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If $X$ is a topological space and $Y\subset X$ is a subspace, and if there is an equivalence relation on $X$, then we have the map $ Y/\sim\to X/\sim$ that sends an equivalence class of $y\in Y$ to itself; it's one-to-one and continuous (w.r.t. the quotient topologies on both spaces). If we further assume it's onto, does it follow that it's a homeomorphism? I couldn't prove it, but I couldn't come up with a counterexample either.

(I know that in general, continuous bijections need not be homeomorphisms, but it might be the case quotient spaces have some properties that make any continuous bijection between them be a homeomorphism.)

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Yes. Let $Y$ be $[0, 1]$ with the discrete topology, let $Y'$ be $[0,1]$ with the Euclidean topology, and let $X = Y \sqcup Y'$. Let the equivalence relation on $X$ be the least equivalence relation such that $a \in Y$ is equivalent to $a \in Y'$. Then $\sim$ on $Y$ is just the identity relation, so $Y/\sim$ is just $Y$; but ${X/\sim} \cong Y'$. Clearly the identity $Y \to Y'$ is not a homeomorphism.

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  • $\begingroup$ How does one prove $X/\sim \simeq Y'$? The map I guess is given by $Y\sqcup Y'\supset Y \ni t \mapsto [t]$. Suppose we had proven it's bijective. How to see it's a homeomorphism? Consider an open set in the quotient. It is open iff its inverse image in the disjoint union is open (this by definition means that its preimage is a union of open sets of $Y$ and $Y'$). But why is it open in $Y'$? The preimage may be a union of open sets that are open in $Y$ but not in $Y'$, and such a union is not an open subset of $Y$. $\endgroup$ – user531587 Sep 20 '18 at 17:31

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