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Suppose that we have a continuous open and closed surjection $f\colon X\to Y$ of a Polish space $X$ to $Y.$ The closeness of $f$ implies that $Y$ is a metric space.

But i do not know how to use that $f$ is continuous and open to prove that $Y$ is Polish, is there some result that implies this?

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  • $\begingroup$ Well, showing $Y$ is separable is pretty straightforward, just using continuity. Let $E$ be a countable dense subset of $X$ and consider $f(E)$. What is $f^{-1}\left(\overline{f(E)}\right)$? Now note that $f$ is surjective. $\endgroup$
    – Nate Eldredge
    Commented Sep 13, 2015 at 4:46
  • $\begingroup$ The problem was not the separability, i was stuck in trying to prove that $Y$ is completely metrizable. $\endgroup$
    – Hector Pinedo
    Commented Sep 13, 2015 at 12:28

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The following list of results will show that $Y$ is Polish.

  • It is basic that the continuous image of a separable space is separable.

  • Also basic is that the open image of a first-countable space is first-countable.

  • Recall the Hanai-Morita-Stone Theorem:

    Let $X$ be a metrizable space, and let $f : X \to Y$ be a closed continuous surjection. Then the following are equivalent.

    1. $Y$ is metrizable.
    2. $Y$ is first-countable.
    3. $\partial f^{-1} \{ y \}$ is compact for each $y \in Y$ (where $\partial A$ denotes the boundary of $A$).

  • (The two previous results imply that that the closed-and-open image of a metrizable space is metrizable.)

  • It is a result of I.A. Vaĭnšteĭn (see Problem 4.5.13(e), p.293, in Engelking's General Topology, revised and completed ed.) that the metrizable closed image of a completely metrizable space is completely metrizable.

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  • $\begingroup$ Is it true that if we only require $f$ open and $Y$ metric, then $Y$ is Polish? $\endgroup$
    – Hector Pinedo
    Commented Sep 28, 2015 at 14:31
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    $\begingroup$ @HectorPinedo Yes, that would also work. (It is a result of Hausdorff that the metrizable open continuous image of a completely metrizable space is completely metrizable. This is also an exercise in Engelking.) $\endgroup$
    – user642796
    Commented Sep 28, 2015 at 16:14
  • $\begingroup$ You mean Excercise 5.5.8 (d), in page 341 right? $\endgroup$
    – Hector Pinedo
    Commented Sep 29, 2015 at 2:10

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