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This question already has an answer here:

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.

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

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  • $\begingroup$ What about taking as $X$ two lines and as $Y$ the line with two origins and as $f$ the projection? $\endgroup$ – Onil90 Mar 5 at 16:15
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    $\begingroup$ @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. $\endgroup$ – Rhys Steele Mar 5 at 16:32
  • $\begingroup$ @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 $f(\{0\} \times \mathbb{R}) \cup f(1,0) = Y$. $\endgroup$ – Rhys Steele Mar 5 at 16:33
  • $\begingroup$ @RhysSteele You're totally right! $\endgroup$ – Onil90 Mar 5 at 19:50
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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.

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  • $\begingroup$ Why doesn't a $T_3$ space suffice? $\endgroup$ – YuiTo Cheng Mar 5 at 16:43
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Mar 5 at 16:46
  • $\begingroup$ Just for clarification. +1! $\endgroup$ – YuiTo Cheng Mar 5 at 16:47

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