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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.

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    $\begingroup$ The most natural sufficient condition that comes to mind is that $f$ be a $C^1$ submersion. $\endgroup$
    – Georges Elencwajg
    Commented Dec 3, 2013 at 13:46
  • $\begingroup$ One remark: the discreteness above should be replaced by lightness. Discreteness means that the fiber of f does not have accumulation points. $\endgroup$
    – Changyu Guo
    Commented Dec 3, 2013 at 13:59
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    $\begingroup$ There is classical nice sufficient condition (Brouwer): a continuous injective map from an open subset of $\mathbf R^n$ to $\mathbf R^n$ is open. However, it looks like it is superseded by the theorem of Titus-Young you mention. $\endgroup$
    – ACL
    Commented Mar 6, 2014 at 13:01
  • $\begingroup$ To ACL: the result you mentioned is the invariance of the domain, which is a simply application of degree theory. Here I am looking for more analytic assumptions on $f$, instead of strong topological assumptions, like you mentioned local injectivity. $\endgroup$
    – Changyu Guo
    Commented Mar 6, 2014 at 14:00

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The zero one-dimensional Hausdorff measure assumption on fibers is NOT a sufficient condition for openness. The smooth map (x,y) to (x square, y) on the plane is a counterexample. According to Titus and Young, one has to impose another assumption.

The zero one-dimensional Hausdorff measure of critical points is a sufficient condition when n is greater than or equal to 2. For details, see Titus and Young’s 1952 paper.

Liangpan Li @ 2021

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