Is compactness (without Hausdorff) enough to get a closed quotient map?

This is some attempt to generalise the classical result for a compact Hausdorff space $$D$$, if $$E\subseteq D\times D$$ is closed, then the quotient map $$p:D\to D/{E}$$ is a closed equivalence relation. Below $$D$$ is not Hausdorff anymore (but (quasi)-compact and $$D\times D$$ is considered with topology that is finer than product topology in terms of closed sets).

Let $$D$$ be a Noetherian topological space. Let $$E\subseteq D\times D$$ be a closed equivalence relation. Note that the topology on $$D\times D$$ is not necessarily the product topology, but it is a Noetherian topology which is finer than the product topology with respect to closed sets (e.g. as for algebraic varieties, I am aware that this is a rather vague statement. I am thinking on it to clarify further).

Consider $$T := D/{E}$$ with the natural quotient topology, and let $$p:D\to T$$ be the quotient map. Is it true that $$p$$ is a closed function, i.e. is the image of a closed set under $$p$$ closed?

Note that $$D$$ is not necessarily Hausdorff, but it is compact (or quasi-compact in algebraic geometry terminology) in the sense that every open cover has a finite subcover.

• What is the topology on $D\times D$? Any finer? For example discrete? If so then the condition "$E\subseteq D\times D$ is closed" is automatically satisfied. And I suspect that $D$ being noetherian is not enough in this case. Although I don't have a counterexample. – freakish Mar 8 at 14:54
• Good point! Thanks. It's certainly not discrete, in fact it'll we Noetherian too. I'll try to clarify the topology in the product further. – ugur efem Mar 8 at 14:57
• @freakish the product topology, of course. The only natural chocie. – Henno Brandsma Mar 9 at 6:18
• What is the Noetherian topology on $D \times D$?? It's true for the product topology just in general topology. – Henno Brandsma Mar 9 at 6:20
• @HennoBrandsma the OP clearly says that the topology is not necessarily the product topology. Please read the question again. – freakish Mar 9 at 9:37

A classical theorem by Kuratowski: if $$K$$ is compact (quasi-compact in Bourbaki parlance, so no separation axioms) and $$X$$ is any space, $$\pi_X: X \times K \to X$$ is closed (where $$X \times K$$ has the product topology and $$\pi_X$$ is the projection onto the first coordinate), which is often formulated as "a projection along a compact space is closed". This property in fact characterises compactness : if $$K$$ is a space such that for any space $$X$$ this $$\pi_X$$ is closed, then $$K$$ is compact.

Now:

If $$X$$ is compact and $$R \subseteq X \times X$$ is a closed equivalence relation (where $$X \times X$$ has the product topology) the natural map $$q: X \to X/{R}$$ is a closed map, where $$X/{R}$$ has the quotient topology wrt $$q$$.

Proof Let $$F$$ be closed in $$X$$. Then a moment's thought shows that (letting $$\pi_1: X \times X \to X$$ be the projection onto the first coordinate):

$$q^{-1}[q[F]] = \pi_1[((X \times F) \cap R]$$

and so $$q^{-1}[F]$$ is closed in $$X$$ as $$\pi_1$$ is a closed map by the Kuratowski theorem and $$(X \times F) \cap R$$ is closed in $$X \times X$$ as $$R$$ is a closed set in the product. So $$q[F]$$ is closed in $$X/{R}$$ as the latter set has the quotient topology wrt $$q$$. So $$q$$ is closed. QED.

• Typo in the second line (compact). – Paul Frost Mar 9 at 10:19