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Suppose $(Y,\tau')$ is Hausdorff and a function $f:(X,\tau)\to (Y,\tau')$ is a bijection such that $f^{-1}$ is continuous.

Can you show that $(X, \tau)$ is Hausdorff?

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    How is $\tau$ a topology on both $Y$ and $X$?2012-11-10
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    @ChrisEagle: $\tau'$ is the topology on $Y$.2012-11-10
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    @Asaf: what is $(Y,\tau)$ then?2012-11-10
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    @Chris: A typo?2012-11-10
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    Hm. I just noticed that indeed $(Y,\tau)$ appears in the title; and that $f^{-1}$ is required to be continuous rather than $f$. Is that another typo? Otherwise why not just take $g\colon(Y,\tau')\to(X,\tau)$ where $g=f^{-1}$?2012-11-10
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    @Asaf: I suspect that it’s a typo in both places, and the making $f^{-1}$ continuous instead of making $f$ open is just a way to make the problem superficially more difficult.2012-11-10
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    @Brian: Take $X=Y$ and $f=id$, then this is essentially to say that $\tau\subseteq\tau'$. If you *add* open sets to Hausdorff you are still Hausdorff, but if you *remove* open sets from a Hausdorff you are no longer Hausdorff. You need the function continuous and not its inverse. For example $(\mathbb R,\tau_{\text{cofinite}})\to(\mathbb R,\tau_{\text{standard}})$ has the property that for $f=id$ we have $f^{-1}$ continuous, but cofinite topologies are rarely Hausdorff.2012-11-10
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    Sorry its supposed to be ($Y$, $\tau$') is Hausdorff but $f^{-1}$ is not typo.2012-11-10
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    @Asaf: You’re right, of course. (I just woke up, and the mental gears are still sluggish.)2012-11-10
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    @Brian: Good, because for a moment there the instinct to trust a topologist was kicking in. Then I remembered that I am capable of producing a counterexample or prove this on my own. Sure enough, there was a counterexample. :-)2012-11-10
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    @AllenShoemaker: See my previous comment, and the edit to my answer. You might have to assume that $f^{-1}$ is *open* but this is the same as assuming $f$ is continuous, of course.2012-11-10
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    @Asaf: Oh, you can trust us set-theoretic topologists. It’s the algebraic topologists that you have to watch out for: they commit *category theory*! :-)2012-11-10
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    @Brian: That's a very *categorical* judgment on your side! :-)2012-11-10

2 Answers 2

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Recall that:

  • $f$ is continuous if $f^{-1}(U)\in\tau$ for all $U\in\tau'$;
  • and that $(X,\tau)$ is Hausdorff if and only if for every $a,b\in X$ there are $U,V\in\tau$ such that $a\in U, b\in V$ and $U\cap V=\varnothing$.

Hint:

Now, take $a,b\in X$. Since $f$ is a bijection $f(a)\neq f(b)$. In $Y$ find two disjoint open sets one which contains $f(a)$ and the other containing $f(b)$, and show that their preimage have to be disjoint, and open by continuity.


As stated with $f^{-1}$ continuous rather than $f$ continuous this is false.

Consider $X=Y=\mathbb N$, $\tau=\{\varnothing, X\}$ and $\tau'=\mathcal P(X)$, and take $f(x)=x$ the identity function. Certainly $f$ is a bijection, and certainly $f^{-1}$ is continuous, because there are only two open sets in $\tau$ and both are open in $\tau'$.

However $(X,\tau)$ is as far from Hausdorff as possible, and $(Y,\tau)$ is as Hausdorff as it gets.

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Corrected: Let’s see what happens if you follow your nose and use the definitions. To show that $\langle X,\tau\rangle$ is Hausdorff, you need to show that if $p,q\in X$ and $p\ne q$, then there are disjoint sets $U_p,U_q\in\tau$ such that $p\in U_p$ and $q\in U_q$. The only possible source of information about $X$ is $Y$, so look at $f(p)$ and $f(q)$; $f$ is a bijection, so $f(p)\ne f(q)$, and $\langle Y,\tau'\rangle$ is Hausdorff, so there are disjoint $V_p,V_q\in\tau'$ such that $f(p)\in V_p$ and $f(q)\in V_q$. Now you’d like to use $V_p$ and $V_q$ to get $U_p$ and $U_q$. Unfortunately, the only obvious way to try to do this is to set $U_p=f^{-1}[V_p]$ and $U_q=f^{-1}[V_q]$, and it doesn’t work: we know that $f^{-1}$ is continuous, but we have no reason to think that it is an open map, meaning one that takes open sets to open sets.

Of course it’s conceivable that the result is true even though the obvious, natural approach doesn’t work, but in fact it’s easy to come up with counterexamples. The simplest one that I can think of is to let $X=Y=\{0,1\}$, $\tau=\{\varnothing,X\}$, and $\tau'=\{\varnothing,\{0\},\{1\},X\}$, and to let $f$ be the identity map: $f(0)=0$ and $f(1)=1$. You can easily check that all of the hypotheses are satisfied, but that $\langle X,\tau\rangle$ is definitely not Hausdorff.

If you replace the condition that $f^{-1}$ is continuous with the condition that $f$ is continuous, or with the equivalent condition that $f^{-1}$ is open, the result is true. (Note that these conditions are equivalent only under the assumption that $f$ is a bijection, not in general.)