My question is that from above. Here are my approaches so far:

I know that there is no homeomorphism between an open set of $\mathbb R^n$ and an open set of $\mathbb R^m$. So if there is an open set where $ f $ is injective we get a contradiction.
Furthermore, since $ f $ is surjective there are right-inverses of $ f $. If there was a continuous right-inverse, we would also get a contradiction since this right-inverse would be a continuous injection from $ \mathbb R^m $ to $ \mathbb R^n $ which cannot exist by Borsuk-Ulam.

Unfortunately, I was not able to use one of these two approaches to give an answer to my question.

If the answer is yes, I would also be interested in stronger assumptions on $ f $ to make the answer no. I wonder if uniform continuity does the job, since for Hoelder continuity and large enough $ m $ the answer is no even if we drop the openness of $ f $ (This one can prove using Hausdorff-Dimension and how Hoelder continuous maps preserve them.)

Thanks for your help!

  • 1
    $\begingroup$ I am not so experienced with this, but one observation is the following: A continuous, surjective, open map is a quotient map. So a more "physical" way of looking at it is saying "is $\mathbb{R}^m$ a quotient of $\mathbb{R}^n$? Then one idea might be exploiting well understood things, like if the quotient is to be Hausdorff, given two convergent sequences $a_1,a_2,\dots$, $b_1.b_2,\dots$ so that $a_i \sim b_i$ the $lim\: a_i ~ lim\:b_i$. Then if much is known about preimages of points of space-filling curves, you might be able to show that the quotient can't be Hausdorff. $\endgroup$ – Connor Malin Mar 1 at 5:16
  • 1
    $\begingroup$ I am not shure how to use properties of space filling curves (maybe I do not know right one) but your comment @ConnorMalin made me to think first about the case $n=1$. And here the answer is in fact no, since I can compose $ f $ with a projection onto the first component. Then this is an open continuous and surjective map from $ \mathbb R $ to $ \mathbb R $ wich has then to be a homeomorphism. But then $ f $ itself can not be surjectve anymore. Thank you! $\endgroup$ – Nemesis Mar 1 at 16:56
  • 1
    $\begingroup$ math.stackexchange.com/questions/1692266/… $\endgroup$ – Moishe Kohan Mar 1 at 17:10
  • $\begingroup$ This is essentially a duplicate of the linked question. The only difference is the surjectivity requirement which can be easily done. $\endgroup$ – Moishe Kohan Mar 1 at 21:44
  • $\begingroup$ I see the analogy. But what do you mean by "which can be easily done"? I think surjectivity is a quite strong property, so it should make a difference. And all the cited articles speak of cubes.How do I translate the results to the whole $\mathbb R^n$ and $\mathbb R^m$? I would be thankful for a detailed answer, since I am not that experienced in this field. $\endgroup$ – Nemesis Mar 2 at 13:06

Theorem 1. For every $n> m\ge 3$ there exists a continuous open mapping $f: R^m\to R^n$.

Proof. I will give a proof which is a variation on my answer to this question.

The key result is a rather nontrivial theorem due to John Walsh (he proved something stronger, I am stating a special case):

Theorem 2. Fix $n, m\ge 3$. Then for any pair of compact connected triangulated manifolds (possibly with boundary) $M, N$ of dimensions $m, n$ respectively, every continuous map $g: M\to N$ inducing surjective map of fundamental groups $\pi_1(M)\to \pi_1(N)$ is homotopic to a surjective open continuous map $h: M\to N$.

See corollary 3.7.2 of

J. Walsh, Monotone and open mappings on manifolds. I. Trans. Amer. Math. Soc. 209 (1975), 419-432.

This deep theorem is a generalization of earlier results on existence of open continuous dimension-raising maps from $m$-cubes to compact triangulated manifolds due to Keldysh and Wilson.

The next part of the proof uses some basic algebraic topology covered, say, in Hatcher's "Algebraic Topology".

Take $N=T^n$, the $n$-dimensional torus ($n$-fold product of circles). Its fundamental group is ${\mathbb Z}^n$. Let $S$ be a compact connected oriented surface of genus $n$. Its fundamental group admits a surjective map to ${\mathbb Z}^{2n}$ (given by the abelianization) and, hence, to ${\mathbb Z}^{n}$. Consider the manifold $M$ which is the product $S\times T^{m-2}$. Its fundamental group admits an epimorphism to ${\mathbb Z}^{n}$. The universal covering spaces of the manifolds $M$ and $N$ are homeomorphic to ${\mathbb R}^m$ and ${\mathbb R}^n$ respectively.

Since the manifold $N$ is $K( {\mathbb Z}^n, 1)$, Whitehead's theorem implies that the epimorphism $$ \pi_1(M)\to \pi_1(N) $$ is induced by a continuous map $g: M\to N$. Applying Walsh's theorem, we obtain that $g$ is homotopic to an open map $h: M\to N$. Lifting $h$ to the universal covering spaces we obtain a continuous open map $\tilde{h}: {\mathbb R}^m\to {\mathbb R}^n$. I claim that $\tilde{h}$ is a surjective map. Indeed, the map $h$ is surjective (since otherwise the image $h(M)$ is a proper closed and open subset of $N$ contradicting connectivity of $N$). Since the map $\tilde{h}$ is equivariant with respect to the actions of the fundamental groups of $M, N$ on the respective universal covering spaces, the image $\tilde{h}({\mathbb R}^m)$ is invariant under the covering group $\Gamma$ of the universal covering ${\mathbb R}^n\to T^n$. Therefore, surjectivity of $h$ implies surjectivity of $\tilde{h}$.

Theorem 1 follows. qed

  • $\begingroup$ Thank you very much for your work! $\endgroup$ – Nemesis Mar 4 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.