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Let $f:X\to Y$ be a continuous bijection.

If $Y$ is Hausdorff, then can we conclude that $X$ is also Hausdorff?

What I learned so far is that:
- If $X$ is compact, then $f(X)$ is also compact;
- If, $X$ is compact and $Y$ is Hausdorff, then $f^{-1}$ is continuous.

I've no idea how to approach the problem. Is there any counterexample for the statement above?

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Yes! ${}{}{}{}{}$ – leo Nov 10 '12 at 2:58
up vote 4 down vote accepted

Let $x_1, x_2 \in X$, $x_1 \ne x_2$. As $f$ is one-to-one, $f(x_1) \ne f(x_2)$ and as $Y$ is Hausdorff, there are disjoint open neighbourhoods $U_1 \ni f(x_1)$, $U_2 \ni f(x_2)$. This gives $x_1 \in f^{-1}(U_1)$, $x_2 \in f^{-1}(U_2)$. As $f$ is continuous, $f^{-1}(U_i)$ are open neighbourhoods of $x_i$, $i = 1,2$ moreover they are disjoint. So $X$ is Hausdorff.

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Indeed, that's the thing – leo Nov 10 '12 at 3:00
Thank you for your answer! Why are $f^{-1}(U_i)$ disjoint? – Goku Nov 10 '12 at 3:09
As inverse images of disjoint sets are always: Suppose $x \in f^{-1}(U_1)\cap f^{-1}(U_2)$, then $f(x) \in U_1 \cap U_2$ ... – martini Nov 10 '12 at 3:11
Fair enough. Thank you! I didn't notice that the inverse preserves intersections. – Goku Nov 10 '12 at 3:13

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