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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.

J126
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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 $y$.

J126
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  • You still need to show $f^{-1}(V_1\cap\dots\cap V_k)\subset U_1\cup\dots\cup U_k$. Also, there is not just one covering map; but every covering map is single-sheeted. – Ted Shifrin Jul 05 '14 at 17:48
  • @TedShifrin I think I have fixed it. – J126 Jul 05 '14 at 18:19
  • It is harder than you think: there are local diffeomorphisms which are finite-to-one but not coverings. What you need to use is the that that f is proper. You can check for instance the path lifting property using properness. Or appeal to Ehresmann's theorem. – Moishe Kohan Jul 05 '14 at 18:28
  • @studiosus I do not see why my neighborhood of $y$ is not evenly covered. – J126 Jul 05 '14 at 18:34
  • That's because you did not write a complete argument. As I said, the data you are using is not enough to conclude that your map is a covering. – Moishe Kohan Jul 05 '14 at 18:43
  • @studiosus: You're making this too complicated. It follows immediately from compactness of the domain. The OP has fixed the argument. Now the preimage of the set in the range is contained in $U_1\cup\dots\cup U_k$, hence is evenly covered. (BTW, you could also write open set more simply as this as $W=S^n - f\big(S^n-(U_1\cup\dots\cup U_k)\big)$. This is a standard undergraduate exercise (cf. the Stack of Records Theorem in Guillemin and Pollack) to show that a local homeomorphism from a compact space onto a Hausdorff space is a covering map. – Ted Shifrin Jul 05 '14 at 18:56
  • @TedShifrin: Yes, indeed, this is very standard, but the argument written above is incomplete. (Tell me where OP used the compactness assumption.) – Moishe Kohan Jul 05 '14 at 19:05
  • @studiosus: To say that the set $W$ in the range is open!! – Ted Shifrin Jul 05 '14 at 19:06
  • Oh, I see, then it is all fine. – Moishe Kohan Jul 06 '14 at 12:15