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Let $(X,\tau)$ be a topological space. We define the "moving" relation by setting $$ x \simeq_m y \text{ iff there is a homemomorphism }\varphi: X\to X \text{ such that } \varphi(x) = y.$$

Clearly $\simeq_m$ is an equivalence relation. We call a space "immovable" if $\simeq_m$ is the diagonal $\Delta_X=\{(x,x):x\in X\}$.

If $(X,\tau)$ is a topological space, is $X/\simeq_m$ always immovable?

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    $\begingroup$ The usual name for "immovable" is "rigid". $\endgroup$
    – Andreas Blass
    Commented Oct 7, 2015 at 7:59

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No. Let $X$ be the disjoint union of the real line and one isolated point. The quotient by "movability" collapses the real line to one point, so the quotient is a discrete space of two points, which is not immovable.

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  • $\begingroup$ Is it true that given any topological space $X$ there is a space $Y$ such that $Y/ \simeq_m = X$? $\endgroup$
    – Lajos Soukup
    Commented Oct 11, 2015 at 11:33

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