Example of continuous bijective function s.t. the inverse in not continuous.

Question

I saw an exercise today that says that : Let $f:\mathbb R\to \mathbb R$ a continuous and injective function. Show that the inverse $g:f(\mathbb R)\to \mathbb R$ is continuous. And under, there is a remark that says that it's not true in higher dimension. Could someone give me an example ? If $f:\mathbb R^n\to \mathbb R^n$ is injective and continuous, I know by Brouwer invariance domain that $f:\mathbb R^n\to f(\mathbb R^n)$ is a homeomorphism, so it should be a weirder function. Unfortunately, I can't imagine it. Could someone give me an example of such function ? May be it's a function of $$\mathbb R^n\supset \Omega \to f(\Omega )\subset \mathbb R^m,$$ but I don't think that there is bijective function if $n\neq m$. If such function doesn't exist in finite dimension, may be in infinite dimension it does ?

Answer 1

Consider the map $f:\Bbb R\to\Bbb R^2$, $$f(x)=\begin{cases}\left(\frac{1-x^2}{x^2+1},\frac{2x}{x^2+1}\right)&\text{if }x\ge 0\\ (x+1,0)&\text{if }x<0\end{cases}.$$ It parametrizes injectively the union of the ray $(-\infty,1]\times\{0\}$ with the upper half of the unit circle ($f(\Bbb R)$ is sort of a $P$-shaped set). Notice, however, that $f^{-1}:P\to\Bbb R^2$ cannot be continuous, because $f$ sends the non-compact set $[-2,\infty)$ onto the compact set $$\{(a,b)\in\Bbb R^2\,:\, (a^2+b^2=1\land b\ge 0)\lor(-1\le a\le 1\land b=0)\}.$$

This $f$ specifically turns out not to be differentiable, but it is entirely possible to parametrize a figure like the symbol "$\infty$" by a $C^\infty$ function $f:\Bbb R\to\Bbb R^2$ such that $f'$ has no zeros.