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Let $f:\mathbb R\to\mathbb R$ is bijective and continuous, does $f^{-1}$ is also continuous ?

Does this result hold for $f:U\to\mathbb R$ where $U\subset \mathbb R^n$ ? and for $f:V\to\mathbb R^m$ where $V\subset \mathbb R^n$.

Actually, I ask this question since an homeomorphism is bijective function, continuous and such that the inverse is also continuous. But to me, if $f$ is continuous, the the inverse is also continuous. Therefore, if in the definition we precise $f^{-1}$ continuous, I guess that there is case where $f^{-1}$ is not continuous, but I can't find any example. (I'm only working with $\mathbb R^n$, i.e. don't give me complicate example with strange topology... only $\mathbb R^n$ with usual topology). Do you have an example ?

Martin Sleziak
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idm
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  • There do not exist any continuous injections from an open subset of $\Bbb R^n$ to $\Bbb R^m$ when $m>n$. If $m=n$, then a continuous injection will be an open map. – Stefan Hamcke Nov 30 '15 at 11:05
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    A bijective continuous function from $f: \mathbb{R} \to \mathbb{R}$ is a homeomorphism. – Alvin Jin Nov 30 '15 at 11:07
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    All continuous bijections $\mathbb{R}\rightarrow \mathbb{R}$ are homeomorphisms. http://math.stackexchange.com/questions/59532/bijective-continuous-function-on-mathbb-rn-not-homeomorphism But a good answer would be using only real analysis. – João Victor Bateli Romão Nov 30 '15 at 11:26
  • Duplicate? http://math.stackexchange.com/questions/145639/a-continuous-bijection-f-mathbbr-to-mathbbr-is-an-homeomorphism – João Victor Bateli Romão Nov 30 '15 at 11:36
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    If you don't require your domain to be open, examples of continuous bijections that aren't homeomorphisms are easy to come by. For example a continuous bijection $(-\infty,0) \cup [1,+\infty) \to \mathbb{R}$ exists, but can't be a homeomorphism. If you require the domain to be open, then as Stefan said a continuous bijection is a homeomorphism. That's a nontrivial result (for $n > 1$, it's easy in dimension one), known as "invariance of domain". – Daniel Fischer Nov 30 '15 at 11:50
  • Ok, I got it. Thanks to all of you. – idm Nov 30 '15 at 11:51

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