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Let M be a compact metric space. We know that if $ g:M\to M$ is a continuous bijection then it's a homeomorphism. But I want to know, if I have a continuous bijection $ f:M - \left\{ p \right\} \to M - \left\{ p \right\} $, then it's true that can always be extended to a continuous bijection $M\to M$ or not?

clearly I assume that $ M - \left\{ p \right\} $ it's under the restricted metric of M.

EDITED: Even knowing that M it's the one point compactification and that the open sets of M are all the open sets of M-p , and the complement of compacts of M-p , even with that I can't prove the result. Maybe it's not true. I'm not sure, if you want to use that you are welcome, and maybe it's false and I need a counterexample :/ I'll also change the name of the post

share|cite|improve this question it true that $M$ will be $M-p$'s one point compactification? – uncookedfalcon Jun 5 '12 at 5:27
@uncookedfalcon: yes, and this holds for any compact $M$ (not necessarily metrizable). – Qiaochu Yuan Jun 5 '12 at 5:29
How can I prove this? – Matias Jun 5 '12 at 5:31
Yes the unique option is to send p to p, but how can I prove that this extension is continuous? – Matias Jun 5 '12 at 5:35
Check out munkres thm 29.1 (…) (the answer is yes, it always extends) – uncookedfalcon Jun 5 '12 at 5:48
up vote 0 down vote accepted

WARNING: I just noticed there are two big problems with my answer. The closed upper half plane is not homeomorphic to the space $X$ obtained by removing an open interval from the boundary of a closed 2-disk. It is true that $X$ has a continuous self bijection which is not a homeomorphism given by folding the boundary in on itself, but $X$ is not locally compact so my example does not solve your problem. However, my link is still relevant because Jim Belk's answer on the same page exhibits a continuous self-bijection of a locally compact space.

As has been pointed out, the spaces obtained by subtracting a point from a compact Hausdorff space are precisely the locally compact Hausdorff spaces. So your question can be rephrased as follows.

Does there exist a continuous self-bijection of a (necessarily noncompact) locally compact Hausdorff space which is not a homeomorphism?

This is possible even for as nice a space as the upper half plane $\mathbb{H}^2 = \{ (x,y) \in \mathbb{R}^2 : y \geq 0\}$ (a manifold with boundary!).

We can also identify $\mathbb{H}^2$ with a closed 2-disk minus a point on the boundary. Or, more usefully, a closed 2-disk with an open interval subtracted from the boundary. Now there is a clear self-bijection given by tucking the boundary in on itself and this is not a homeomorphism. For a picture, see my answer here.

Are continuous self-bijections of connected spaces homeomorphisms?

share|cite|improve this answer

I will denote by $C(\omega)=\{0,1,2,\dots\}\cup\{\omega\}$ the one-point compactification of the discrete space on the countable set $\{0,1,2,\dots\}$. I.e. all points different from $\omega$ are isolated and neighborhoods of $\omega$ are precisely complements of finite subsets of $\{0,1,2,\dots\}$.
(This space is homeomorphic to the space $\{0\}\cup\{\frac1{n+1}; n=0,1,2,\dots\}$ with the topology inherited from real line. So this is simply a convergent sequence.)

Now I take $M=\{0,1\}\times C(\omega)$, where $\{0,1\}$ has the discrete topology. (I.e., $M$ is the topological sum of two copies of $C(\omega)$.) And I choose $p=(0,\omega)$.

It is easy to see that $M$ is a compact metric space.
(E.g. it is homeomorphic to $\{0,1\}\times(\{0\}\cup\{\frac1{n+1}; n=0,1,2,\dots\})$ with the topology induced by the usual Euclidean metric on $\mathbb R^2$.)
Note that every subset of $\{0\}\times C(\omega)$ is open in $M\setminus\{p\}$.

Let us define a map $f \colon M\setminus\{p\} \to M\setminus\{p\}$ by putting $$f(0,2n)=(1,2n)\\ f(0,2n+1)=(0,n)\\ f(1,n)=(1,2n+1)\\ f(1,\omega)=(1,\omega).$$

This map is bijective and continuous, but the extension $\overline f\colon M\to M$ with $\overline f(0,\omega)=(0,\omega)$ is not continuous, since the sequence $x_n=(0,2n)$ converges to $(0,\omega)$ in $M$ but $\overline f(x_n)=(1,2n)$ converges to $(1,\omega)\ne \overline f(0,\omega)$.

Here's my attempt to sketch the above map (the two big arrows indicate the direction in which the sequences $C(\omega)\times\{0\}$ and $C(\omega)\times\{1\}$ converge):

Two copies of C(omega)

It is perhaps worth mentioning that $f^{-1}$ is not continuous. The sequence $(1,2n)$ converges to $(1,\omega)$ but $f^{-1}(1,2n)=(0,2n)$ does not converge.

So this does not contradict Theorem 29.1 from Munkres, which was mentioned in comments.

Theorem 29.1. Let $X$ be a space. Then $X$ is locally compact Hausdorff if and only if there exists a space $Y$ satisfying the following conditions:

  • $X$ is a subspace of $Y$.
  • The set $Y-X$ consists of a single point.
  • $Y$ is a compact Hausdorff space.

If $Y$ and $Y'$ are two spaces satisfying these conditions, then there is a homeomorphism of $Y$ with $Y'$ that equals the identity map on $x$.

