# Continuous and open image of a Polish space

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?

• 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. Sep 13 '15 at 4:46
• The problem was not the separability, i was stuck in trying to prove that $Y$ is completely metrizable. Sep 13 '15 at 12:28

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.

• Is it true that if we only require $f$ open and $Y$ metric, then $Y$ is Polish? Sep 28 '15 at 14:31
• @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.) Sep 28 '15 at 16:14
• You mean Excercise 5.5.8 (d), in page 341 right? Sep 29 '15 at 2:10