Given a continuous mapping $f$ between Euclidean domains (or domains in topological manifolds) of the same (topological) dimension, what are the natural assumptions to conclude that $f$ is open? Here, open means that f maps open sets to open sets.

The only way I know is imposing certain regularity assumption on $f$ to
conclude that the fiber of $f$ has zero one-dimensional Hausdorff measure, i.e.,
$H^1(f^{-1}(y))=0$ for all $y$. Then one invokes a result of C. Titus, G. Young, *The extension of interiority, with some applications*, Trans. Amer. Math. Soc. 103 (1962) 329–340 to conclude that $f$ is open. Actually, the latter theorem implies that $f$ is both discrete and open. Here discreteness means that the fiber of $f$ does not have accumulation points.

Some update: In the following paper of Bonk and Kleiner, it was proved that a mapping of bounded multiplicity from $X$ to $\mathbb{R}^n$ is open, provided that $X$ is a compact metric space such that every non-empty open subset of $X$ has topological dimension at least $n$.

Bonk, Mario; Kleiner, Bruce Rigidity for quasi-Möbius group actions. J. Differential Geom. 61 (2002), no. 1, 81–106.

sufficientcondition that comes to mind is that $f$ be a $C^1$ submersion. $\endgroup$ – Georges Elencwajg Dec 3 '13 at 13:46