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Some caveats: I am looking for a solution that uses smoothness. I know a way to prove it assuming only Borsuk Ulam with only continuity assumed. I was hoping for a more elementary solution that uses differentiability specifically.

Some things I have tried: I wanted to relate the problem to linear algebra somehow, maybe with some sort of partial converse to the inverse function theorem. In the examples I can think of where this would fail (say $f(x)=x^3$ with trivial linearization at $0$) there aren't many "bad" points. I am not sure how to pick the good ones however.

Next, I tried something using Sard's theorem to find some regular values, and an $m-n$ dimensional manifold, after which the conclusion would be immediate. However, there is no reason to believe that these regular points should be in the image of $f$, i.e. a map $$ f:\mathbb{R}^3\to \mathbb{R}^2 $$ taking everything to the real axis.

Any thoughts and help would be appreciated.

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    $\begingroup$ Is it true that an injective smooth map is a local immersion? $\endgroup$ – Randall Nov 17 '17 at 4:11
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    $\begingroup$ @Randall I would think no by the $x\to x^3$ example? $\endgroup$ – operatorerror Nov 17 '17 at 4:21
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This file gives a short proof of "smooth invariance of domain" (really it is "$C^1$ invariance of domain").

Your result follows immediately. For suppose $f$ were injective, and compose it with $\phi$, the canonical smooth immersion of $\mathbb{R}^n$ into $\mathbb{R}^m$.

Then $\phi\circ f$ is a smooth injection from $\mathbb{R}^m$ into itself. Smooth invariance of domain implies the image must be open, but this is impossible.

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  • $\begingroup$ Someone had mentioned this to me and I thought it seemed a rather big gun. The linked proof isn't too bad though. $\endgroup$ – operatorerror Nov 17 '17 at 5:39
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    $\begingroup$ I think you need some such big gun for the general case. When $m$ and $n$ are at most $3$, you might have better luck finding simpler, more geometric arguments. (I assume you know the case $m=2$ and $n=1$ is very easy.) $\endgroup$ – symplectomorphic Nov 17 '17 at 5:59
  • $\begingroup$ @qbert: Sard's theorem is a much bigger gun than the inverse function theorem for smooth maps (which is all what's required in this proof). $\endgroup$ – Moishe Kohan Nov 25 '17 at 13:49
  • $\begingroup$ @MoisheCohen fair point. I accepted the above $\endgroup$ – operatorerror Nov 25 '17 at 17:28
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    $\begingroup$ The file states that an injective map has an injective derivative, which is not correct at all. $\endgroup$ – Shaked Feb 1 '18 at 21:00

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