# Open Quotient Map and open equivalence relation.

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?

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.

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.

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