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Let $S$ be a set and $\vartheta$ be an equivalence relation on $S$. We say that $\vartheta$ is proper if there are $x\neq y\in S$ with $(x,y)\in\vartheta$.

Is there an infinite Hausdorff space $(X,\tau)$ such that for every proper equivalence relation $\vartheta$ on $X$ we have $X\not\cong X/\vartheta$?

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  • $\begingroup$ A strongly rigid infinite Hausdorff space would do, right? $\endgroup$
    – jmc
    Commented Mar 21, 2015 at 13:34
  • $\begingroup$ Oh, possibly, have to think about it. Please post your example as an answer $\endgroup$
    – Dominic van der Zypen
    Commented Mar 21, 2015 at 16:20
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    $\begingroup$ Well, you know more about strongly rigid spaces then I do. Doe strongly rigid infinite Hausdorff spaces exist? If so, let $X$ be one. If $f \colon X/\vartheta \to X$ is a homeomorphism, compose it with the quotient map $q \colon X \to X/\vartheta$, to obtain a continuous map $(f \circ q) \colon X \to X$. Since $X$ is strongly rigid, the composition is constant or the identity. Consequently $\vartheta$ is trivial. $\endgroup$
    – jmc
    Commented Mar 21, 2015 at 16:51

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Let $X$ be a strongly rigid infinite Hausdorff space. Let $\vartheta$ be an equivalence relation on $X$. Let $q \colon X \to X/\vartheta$ be the quotient map. If there is a homeomorphism $f \colon X/\vartheta \to X$, then $(f \circ q)$ is a continuous map from $X$ to itself; hence constant or the identity. Since $X$ is infinite, we deduce that $\vartheta$ is trivial.

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