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Possible Duplicate:
Functions which are Continuous, but not Bicontinuous

If $f$ is a continuous map from a subset of $\mathbb{R}^n$ to another subset of $\mathbb{R}^n$, must it have a continuous inverse? (in usual topology) Is the same true of metric spaces? When is it true/not true?

Requesting example if not.

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    @Swaprava : Sorry, I only noticed your comment now. For arbitrary topological spaces, the first two of your conditions do not imply the third. However, for manifolds they do. This is a fairly nontrivial theorem called "invariance of domain".2013-11-22

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My favorite example. Domain is the interval $[0,2\pi[\;$ in $\mathbb R$, range is in $\mathbb R^2$, formula is $f(\theta) = (\cos \theta, \sin\theta).\;$ This is a continuous map of that interval one-to-one onto a circle. But the inverse is discontinuous.

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    @GEdgar I haven't seen that interval notation for years. It's good to see it's well applied and not forgotten.2013-05-30
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The continuous function $f$ given by $f(x)=x^2$ is a counterexample. It doesn't have an inverse, let alone a continuous inverse.

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    @very Now you are confusing "or" with "and".2012-02-04
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Yes, if open subsets. http://en.wikipedia.org/wiki/Invariance_of_domain

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    I upvoted this, because despite @guy's objections, valid though they may be, this comment led me to the answer of a question I had been asking.2013-07-16