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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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marked as duplicate by Jonas Meyer, Rahul, Asaf Karagila, Henning Makholm, Zev Chonoles Feb 5 '12 at 7:25

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No: Several examples appear at Functions which are Continuous, but not Bicontinuous. –  Jonas Meyer Feb 4 '12 at 5:50
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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 4 '12 at 7:32
    
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 4 '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 5 '12 at 6:34
    
@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. –  Swaprava Oct 11 '13 at 5:50
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3 Answers 3

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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Do you mean $[0,2\pi)$? –  katari Oct 5 '12 at 2:16
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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 5 '12 at 3:26
    
@GEdgar I haven't seen that interval notation for years. It's good to see it's well applied and not forgotten. –  Alberte Romero May 30 '13 at 12:30
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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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+-sqrt(x)?????? –  veryveryverycoolusername Feb 4 '12 at 6:05
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@veryveryverycoolusername, that's not a function. A function maps an input $x$ to a unique output. –  Rahul Feb 4 '12 at 6:08
    
sqrt x for x>zero, else -sqrtx –  veryveryverycoolusername Feb 4 '12 at 6:12
    
@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. –  alex.jordan Feb 4 '12 at 6:23
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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. –  alex.jordan Feb 4 '12 at 6:32
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Yes, if open subsets. http://en.wikipedia.org/wiki/Invariance_of_domain

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We do need to assume that $f$ is a bijection, of course. –  Henno Brandsma Feb 4 '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 4 '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 4 '12 at 18:12
    
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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