It is well-known that continuous and bijective function from a compact space onto a T2 space is always a homeomorphism. What I'm trying to do is to show that this isn't true if we relax, even a little, the separation property. In particular, I'm trying to construct a continuous and bijective map from a compact T1 space onto itself such that it is not a homeomorphism. Well, it seems to be much more difficult than it looked at the beginning. I'm not even able to choose a proper space.

The canonical examples of T1 spaces not T2 (of course, what we do know because of the well-known result is that the space cannot be T2) are sets equipped with cofinite or cocountable topology,and they are indeed compact. But any bijection from one of this sets onto itself is bicontinuous.

I tried using a closed interval of the line with two origins. This is a T1 non T2 space, more or less easy to work with and I could find compact subsets not closed (which is the true reason why we need a non T2 space). But a continuous bijective function from this space onto itself must send fix the origins or swap them, neither of which it's useful for what I need (I could explain it with more detail, but I think you catch the idea. Please, tell me if I'm making a mistake)

Something much similar to this happens with the uncountable Fort Space.

As for the one point compactification of the rationals, although it is,again, T1, not T2 and compact; we can prove easily that every continuous function from this space onto itself is closed, so we cannot use it either.

In the last hours, I've been trying with the product of $[0,1]$ with usual topology and $R$ with cofinite topology. In this space there are much more possibilities and I may be able to found the map I'm looking for. However, it is so difficult to build continuous functions in this space ( or maybe I'm not used enough to working with the product)

Please tell me if my reasoning until now was correct and in which direction should I proceed, if you know where could an undergraduate student find "easily" a map like this.

  • $\begingroup$ It cannot be a kc-space (compact sets are closed). $\endgroup$ Apr 24, 2020 at 20:29
  • $\begingroup$ There are much easier examples if you consider continuous bijections between two different compact $T_1$ spaces, rather than from a single space to itself. $\endgroup$ Apr 24, 2020 at 22:08

2 Answers 2


A very easy example is $\Bbb Z$ with the following topology: negative integers are isolated, and the nbhds of an $n\ge 0$ are the cofinite subsets of $\Bbb Z$ containing $n$. This space is clearly compact and $T_1$, and the map $f:\Bbb Z\to\Bbb Z:n\mapsto n+1$ is a continuous bijection whose inverse is not continuous at $0$.

This comes from the observation that if $\langle X,\tau\rangle$ is a compact $T_1$ space, and $f:X\to X$ is a continuous bijection that is not a homeomorphism, $\tau_f=\{f^{-1}[U]:U\in\tau\}$ must be a topology on $X$ strictly coarser than $\tau$. Fix a point $x_0\in X$, and for $n\in\Bbb Z$ let $x_n=f^{n}(x_0)$, so that $Y=\{x_n:n\in\Bbb Z\}$ is the orbit of $x_0$ under $f$. For each $n\in\Bbb Z$ let

$$\mathscr{U}_n=\{U\in\tau:x_n\in U\}\;;$$

then for each $n\in\Bbb Z$ we must have $\{f^{-1}[U]:U\in\mathscr{U}_n\}\subseteq\mathscr{U}_{n-1}$. If $Y$ is compact and there is at least one $n\in\Bbb Z$ for which the inclusion is strict, then $Y$ and $f\upharpoonright Y$ are also an example. The example in the first paragraph is about the simplest space that one can construct along these lines.


It is perhaps worth observing that such a space must be "infinite" in some sense. In particular, if $f:X\to X$ is a continuous bijection which is not a homeomorphism, and $T$ is the topology on $X$, then $f^{-1}:T\to T$ is an injection that is not surjective. So, you might take some inspiration from the ways you can construct injections that are not surjections from an infinite set to itself.

Another helpful observation is that it is much easier to find an example of a continuous bijection $f:X\to Y$ between two different compact $T_1$ spaces which is not a homeomorphism. For instance, you could take $X$ to be any infinite compact Hausdorff space, $Y$ to be $X$ with the cofinite topology, and $f$ to be the identity.

With this in mind, then, here is a way you could construct the counterexample you're searching for. Start with a continuous bijection $f:X\to Y$ between two compact $T_1$ spaces which is not a homeomorphism. Now let's enlarge our space with infinitely many copies so that we can make $f$ into a map from a single space to itself. Specifically, let $Z$ be the disjoint union $X\times\mathbb{N}\coprod Y\times\mathbb{N}$. We can then take $g:Z\to Z$ which maps one of the copies of $X$ to $Y$ via $f$, and then maps all the other copies of $X$ to each other homeomorphically, and similarly for the other copies of $Y$. Very explicitly, you can define $g$ by $g(x,0)=(f(x),0)$ for $x\in X$, $g(x,n)=(x,n-1)$ for $x\in X$ and $n>0$, and $g(y,n)=(y,n+1)$ for $y\in Y$.

Now, this $Z$ is not compact anymore, since we've taken these infinitely many disjoint copies. That's easy to fix, though: just take a one-point compactification. Explicitly, let $Z^*=Z\cup\{\infty\}$ where the inclusion $Z\to Z^*$ is an open embedding and neighborhoods of $\infty$ are open subsets of $Z$ that contain $X\times\{n\}$ and $Y\times\{n\}$ for all but finitely many values of $n$. The map $g:Z\to Z$ then extends continuously to $g^*:Z^*\to Z^*$ by mapping $\infty$ to itself, $Z^*$ is a compact $T_1$ space, and $g$ is a continuous bijection but not a homeomorphism.


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