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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!

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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

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

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