2
$\begingroup$

I am trying to produce closed quotient maps, as they allow a good way of creating saturated open sets (as in this question).

A map $f:X\rightarrow Y$is called proper, iff preimages of compact sets are compact. It is called quotient map, iff a subset $V\subset Y$ is open, if and only if its preimage $f^{-1}(V)$ is open. And it is called closed, iff it maps closed sets to closed sets.

So the question is, whether a proper quotient map is already closed.

Note that, I am particular interested in the world of non-Hausdorff spaces.

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ By your definition, a quotient map does not have to be onto. Is this deliberate? (If so, the answer to your question is “no”.) $\endgroup$ – Harald Hanche-Olsen Mar 19 '10 at 13:19
  • $\begingroup$ Never mind that. See my answer below. $\endgroup$ – Harald Hanche-Olsen Mar 19 '10 at 13:29
  • $\begingroup$ It is sufficient to assume that the codomain is locally compact. See ncatlab.org/nlab/show/… $\endgroup$ – Jeremy May 2 '19 at 13:44
3
$\begingroup$

No. Let X={1,2,3} and Y={1,2}. Let f map 1 to 1, 2 and 3 to 2. Let the topology on X be {∅,{2},{1,2},{2,3},{1,2,3}} and that on Y be {∅,{2},{1,2}}. f maps the closed set {3} onto the non-closed set {2}.

| cite | improve this answer | |
$\endgroup$

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.