If $\varphi:\Omega \to \mathbb R^n$ and $\Omega\subset \mathbb R^n $ not supposed open, does $\varphi : \Omega \to \varphi(\Omega )$ a homeomorphism?

Question

Let $\Omega \subset \mathbb R^n$ and $\varphi:\Omega \to \mathbb R^n$ an injective and continuous function. I know that if $\Omega $ is open then $\varphi(\Omega )$ is open an $\varphi:\Omega \to \varphi(\Omega )$ is an homeomorphism (by Brouwer invariance domain).

What happen if $\Omega $ is not supposed open ? Will $$\varphi:\Omega \to \varphi(\Omega )$$ also be a homeomorphism or not ? If not, could you give a counter example ?

Answer 1

How about $n=2$, $\Omega=\{(x,0):0\le z<2\pi\}$ (neither open nor closed) and $f(x,0)=(\cos x,\sin x)$? This takes a non-compact interval continuously and injectively to a circle (which is compact).

Answer 2

It is not a homeomorphism in general. Take$$\begin{array}{rccc}f\colon&\mathbb{Z}^+&\longrightarrow&\mathbb{R}\\&n&\mapsto&\begin{cases}0&\text{ if }n=0\\\frac1n&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is injective, but not a homeomorphism onto its image.

Answer 3

Consider $$f(x)=\left\{\begin{array}{lc}e^x &-\infty<x\le -1\\ 0 & x=0\end{array}\right.$$

$f$ is continuous and injective. But $\Omega=(-\infty,-1]\cup\{0\}$ is not homeomorphic to $f(\Omega)=[0,1/e].$