$$
[(x,t)]\longmapsto
\bigl((R+t\cos(x/2))\cos x,
(R+t\cos(x/2))\sin x,
t\sin(x/2)\bigr)
$$
defines the standard smooth embedding of the open <Möbius band> into $\mathbb R^3$. Replacing $(x,t)$ by $(x+2\pi,-t)$ leaves the displayed point unchanged. The image is the interior of a compact Möbius strip, so the embedding fails to be a <proper map>; this is why it does not contradict part (d).