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?
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.