Proving a Certain Smooth Map $S^n\rightarrow S^n$ is a Diffeommorphism

Question

I am given a smooth map $f:S^n\rightarrow S^n$, for $n\geq 2$, whose differential is injective at each point. I am asked to prove that it is a diffeomorphism. Since the differential is injective between manifolds of the same dimension, it is also surjective. This makes $f$ a submersion. Submersions are open and $S^n$ is both compact and Hausdorff. Thus the image of $f$ is both open and closed. Since $S^n$ is connected, this makes $f$ surjective. The problem I am having is proving that $f$ is injective. Any help is appreciated.

Answer 1

I think I have it. There is only one covering map $S^n\rightarrow S^n$, up to isomorphism, for $n\geq 2$. If I can show that $f$ is a covering map I will be done. Since every value $y\in S^n$ is regular, the set $f^{-1}(y)$ is finite, say $\{x_1\ldots,x_k\}$. By the Inverse Function Theorem and the fact that $S^n$ is Hausdorff, I can find pairwise disjoint neighborhoods $x_i\in U_i$ that map homeomorphically to some neighborhood $y\in V_i$. Then $V_1\cap \cdots\cap V_k-f(S^n-(U_1\cup\cdots\cup U_k))$ is an evenly covered neighborhood of $y$.