# Are continuous self-bijections of connected spaces homeomorphisms?

I hope this doesn't turn out to be a silly question.

There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms. But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces.

My Question: Is there a connected topological space $X$ and a continuous bijection $X\to X$ that is not a homeomorphism?

For the record, my example of a continuous bijection $X\to X$ that is not a homeomorphism is the following. Roughly, the idea is to find an ordered family of topologies $\tau_i$ ( $i\in \mathbb Z$) on a set $S$ and use the shift map to create a continuous bijection from $\coprod_{i\in \mathbb Z} (S, \tau_i)$ to itself. Let $S = \mathbb{Z} \coprod \mathbb Z$. The topology $\tau_i$ is as follows: if $i<0$, then the left-hand copy of $\mathbb Z$ is topologized as the disjoint union of the discrete topology on $[-n, n]$ and the indiscrete topology on its complement, while the right-hand copy of $\mathbb Z$ is indiscrete. The space $(S, \tau_0)$ is then indiscrete. For $i>0$, the left-hand copy of $\mathbb Z$ is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on $[-n, n]$ with the discrete topology on its complement. Now the map $\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)$ sending $(S, \tau_i) \to (S, \tau_{i+1})$ by the identity map of $S$ is a continuous bijection, but not a homeomorphism.

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There is a thread on Mathoverflow devoted to a similar question: mathoverflow.net/questions/30661/… –  Andrey Rekalo Feb 8 '11 at 1:15

Here's a nice geometric example. Let $X\subset\mathbb{R}^2$ be the union of the $x$-axis, the line segments $\{n\}\times[0,2\pi)$ for $n\in \{\ldots,-3,-2,-1,0\}$, and circles in the upper half plane of radius $1/3$ tangent to the $x$-axis at the points $(1,0),(2,0),\ldots$.

Note that $X$ is connected.

Define a map $f\colon X\to X$ by $$f(x,y) \;=\; \begin{cases}(x+1,y) & \text{if }x\ne 0 \\ \left(1+\frac{\sin y}{3},\frac{1-\cos y}{3}\right) & \text{if }x=0\end{cases}.$$ That is, $f$ translates most points to the right by $1$, and maps the line segment $\{0\}\times[0,2\pi)$ onto the circle that's tangent to the $x$-axis at the point $(1,0)$. Then $f$ is continuous and bijective, but is not a homeomorphism.

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Very nice example! –  Qiaochu Yuan Feb 8 '11 at 1:36
That's great! It builds very nicely on the example of the continuous bijection from a half-open interval to the circle. Thanks, Jim! –  Dan Ramras Feb 8 '11 at 1:40

Zipping up halfway gives a continuous bijection from your pants with the fly down to your pants with the fly at half mast and this is not a homeomorphism. However, the two spaces are homeomorphic no?

One can well-imagine this phenomena persists for various other "manifolds with tears" - even in higher dimensions.

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I think this is a great example. I don't know why it only has one upvote. –  Grumpy Parsnip Mar 31 '11 at 13:43
I did not understand your example Mike D: how can I see that this function is continuous ( geometrically clearly) –  Daniel Aug 31 '11 at 2:37
@Daniel: Well you can't check continuity since I didn't specify the function. An explicit example along these lines might go something like this: let $A=(1,1),B=(1,-1),C=(-1,-1),D=(-1,1),P=(1,1/2),Q=(1,-1/2),R=(1,0),O=(0,0) \in \mathbb{R}^2$ (now's a good time to draw a picture). Take $X$ to be the square $ABCD$ minus the interior of the line segment $AR$. –  Mike F Sep 6 '11 at 22:32
Note each point of $X$ is in at least one of the five triangles $APD,PRD,DRC,RQC,QBC$. Define $g:\{A,B,C,D,P,Q,R\} \to \mathbb{R}^2$ by having $g$ fix the corners of our square, send $P,Q \to R$, and send $R \to O$. Finally, define $f:X \to X$ by sending a point $x \in T \cap X$ where $T$ is one of the aforementioned triangles to the image of $x$ under the unique affine map which extends the restriction of $g$ to the vertices of $T$. It should be possible (but probably somewhat painful) to verify that $f$ is a well-defined, continuous, bijective map $X \to X$. –  Mike F Sep 6 '11 at 22:33

Yes (to your body question, not your title question; it is confusing when people do this). Take $X = \mathbb{Z}$ with the topology generated by an open set containing $n$ for every positive integer $n$. (This space is connected because the smallest open set containing a non-positive integer is the entire space.) Consider the continuous bijection given by sending $x$ to $x - 1$.

Here is what might be a Hausdorff example: take $X = \mathbb{R}$ with the topology generated by the usual topology together with the open set $(0, \infty) \cap \mathbb{Q}$, and again consider the continuous bijection $x \to x-1$. Unfortunately I am not sure if $X$ is connected.

The most general situation I know where a continuous bijection $X \to Y$ is automatically a homeomorphism is if $X$ is compact and $Y$ is Hausdorff. This is a nice exercise and extremely useful.

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The closure of every negative integer is the non-negative integers? –  Soarer Feb 8 '11 at 1:21
@Soarer: whoops. Fixed. –  Qiaochu Yuan Feb 8 '11 at 1:29
Thanks for the example. I accepted Jim's answer since it is so geometric. –  Dan Ramras Feb 8 '11 at 1:43
Oh, and thanks for pointing out that I inverted the title question... I'll remember to watch for that in the future (at this point, it's probably least confusing to leave it as is). –  Dan Ramras Feb 8 '11 at 1:46
And as a 3rd and final comment, I'll mention that my favorite version of the exercise you state is: a continuous injection from a compact space to a Hausdorff space is a homeomorphism so long as its image is dense. This comes up when working with profinite groups. –  Dan Ramras Feb 8 '11 at 1:49