Open mapping and space filling

Question

Suppose $n<m$. Is there a continuous function $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ that is also an open mapping (maps open sets to open sets)? For example, are any of the standard space-filling curves open mappings?

Answer 1

A continuous mapping $f: R\to R^m, m>1$, cannot be open. Indeed, the image $f([0,1])$ is a compact $C$ with continuum of frontier points (unless $C$ is a point); hence, $f((0,1))$ cannot be open.

On the other hand, for each $m>3$ there exists a continuous open map $f: R^3\to R^m$, see

L. V. Keldysh, A monotone mapping of the cube onto the cube of greater dimension, Matem. Sb., 43:2 (1957), 129–158

and the follow-up paper where the construction was completed:

L. V. Keldysh, Transformation of a monotone irreducible mapping into a monotone-open one, and monotone-open mappings of the cube which raise the dimension. Mat. Sb. N.S. 43 (85) 1957 187–226.

A proof in English can be found in

D. Wilson, Open mappings on manifolds and a counterexample to the Whyburn conjecture, Duke Math. J. 40 (1973), 705-716.

In these papers the authors construct open mappings $f_m$ of the 3-dimensional cube $I^3$ onto $m$-dimensional cube $I^m$ for all $m\ge 4$. In order to get similar maps $R^3\to R^m$ one proceeds as follows:

Restrict $f_m$ to a small open ball $B^3$ in the interior of $I^3$, whose image is disjoint from the boundary of $I^m$ (such $B^3$ exists due to continuity of $f_m$). Then $f_m(B^3)$ is an open domain in $R^m$. Now, use the fact that $B^3$ is homeomorphic to $R^3$. You can also promote this to a surjective map $R^3\to R^m$ using the fact that every open nonempty subset in $R^m$ admits an open map to $R^m$. (First prove it for $m=2$ using the maps of the form $z\mapsto z^2$, $z\in {\mathbb C}$.)

I do not know about open maps $R^2\to R^m$. There are reasons to expect that they do not exist.