Continuous function whose inverse is not continuous

Question

Suppose $f:\mathbb{R}^n\to\mathbb{R}^n$ is bijective and continuous. Is it possible that $f^{-1}$ is not continuous?

I can prove that for $n=1$ it is not possible, i.e. if $f:\mathbb{R} \to \mathbb{R}$ is bijective and continuous then $f^{-1}$ has to be continuous.

I also know that one can easily construct an example of a function $f : \mathbb{R}^2 \to S$, for some $S\subset\mathbb{R}^2,$ such that $f$ is bijective and continuous but $f^{-1}$ is not continuous. But this is different from what the above question is demanding.

Do you have such an example? If $S$ is a convex open subset of $\Bbb R$, then we could exploit this to answer your question. — Ben Grossmann, Jan 17 '20 at 14:37
@above, see this: https://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous — Aditya Ghosh, Jan 17 '20 at 14:39
@AdityaGhosh: No, this is not possible due to the invariance of domain theorem (see https://en.wikipedia.org/wiki/Invariance_of_domain). — levap, Jan 17 '20 at 14:41
Does this answer your question? bijective continuous function on $\mathbb R^n$ not homeomorphism? — Alejandro Tolcachier, Jan 17 '20 at 14:48

score 1 · Accepted Answer · answered Jan 17 '20 at 14:57

No, the invariance of domain theorem tells us that a bijective map from an open $U \subseteq \Bbb R^n \to \Bbb R^n$ is a homeomorphism between $U$ and $f[U]$. (an open subset used to be called a "domain", hence the name). In particular $f$ is open (which for a bijection this implies continuity of $f^{-1}$).

This is quite typical for $\Bbb R^n$ though, and quite non-trivial to show.

Continuous function whose inverse is not continuous

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