10
$\begingroup$

What is the topological version of the following theorem?

Open Mapping Theorem. Let $\Omega \subset \mathbb{R}^n$ be open and $f: \Omega \rightarrow \mathbb{R}^m$ a continuously differentiable function. If for every $x\in\Omega$ the derivative $f'(x)$ is surjective and $U\subset \Omega$ is open, then the image $f(U)$ is open in $\mathbb{R}^m$.

So previous derivative matrix is surjective. For a weaker version where $n=m$ and derivative matrix is bijective, i.e. $\det f'(x) \neq 0$, we have the corresponding Invariance of Domain Theorem by Brouwer that is based on injectivity. But is there any similar results for that stronger form of the Open Mapping Theorem?

EDIT: Is a (continuous) linear injection only suitable topological map?

$\endgroup$
1
  • 1
    $\begingroup$ I do not know, but here is one conjecture. Let $f: R^n\to R^m$ be a continuous map with Chech-acyclic fibers of topological dimension $n-m>0$. Then $f$ is a open. (Note that some form of acyclicity of fibers is needed, it is an analogue of the condition that in Brouwer's theorem the map is injective, rather than has discrete fibers, which is not enough to ensure openness.) Such a theorem would be a form of the Vietoris-Begle theorem, see Iverson's book "Cohomology of sheaves". $\endgroup$ – Moishe Kohan Aug 8 '17 at 19:06
3
$\begingroup$

This was too long for a comment, it is not an answer.

I don't know of any analogues in some arbitrary topological space. here is a list of counter-examples for relatively well-behaved spaces.

One really needs some kind of injectivity requirement, which gives the well-known result: for $X$ compact and $Y$ Hausdorff, a bijective continuous map $f:X \to Y$ is open (and hence a homeomorphism.)

If you add some additional algebraic structure, you could recover the open mapping theorems for functional analysis and sufficiently nice topological groups.

A lukewarm observation: if we let $U \subset \mathbb R^n$ be open, then if $f: U: \to \mathbb R^m$ factors through projection $\pi:\mathbb R^n \to \mathbb R^m$ so that the induced map $\tilde{f}:U \to \mathbb R^m$ is injective, then $f$ is open since it is the composition of open maps.

Hopefully someone else can provide a better answer.

$\endgroup$
1
  • $\begingroup$ Corollary of your last statement: suppose U Is homeomorphic to a ball. Then f is open. Indeed, one can easily extend the function to the whole R^n! $\endgroup$ – Andrea Marino Sep 17 '19 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.