# Does there exist a continuous, open, and surjective map from $f\colon\mathbb{R}^n\to\mathbb{R}^m$ for $m>n$?

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.)

• 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. – Connor Malin Mar 1 at 5:16
• 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! – Nemesis Mar 1 at 16:56
• math.stackexchange.com/questions/1692266/… – Moishe Kohan Mar 1 at 17:10
• This is essentially a duplicate of the linked question. The only difference is the surjectivity requirement which can be easily done. – Moishe Kohan Mar 1 at 21:44
• 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. – 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

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