# Counterxample: A continuous closed surjective map doesn't preserve Hausdorff property [duplicate]

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I'm looking for counterexamples such that

$$f: X \rightarrow Y$$ is a continuous closed surjective map with $$X$$ Hausdorff and $$Y$$ non-Hausdorff

I know perfect maps preserve Hausdorff property. The only extra assumption that is crucial in the proof is $$f^{-1}\{y\}$$ is compact for all $$y\in Y$$. That's why I try to drop this assumption. The closest question I can find is Is Hausdorffness preserved under continuous surjective open mappings?, which almost excites me except for the open. I looked at the answers in it and tried them one by one. Unfortunately, none of them work or can be mimicked. I also tried the quotient map with $$Y$$ being the "line with infinitely many origins". Also failed. Any hint would be appreciated.

## marked as duplicate by YuiTo Cheng, Community♦Apr 22 at 8:05

• What about taking as $X$ two lines and as $Y$ the line with two origins and as $f$ the projection? – Onil90 Mar 5 at 16:15
• @Onil90 Is this projection closed? If we write for concreteness $X = \{0,1\} \times \mathbb{R}$ and $Y = (\mathbb{R} \setminus \{0\}) \cup \{f(1,0),f(0,0)\}$ then unless I'm mistaken, for example $U = \{0\} \times \mathbb{R}$ is closed in $X$ but $f(U) = Y \setminus \{f(1,0)\}$ and since the singleton $\{f(1,0)\}$ is not open in the line with two origins, $Y = f(U)$ is not closed. – Rhys Steele Mar 5 at 16:32
• @YuiToCheng The projection is clearly surjective, since it is the quotient map arising in the definition of the line with two origins. With $X$ as in my comment above, we even have $f(\{0\} \times \mathbb{R}) \cup f(1,0) = Y$. – Rhys Steele Mar 5 at 16:33
• @RhysSteele You're totally right! – Onil90 Mar 5 at 19:50

Let $$X$$ be a Tychonoff ($$T_{3\frac12}$$) space that is not normal (like the Sorgenfrey plane $$\mathbb{S}^2$$, e.g.), and let $$A$$ and $$B$$ be two disjoint closed sets that cannot be separated by open sets.
Define an equivalence relation $$R$$ on $$X$$ by specifying its classes: $$A$$, $$B$$, and $$\{x\}, x \notin A \cup B$$, and give $$Y=X / R$$ the quotient topology wrt the canonical map $$q: X \to X/R$$ sending $$x$$ to its class in $$X/R$$.
Then $$q$$ is a closed continuous surjective map, $$X$$ is Hausdorff and $$Y$$ is not, as we cannot separate the points $$A$$ and $$B$$ of $$X/R$$, or their inverse images would have separated $$A$$ and $$B$$ which was impossible.
We cannot strengthen the example further, as the closed continuous image of a $$T_4$$ (normal and $$T_1$$) space is again $$T_4$$, so a fortiori Hausdorff.
• Why doesn't a $T_3$ space suffice? – YuiTo Cheng Mar 5 at 16:43
• @YuiToCheng it does, and we identify a closed set to a point. But the example gets better with nicer $X$ and non normal Tychonoff examples are better known than nonregular Hausdorff ones. – Henno Brandsma Mar 5 at 16:46