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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.

Community
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veryveryverycoolusername
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    No: Several examples appear at Functions which are Continuous, but not Bicontinuous. – Jonas Meyer Feb 04 '12 at 05:50
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    In general, it's very false, as others pointed out; but if $f$ is a continuous bijection (a necessary condition for having an inverse) between open subsets of $\mathbb{R}^n$, then $f$ does have a continuous inverse: this is known as the invariance of domain theorem (domain being an old name for open subset of $\mathbb{R}^n$), due to Brouwer, and quite hard to prove from first principles. – Henno Brandsma Feb 04 '12 at 07:32
  • Not only is it false that a continuous map between Euclidean subspaces must have a continuous inverse,this is not even true in general if THE MAP IS A CONTINUOUS BIJECTION BETWEEN THE SPACES! – Mathemagician1234 Feb 04 '12 at 17:11
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    @Mathemagician1234 : You are wrong (and there is no need to shout). A continuous bijection between two manifolds automatically has a continuous inverse. Go read the wikipedia article on the invariance of domain theorem. – Adam Smith Feb 05 '12 at 06:34
  • @AdamSmith: I also want to understand this thing. Let us look at the definition of homeomorphism in this page: it says $f$ is a homeomorphism if it is continuous, a bijection, and inverse is also continuous. If the first two implied the third, why did we need the third one in a definition? I guess there are examples which are continuous bijections but inverse isn't continuous. – Pagol Oct 11 '13 at 05:50
  • @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". – Adam Smith Nov 22 '13 at 22:36

3 Answers3

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My favorite example. Domain is the interval $[0,2\pi[\;$ in $\mathbb R$, codomain 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.

GEdgar
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    Do you mean $[0,2\pi)$? – katari Oct 05 '12 at 02:16
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    Yes, but an alternate notation is $[0,2\pi[$. To avoid confusion, Bourbaki style. Note that (in the same sentence) $(\cos \theta, \sin\theta)$ is not an open interval but an ordered pair. – GEdgar Oct 05 '12 at 03:26
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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. – J. A. Corbal May 30 '13 at 12:30
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Yes, if open subsets. http://en.wikipedia.org/wiki/Invariance_of_domain

veryveryverycoolusername
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    We do need to assume that $f$ is a bijection, of course. – Henno Brandsma Feb 04 '12 at 13:57
  • Bicontinuity is probably the single strongest condition one can impose on an arbitrary function. It's why homeomorphisms are such an important class of functions in both analysis and topology-topological invariants are cleanly carried over from the domain space to the range space. – Mathemagician1234 Feb 04 '12 at 17:15
  • I downvoted this, and I guess I should have explained since it looks like someone upvoted to undo it. First, the OP just copied a comment, without the full requirement of the theorem that $f$ be a bijection between the open sets (and so apparently didn't fully read the theorem he was referred to) and posted it as an answer. Second, OP is being overly dismissive of the counterexample of $f(x) = x^2$ posted above, claiming that $\pm \sqrt{x}$ is a function, so part of the downvote was for that. Apparently he doesn't appreciate the need for the function to be one-one in order to admit an inverse. – guy Feb 04 '12 at 18:12
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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. – Eric Auld Jul 16 '13 at 14:46
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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.

2'5 9'2
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  • +-sqrt(x)?????? – veryveryverycoolusername Feb 04 '12 at 06:05
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    @veryveryverycoolusername, that's not a function. A function maps an input $x$ to a unique output. –  Feb 04 '12 at 06:08
  • sqrt x for x>zero, else -sqrtx – veryveryverycoolusername Feb 04 '12 at 06:12
  • @very If $f(x)=x^2$, then $f(2)=f(-2)$, no? So if an inverse function existed, we could apply it, and then $f^{-1}(f(2))=f^{-1}(f(-2))$. This would imply $2=-2$. Now consider your proposed inverse function. When it applies to $4$ (which is $f(2)$ and is also $f(-2)$), you do not get to sometimes choose the output to be $2$ and other times choose the output to be $-2$ in order to get the results that you like. Your inverse function applied to $4$ needs to be one number no matter what the context is. – 2'5 9'2 Feb 04 '12 at 06:23
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    Not to beat a dead horse, but "$\pm\sqrt{}$" is not actually a function. When someone says something like $x=\pm\sqrt{5}$, they are using a lazy shorthand to say that "either $x=\sqrt{5}$ or $x=-\sqrt{5}$". As a math teacher, I actively work to stop this use of $\pm$ at pre-calculus levels, where students still don't have a firm grasp on the concept of a function. – 2'5 9'2 Feb 04 '12 at 06:32
  • Actually it is a function. f(x)=x*2 means "either f(1)=1 or f(2)=4 or f(0.3)=0.009 or..." – veryveryverycoolusername Feb 04 '12 at 13:44
  • @very, what unique real number is $\pm\sqrt{5}$? – guy Feb 04 '12 at 18:08
  • @very Now you are confusing "or" with "and". – 2'5 9'2 Feb 04 '12 at 18:28