The OP indicated that he was originally thinking about the case that $M=S^n$. Since $S^n$ is the one-point compactification of $\mathbb R^n$, from Theorem 29.1 and from the fact that every continuous bijection $\mathbb R^n\to\mathbb R^n$ is a homeomorphism (which was shown in this question) we get that in the case $M=S^n$ the result from the question is true.

share|cite|improve this answer
Very nice. I spent quite a while trying to come up with an example but didn't succeed. – Nate Eldredge Jun 7 '12 at 13:06
@Martin Sleziak Thanks Martin! I wanted to know if this was true, because if the result was true, I could prove that every continuous bijection from $ R^n \to R^n$ is a homeomorphism (working in $S^n$ the one point compactification of $R^n$). Do you think that would help, to prove that result working in $S^n-p$ instead of $R^n$ ? And how can I give you the reward of $100$ points? I don't know – Matias Jun 7 '12 at 13:32
@Matias If I were you, I would wait a little before awarding a bounty - perhaps you'll get more answer and it is possible that someone will spot a mistake in my answer. Anyway, I think that anything you need to know about bounties, you can find at faq and meta. I've tried to add something about $S^n$, I am not sure whether this is what you wanted to see. – Martin Sleziak Jun 7 '12 at 13:43
@MartinSleziak Thanks again for the answer! But I'm not sure if the proofs there are good. – Matias Jun 7 '12 at 13:49
@Matias Just let me know what seems to be a problem - I'll try to work out the details. (Another possibility - we will find out that there is a mistake.) – Martin Sleziak Jun 7 '12 at 13:55

I'll add another example which I found at Ask A Topologist forum. This example is, to some extent, similar to the example above, but it might be easier to visualize.

I'll first put here a LaTeX-ed version of the post I linked:

This is an example of a locally compact space $S$ and a continuous bijective function $f\colon S\to S$, which is not a homeomorphism.

Let $S = \bigcup\limits_{n \in \mathbb Z} S_n \cup \{0\}$, where $S_n$ is the circle centered at the origin with radius $2^n$.
$S$ is a locally compact space as a closed subset of $\mathbb C$.
Define, for $m$ and $n$ in $\mathbb Z$, $f_{m, n} \colon S_m \to S_n$ homeomorphism (for example the multiplication by $2^{n-m}$).
Then define a bijection $g \colon \mathbb Z \to \mathbb Z$ such that $g (n) \to - \infty$ when $n \to -\infty$ and $g (n)$ is not bounded above nor bounded below when $n \to +\infty$.
For example, $g$ defined by $g (n) = n/2$ if $n$ is positive and even, $g (n) = -n$ if $n$ is positive and odd, and $g (n) = -2n$ if $n$ is negative should work.

Now, consider $f \colon S \to S$ $$ x \mapsto f_{n, g (n)} (x),\\ 0 \mapsto 0$$

$f$ is an bijection from $S_n$ to $S_{g (n)}$ for every $n$ and since $g$ is permutation of $\mathbb Z$, $f$ is a bijection from $S - \{0\}$ to $S - \{0\}$.
$f$ sends $0$ to $0$, therefore $f$ is a bijection from $S$ to $S$.
$f$ is continuous on each $S_n$, and is continuous at $0$ since $g (n) \to -\infty$ when $n \to -\infty$.
But $f^{-1}$ is not continuous at zero, because $g^{-1} (n) \not\to -\infty$ when $n \to -\infty$.

Now let us consider $\mathbb C\cup\{\infty\}$ as the one-point-compactification of plane.

Then we have subspace $M=S\cup\{\infty\}$.

The map $f \colon S\to S$ described above is continuous and bijective, but the extension $\overline f \colon M \to M$ such that $\overline f(\infty)=\infty$ is not continuous. To see this, just choose a sequence $(x_n)_{n=0}^\infty$ such that $x_n \in S_n$ and observe that $x_n\to\infty$ but $\overline f(x_n)=f(x_n)$ does not converge to $\infty$.

Interestingly, when I was Googling and trying to find similar examples online, I found the following sentence in some book:

This finishes the proof of the theorem, since every continuous bijection between locally compact Hausdorff spaces is a homeomorphism (see Theorem 10 on page 139 of [Eng68]).

It seems that [Eng68] refers to the book R. Engelking (1968). Outline of General Topology. translated from Polish. North-Holland, Amsterdam.

I only have the Polish original of this book; which of course has a different page numbering; but I did not find the the chapter on locally compact spaces anything similar to the above claim. So I guess it is a mistake or (more probably) a misquote.

share|cite|improve this answer
The statement you quote in the last part is blatantly wrong. The identity from the discrete reals to the reals with the usual topology is an obvious counterexample. Also, there's a result of Parhomenko saying that every locally compact space admits a 1-1 continuous function onto a compact Hausdorff space (see also exercise 3.3.D in Engelking). – t.b. Jun 16 '12 at 9:08
Thanks for the comment. (I did not know about this result.) Of course, I agree that the claim as quoted is not correct. (Unfortunately, I am not able to follow the proof where it is used, so I am not sure to which extent is this mistake important for the result from that book). Anyway - just out of curiosity - if I am able to get somewhere Engelking's Outline of GT, I'll have a look what exactly is Theorem 10 on p.139. – Martin Sleziak Jun 16 '12 at 9:23

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