# Show that a smooth map $f:\mathbb{R}^m\to \mathbb{R}^n$ for $m>n$ cannot be injective

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.

• Is it true that an injective smooth map is a local immersion? – Randall Nov 17 '17 at 4:11
• @Randall I would think no by the $x\to x^3$ example? – operatorerror Nov 17 '17 at 4:21

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.
• 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.) – symplectomorphic Nov 17 '17 at 5:59