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?



Continuous and open image of a Polish space
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.


