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.

  • 1
    $\begingroup$ 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. $\endgroup$ – freakish Mar 8 '19 at 14:54
  • $\begingroup$ 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. $\endgroup$ – ugur efem Mar 8 '19 at 14:57
  • $\begingroup$ @freakish the product topology, of course. The only natural chocie. $\endgroup$ – Henno Brandsma Mar 9 '19 at 6:18
  • $\begingroup$ What is the Noetherian topology on $D \times D$?? It's true for the product topology just in general topology. $\endgroup$ – Henno Brandsma Mar 9 '19 at 6:20
  • $\begingroup$ @HennoBrandsma the OP clearly says that the topology is not necessarily the product topology. Please read the question again. $\endgroup$ – freakish Mar 9 '19 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.


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.

  • $\begingroup$ Typo in the second line (compact). $\endgroup$ – Paul Frost Mar 9 '19 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.