TITLE: Is there a differentiable map surjective from low to high dimension？
QUESTION [37 upvotes]: Does there exist a map $f:\Bbb R^n \rightarrow \Bbb R^m$, where $n<m$ and $ n,m \in\Bbb N^+$ such that $f$ is surjective and differentiable?

REPLY [10 votes]: I think the image of a differentiable map $f:\mathbb R^n \to \mathbb R^m$ with $n<m$ always has measure $0$. As mentioned my @YCor in another answer, this follows from the local Lipschitz property of $f$ when $f$ is $\mathcal C^1$. Let us try to remove this hypothesis.
First, covering $\mathbb R^n$ by countably many compact sets, we reduce to proving that $f(K)$ has measure $0$ for any compact set $K$.
Fix $\epsilon >0$. For every $x \in K$, Since $f$ is differentiable at $x$, $f(B(x,r))$ is contained in a $o(r_x)$-neighbourhood of a euclidean $n$-dimensional disc of radius $\Vert \mathrm d_x f\Vert r_x$, and thus has measure
$$\Vert \mathrm d_x f\Vert^n r_x^n o(r_x^{m-n})~.$$
We can thus choose $r_x$ sufficiently small so that $f(B(x,r_x))$ has Lebesgue measure at most $\epsilon r_x^n$.
Now, the balls $B(x,\frac{1}{3}r_x)$, $x\in K$ obviously cover $K$ and we can extract a finite cover from it. Then by Vitali's lemma, we can find $x_1,\ldots, x_k$ such that the balls $B(x_i, \frac{1}{3}r_{x_i})$ are disjoint and the balls of radius $r_{x_i}$ cover $K$.
By disjointness of the balls $B(x_i,\frac{1}{3}r_{x_i})$, we get that
$$\sum_i r_{x_i}^n \leq \mathrm{Constant}\cdot \mathrm{Leb}(K)~.$$
Since the balls $B(x_i, r_{x_i})$ cover $K$ we get that
$$\mathrm{Leb}(f(K)) \leq \sum_i \mathrm{Leb}(f(B(x_i,r_{x_i})) \leq \epsilon \sum_i r_{x_i}^n~.$$
Taking $\epsilon$ arbitrarily small, we conclude that $f(K)$ has measure $0$.
Remark: the only regularity we need is that for every $x$, the image of $f(B(x,r))$ has volume $o(r^n)$ when $r$ goes to $0$ (with $o(r^n)$ possibly depending on the point $x$). This is satisfied for instance if $f$ is $(\frac{n}{m}+ \epsilon)$-Hölder for any $\epsilon >0$.