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Does anyone know if there is a bijective and continous function $f: \mathbb{R} \rightarrow \mathbb{R}$ with no continuous inverse?

The only thing I have found about this is the invariance of domain theorem but I don't know if I am ok because this theorem mentions a subspace $U \subset\mathbb{R}$, not the entire space $\mathbb{R}$.

According to this theorem, there is not such function like this i.e. every bijective and continous function from $\mathbb{R}$ to $\mathbb{R}$ is a homomorphism, right?

Jean Marie
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user391120
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  • Related: http://math.stackexchange.com/questions/145639/a-continuous-bijection-f-mathbbr-to-mathbbr-is-an-homeomorphism – Henricus V. Nov 19 '16 at 22:34
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    In the case of $\mathbb{R}$, you don't need as huge a hammer as invariance of domain. If $I\subset \mathbb{R}$ is an interval (could be the whole line, but need not) and $f \colon I \to \mathbb{R}$ is a continuous injective function, then $f$ is strictly monotonic. – Daniel Fischer Nov 19 '16 at 22:34
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    Yes, with a little extra argument. By the intermediate value theorem, $f(J)$ is an interval for every interval $J \subset I$, and it's open if $J$ is open. – Daniel Fischer Nov 19 '16 at 22:42
  • Ok, thanks a lot for your answer. – user391120 Nov 19 '16 at 22:49

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