Finite map $\mathbb{R}^n \rightarrow \mathbb{R}^m$

Question

Suppose $f :\mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous and has all of its fibers being finite. I believe it must follow that $n \leq m $, but I can’t seem to prove it really. The case $m=1$ is kind of easy, since if $f$ is not constant we can choose infinitely many (up to endpoints) disjoint paths from $f(x)$ to $f(y)$ with $f(x) \neq f(y)$ and so by the IVT every value between $f(x)$ and $f(y)$ is now reached infinitely many times. I don’t think this argument can be generalized. Maybe some homology/cohomology/homotopy argument is possible?

Answer 1

It is a nice and classical question, solved by Witold Hurewicz about 90 years ago:

Definition. Let $f: X\to Y$ be a map of topological spaces. Then $\dim(f)$ is defined as the supremum of dimensions of the subsets $f^{-1}(y), y\in Y$. In particular, if $f$ is a finite-to-one map, $\dim(f)=0$.

Theorem. Suppose that $X, Y$ are separable metrizable spaces (say, $R^n$ and $R^m$ respectively) and $f: X\to Y$ is a closed map (i.e. a map sending closed subsets to closed subsets). Then $$ \dim(X)\le \dim(Y) + \dim(f). $$

You can find a proof of this result (which is one of the "Hurewicz formulae") for instance in

Hurewicz, Witold; Wallman, Henry, Dimension theory, Princeton: Princeton University Press. 165 p. (1948). ZBL0036.12501.

Now, suppose that $f: R^n\to R^m$ is a continuous map. Let $X\subset R^n$ be the closed unit ball. Then the restriction $f|_X$ is a closed map (by compactness of $X$). For $f$ which is finite-to-one, Hurewicz formula above would tell us that $$ n=\dim(X)\le m=\dim(Y), $$ since $\dim(f)=0$. Instead of assuming that $f$ is finite-to-one, it sufficies even to assume that $f$ is "light", i.e. point-preimages are totally disconnected.