# Finding counterexamples: bijective continuous functions that are not homeomorphisms

Let $f: X \to Y$ be a bijective continuous function. If $X$ is compact, and $Y$ is Hausdorff, then $f:X \to Y$ is a homeomorphism.

My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively. I want to know if my counter examples are correct:

Let $f: X \to Y$ be the identity function.

(1) Let $X=[0,1]$ with the standard topology and let $Y=[0,1]$ with the indiscrete topology. Then $f: X \to Y$ is continuous and bijective but not a homeomorphism. Since $U=(1/3,1/2)$ is open in $X$, but $(f^{-1})^{-1}(U)=f(U)$ is not open in $Y$.

(2) Let $X=Y=\mathbb{R}$. Let $Y$ have the standard topology and $X$ have the discrete topology. Let $U=\{x\}$ since open sets are singletons. $f(^{-1})^{-1}(x)=f(x)$ is closed in $Y$ since $Y$ has the standard topology. Hence $f$ is not a homeomorphism.

This is the old question
My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively. I want to know if my counter examples are correct:

Let $f$ be the identity function. Then $f: X \to Y$ be a bijective continuous function. If $X=[0,1]$ with the standard topology and $Y=[0,1]$ with the indiscrete topology we have that $f:X \to Y$ is not a homeomorphism.

Let $f: X \to Y$ be a bijective continuous function, $X=(0,1)$, and $Y=[0,1]$ both with the standard topology. It's not a homeomorphism because $f^{-1}([0,1])$ is not connected, but $[0,1]$ is connected.

The first shows why Hausdoff is necessary for $Y$ and the second shows compactness is necessary for $X$.

Are both of the examples correct? I have no confidence in my first example because I'm having problems showing $f^{-1}$ is not continuous. However, I am confident in my second example.

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(2) is a good example, but the proof is incorrect. If $f$ is a homeomorphism, then every subset of the codomain is open, which is false. Just take $\{0\}$. The mere fact that singletons are closed in $Y$ doesn't imply they're not open. –  egreg May 1 '13 at 23:02

Second edit: Your new examples are both correct.

First edit: I wrote that the first example was incorrect. That was a typo. It is actually correct.

Your first example is correct. You need to show that $f^{-1}$ is not continuous, that is, that it is not true that $(f^{-1})^{-1}(U)=f(U)$ is an open set for all open sets $U$. Take any set $U$ which is open in $X$ but which is not open in the indiscrete topology, for example $U=(1/3,1/2)$. Then $f(U)=U$, which is not open in $Y$. Thus $f$ is not a homeomorphism.

Your second example is incorrect because there is no bijective continuous $f$ from $(0,1)$ to $[0,1]$: you must have $f(x)=0$ for some $0<x<1$, so by the intermediate value theorem, there must exist $y<x<z$ such that $f(y)=f(z)=\epsilon$ for some small $\epsilon>0$ (think of the curve of $f$ dipping down to $0$ and then going up again; it must pass through equal values somewhere on opposite sides of the zero).

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What about $X=Y=\mathbb{R}$ but $X$ has the discrete topology and $Y$ the standard topology. Would this illustrate that compactness is necessary. Since an open set in $X$ is a singleton which is not open in $Y$? –  emka May 1 '13 at 22:38
Also...would $f$ have to be the identity function in both examples. –  emka May 1 '13 at 22:40
@emka: Yes, that example works to show that compactness is necessary. (You should also prove that $X$ with the discrete topology is not compact.) –  Samuel May 1 '13 at 22:44
@emka: Are you saying that you meant $f$ to be the identity in your second example in your post? In that case you don't have a bijection since $0$ is not in the image of $f:(0,1)\to[0,1]$. –  Samuel May 1 '13 at 22:45
I edited my question with what I was trying to say and with your correction. Are my new (1) and (2) correct? –  emka May 1 '13 at 22:51

If $X$ is not compact, then the theorem does not work. For example, take $\mathbb R$ with the discrete topology and let $f$ be the identity into $\mathbb R$ with the standard topology.

But Hausdorffness isn't really necessary. It suffices that compact subsets are closed. There exist compact spaces where the compact subsets are exactly the closed subsets, still, the space is not Hausdorff. These spaces are called maximal compact. The reason is that these are exactly the spaces where there is no finer topology on the space $X$ such that the space is still compact. It is relatively easy to show that, if the compact subsets are exactly the closed ones, then no finer topology leaves the space compact. The other direction is more difficult. For a proof and an example of such a space, see this paper.

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Let $f: X \to Y$ be the identity function, where $Y=X$. It is obvious a bijection. If you need $f$ is continuou mapping, not a homeomorphism. It only need the topology on $Y$, we say $\tau_Y$, is strictly weaker that the topology $\tau_X$ on $X$. Then $f$ must be continuous, however it is not open, hence it is not a homeomorphism.
Proof: It is continuous. Let $U \in \tau_Y \subseteq \tau_X$, then $f^{-1}(U)= U \in \tau_X$, which witnesses $f$ is continuous. It is not open. There exist $U \in \tau_X \setminus \tau_Y$, then $f(U)=U \notin \tau_Y$, which shows $f$ is not open.