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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.

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    $\begingroup$ 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, 2010 at 13:19
  • $\begingroup$ Never mind that. See my answer below. $\endgroup$
    – Harald Hanche-Olsen
    Mar 19, 2010 at 13:29
  • $\begingroup$ It is sufficient to assume that the codomain is locally compact. See ncatlab.org/nlab/show/… $\endgroup$
    – Jeremy
    May 2, 2019 at 13:44

1 Answer 1

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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}.

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