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Let $X,\tau$ a topological space and ~ an equivalence relation on $X$. Prove that ~$\subseteq X \times X$ is open (closed) if only if the quotient map is an open (closed) map.

Proof: Let $\pi$, the quotient map.

For the first implication, let $U\in\tau$. P.D. $\pi(U)\in\tau/$~. The result follows given that:

$U\subseteq \pi^{-1}(\pi(U))$.

Is this correct? I'm not sure, because I don't use the fact that the equivalence relation is open.

How to prove the converse?

I suposse that $\pi$ is an open map and prove that ~ is open in $X\times X$. For definition, I want to show that all points of ~ are interior points.

Let $(x,y)\in$ ~, I must prove that exists $V\in\tau_{X\times Y}$, such that

$V\subseteq$ ~.

But I don't know how to continue. Any idea?

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I believe this is wrong. If $\sim = \Delta(X) = \{(x,x) : x\in X\}$ is the diagonal, then the quotient map is a homeomorphism, so it is both open and closed. But the diagonal is in general not open (consider for example $X = \mathbb R$), and it is closed if and only if $X$ is a Hausdorff space.

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I'm not quite sure whether what you're asking for actually holds.

Supppose $\sim$ is open, as a set. Let $U \subseteq X$ be open and we need to show that $\pi[U]$ is open, and as $\pi$ is a quotient map, this is equivalent to showing that $\pi^{-1}[\pi[U]]$ is open in $X$. Now, unraveling the definitions:

$$x \in \pi^{-1}[\pi[U]] \iff \exists y \in U: x \sim y\tag{1}$$

So that given $x \in \pi^{-1}[\pi[U]]$ we have a pair $(x,y) \in \sim$ so by openness of that set, plus the definition of the product topology we have $O_x, O_y$ open in $X$ with $x \in O_x, y \in O_y$ and $O_x \times O_y \subseteq \sim$.

Now, is $O_x \subseteq \pi^{-1}[\pi[U]]$? If $z \in O_x$ then $(z,y) \in \sim$ (as $O_x \times O_y \subseteq \sim$) and so by $(1)$ in reverse direction we have that indeed $z \in \pi^{-1}[\pi[U]]$. So every point of $\pi^{-1}[\pi[U]]$ is interior and $\pi^{-1}[\pi[U]]$ is open, so $\pi[U]$ is open etc.

The reverse ($\pi$ open then $\sim$ open) fails, as (as red_trumpet pointed out) when $\sim = \{(x,x): x \in X\}$, the trivial equivalence relation, $\sim$ is open iff $X$ is discrete while the induced $\pi$ is just a homeomorphism, thus open. So counterexamples galore.

For closedness that direction also fails: if $X$ is not Hausdorff (so infinite and cofinite-topology or indiscrete-topology etc. ) and $\sim$ is trivial, then again $\pi$ is closed (a homeomorphism) and $\sim$ is not (as the diagonal is closed iff $X$ is Hausdorff).

I strongly suspect there is an example of a closed $\sim$ where $\pi$ is not a closed map as well.

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  • $\begingroup$ The reverse fails. See red_trumpet's answer. $\endgroup$ – Paul Frost Jan 2 at 15:26

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