I am working on a problem which asks me to show that for any smooth map $f : S^3 \rightarrow \mathbb{R}^3$, there must exist at least one point $p \in \mathbb{R}^3$ where there fails to exist a local inverse for $f$.
I was hoping that someone could tell me if my reasoning works:
Suppose, FSOC, that for all $p \in \mathbb{R}^3$ there is a local inverse for $f$. By the inverse function theorem, this means that $f$ is a local diffeomorphism. Since $S^3$ is compact, this means that $f$ is a proper local diffeomorphism. Hence $f$ is a covering map. Since $S^3$ is simply connected, $f$ must exhibit it as a universal cover of $\mathbb{R}^3$. But $\mathbb{R}^3$ is also simply connected and so is a universal cover of itself. We know that universal covers of any given space are homeomorphic, and so we have reached a contradiction since $S^3$ and $\mathbb{R}^3$ are certainly not homeomorphic. Thus there must exist some point $p \in \mathbb{R}^3$ about which there is no local inverse for $f$.
Any thoughts or guidance on this problem/my reasoning would be greatly appreciated. In particular, I would be very interested to see alternative solutions. Thanks.
Yes, this is a valid proof (and the standard one to this sort of problem). Another way to achieve the desired contradiction is to note that since $\mathbb R^3$ is noncompact, any covering space of it must be noncompact.