# 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.

• There is a thread on Mathoverflow devoted to a similar question: mathoverflow.net/questions/30661/… Feb 8, 2011 at 1:15
• For every $(X,f)$, $X$ topological space and $f$ bijective continuous non-open self map of $X$, we can obtain a another connected example taking the cone. So all non-connected examples (see e.g. math.stackexchange.com/questions/1702979) here can be used to provide examples here .
– YCor
Jun 7, 2017 at 8:27

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.

• Very nice example! Feb 8, 2011 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! Feb 8, 2011 at 1:40
• Why is f^{-1} not continuous? I'm pretty sure it has to do with the circle being mapped to the line segment, but I can't figure it out. Jan 25, 2018 at 12:20
• @PhysicsEnthusiast take a sequence of points converging to the bottom of the leftmost circle from the left-hand-side. Their inverse images under f have no accumulation point in $X$, so $f^{-1}$ can't be continuous. Mar 19, 2021 at 16:28

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.

• I think this is a great example. I don't know why it only has one upvote. Mar 31, 2011 at 13:43
• I did not understand your example Mike D: how can I see that this function is continuous ( geometrically clearly) Aug 31, 2011 at 2:37

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.

• The closure of every negative integer is the non-negative integers?
– user325
Feb 8, 2011 at 1:21
• @Soarer: whoops. Fixed. Feb 8, 2011 at 1:29
• Thanks for the example. I accepted Jim's answer since it is so geometric. Feb 8, 2011 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). Feb 8, 2011 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. Feb 8, 2011 at 1:49

Consider the topology on $$\mathbb R$$ in which the open sets are $$\varnothing,\mathbb R$$ and $$(a,\infty)$$ such that $$a\geq 0$$.

Is this a topology?

$$\varnothing$$ and $$\mathbb R$$ are open.

If we take $$(a,\infty)$$ and $$(b,\infty)$$ with $$a,b\geq 0$$ without loss of generality $$a and so $$(a,\infty) \cap (b,\infty) = (b,\infty)$$. If one of the intersecands is $$\mathbb R$$ or $$\varnothing$$ it is easy. When one of the intersecands is $$\mathbb R$$ or $$\varnothing$$ the situation is easy.

The union of an arbitraty union of sets is only interesting when none of the sets are $$\varnothing$$ or $$\mathbb R$$ and in that case if the intervals are $$(a_i,\infty)$$ our union is $$(\inf a_i, \infty)$$ which is also open.

This set is connected because there is right-interval such that its complement is also open.

Now consider the bijection $$f: \mathbb R \rightarrow \mathbb R$$ given by $$f(x) = (x-1)$$. Notice that the preimage of the interval $$(a,\infty)$$ with $$a\geq 1$$. is the interval $$(a+1,\infty)$$ which is also open. hence the function is continuous. However, the function is not open as $$f((0,\infty)) = (-1,\infty)$$ which is not open.