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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

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