Let $X$ be a topological space, $Y$ be a set and let $q: X \to Y$ be a quotient map. In other words $q$ is surjective and $Y$ is equipped with the final topology induced by $q$. Let $A \subseteq X$ be a subspace of $X$ equipped with subspace (relative) topology and let $p = q|_A$, i.e.: $$p: A \to Y \ ,x \mapsto q(x) $$

Now forget about $X$ and define the quotient space $Y'$ respect to the map $p$.

When $Y$ and $Y'$ are homeomorphic?

First of all we need $Y = Y'$ as sets. We can show that $p$ is surjective, i.e. it covers all its range $Y$. Thus we have to check that $p^{-1}(q(x))$ is not empty for all $x \in X$. In other words $q^{-1}(q(x)) \cap A$ has not to be empty.

Is this sufficient?

Or we have to require topological properties on $A$?

PS: I drastically changed the text of the question. I used quotient maps (as used in topology) instead or equivalence relation in order to simplify the conditions.

PSS: I found the answer in the proposition quoted in this question: A question about the restriction of quotient maps to subsets of domain.

  • $\begingroup$ What do you mean by $A/ \sim$? is it the set of equivalence classes $[a]$ for $a\in A$, where $[a]$ could potentially contain elements in $X\setminus A$? $\endgroup$ – user160738 Nov 12 '16 at 15:21
  • $\begingroup$ @user160738 A relation $\sim$ on $X$ is simply a subset of $X \times X$. Given a subset $A$ of $X$, $\sim_A = \sim \cap A \times A$ is a relation on $A$.Here I identify $\sim_A$ and $\sim$. Then $A/\sim$ is a quotient set by its own $\endgroup$ – EmarJ Nov 12 '16 at 15:28
  • $\begingroup$ If $X=\{1,2,3\}$ and $\sim$ is given by $\{(1,1),(2,3),(3,2),(2,2),(3,3)\}$, then $X/\sim=\{\{1\},\{2,3\}\}$ while for $A=\{1,2\}\subset X$ we have $A/\sim=\{\{1\},\{2\}\}$. But this $X$ and $A$ satisfy your proposition's conditions so that's false $\endgroup$ – user160738 Nov 12 '16 at 15:40
  • $\begingroup$ If $A\neq X$ then actually the underlying sets of $A/\sim$ and $X/\sim$ are not equal. I would rather say that the spaces are homeomorphic under the condition you mention. $\endgroup$ – drhab Nov 12 '16 at 15:41
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    $\begingroup$ @drhab They need not be, something stronger is needed. Let $X = \mathbb{Z}, \, A = \{0,1\}$, and $x\sim y \iff x \equiv y \pmod{2}$. Let the topology on $X$ be $\{ U \subset \mathbb{Z} : U \subset A \lor (X\setminus U) \subset A\}$. Then $X/{\sim}$ is an indiscrete two-point space, and $A/{\sim}$ is a discrete two-point space. $\endgroup$ – Daniel Fischer Nov 12 '16 at 15:50

Your condition is not sufficient.

For example, let $X=\mathbb R$ with the usual topology and $x\sim y$ iff $x-y\in\mathbb Z$.

Then $X/{\sim}$ is a circle $S^1$.

However, let $A=[0,1)$. This satisfies your condition of containing at least one representative of each equivalence class, but the restriction of $\sim$ to $A$ is the identity and therefore $A/{\sim}$ is $[0,1)$ itself, which is distinctly different from the circle.

  • $\begingroup$ Thank you very much! Any ideas on which condition I have to require? Maybe $A$ is closed in $X$? $\endgroup$ – EmarJ Nov 12 '16 at 16:27
  • $\begingroup$ @EmarJ: I can't offhand imagine anything that would be simpler than just restating the desired property itself. $\endgroup$ – Henning Makholm Nov 12 '16 at 16:32

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