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Let $X$ be a topological space and $\sim$ be an equivalence relation defined on $X$. Let $A$ be a union of some of the equivalence classes of $X$.

So we have a well-defined map $X-A~\to~ (X/\sim) -(A/\sim)$ which takes a point in $x\in X-A$ and sends it to its equivalence class $[x]$.

This map factors through $(X-A)/\sim$ to give a bijective map $f:(X-A)/\sim~ \to~ (X/\sim) -(A/\sim)$.

Question. I am wondering if it is necessary that $f$ is a homeomorphism.

If we in addition assume that $A$ is a closed subspace of $X$, then it is clear that the map $X-A \to (X/\sim) -(A/\sim)$ is a quotient map and hence $f$ is a homeomorphism. But I am not able to prove this without the use of this additional hypothesis.

Furthermore, I have been unsuccessful at finding a counterexample.

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  • $\begingroup$ Excuse me, I have a question... why is thw quotient map an homeomorphism if $A$ is closed? $\endgroup$
    – Twink
    Jan 1 '20 at 6:23
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Here's a counterexample. Let $X=\mathbb{R}$, let $\sim$ be the equivalence relation $x\sim y$ iff either $x=y$ or $x,y\in\mathbb{Q}$, and let $A=\mathbb{Q}$. Then $(X-A)/{\sim}$ is just the irrationals with the usual topology, but $X/{\sim}-A/{\sim}$ has a much coarser topology, since every open set in $X/{\sim}$ contains $A/{\sim}$ and hence must be dense with respect to the usual topology on the irrationals.

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