Continuous open metrizable image of a Polish space is Polish

Question

Let $X$ be a Polish space and $Y$ a metrizable space. Assume that there exists a continuous mapping $f \colon X \to Y$ such that $f(X)=Y$ and $f(G)$ is open in $Y$ for every open set $G \subseteq X$.

Under these conditions I'm trying to prove that $Y$ is actually a Polish space. Since $X$ is Polish (and thus separable), it is clear that $Y$ is separable (being a continuous image of a separable space). Unfortunately, I am not able to prove that $Y$ is completely metrizable. I tried the following:

Since $Y$ is assumed to be metrizable, there is a metric $d$ on $Y$ which is compatible with the topology of $Y$. Let $(\hat{Y}, \hat{d})$ be the completion of $(Y,d)$. Then $(\hat{Y}, \hat{d})$ is a separable complete metric space. Therefore, it suffices to show that $Y$ is $G_{\delta}$ in $(\hat{Y}, \hat{d})$. But I don't know how to show this and how to use the openness of $f$.

I will appreciate any help.

Answer 1

I actually needed this fact earlier today to show an unrelated result (namely that the quotient of a Polish group by a closed subgroup is Polish) and I remembered this question since I spent some time thinking about it when it was posted. The answer by Henno Brandsma is quite nice, but I had a feeling there had to be an elementary approach, so I did some digging and I found a proof in Hjorth's book Classification and Orbit Equivalence Relations which I reproduce below with minimal changes. This is theorem 7.5 in the book, where it is attributed to Hausdorff.

Theorem: Let $X$ be Polish and $Y$ metrizable. If $\pi\colon X\to Y$ is open, continuous and onto, then $Y$ is Polish.
Proof: Let $\widehat{Y}$ be the completion of $Y$, which is Polish since $Y$ is separable, so we only need to show that $Y$ is $G_\delta$ in $\widehat{Y}$.

We build a Lusin scheme on $X$, that is a set of open subset of $X$ indexed over $\Bbb N^{<\Bbb N}$ with the following properties:

1. $N_\varnothing=X$.
2. $\mathrm{diam}(N_s)<2^{-|s|-1}$.
3. $N_t=\bigcup_{|s|=|t|+1}\{N_s\mid t\subset s\}$.
4. $\overline{N_s}\subset N_t$ for $t\subset s$.

Now for every $s\in\Bbb N^{<\Bbb N}$ pick $W_s$ open in $\widehat{Y}$ such that $W_s\cap Y=\pi(N_s)$ (here we are using that $\pi$ is an open map). Using that open sets are $F_\sigma$, since we are in a metric space, we can now build open $V_s\subseteq\widehat{Y}$ indexed over $\Bbb N^{<\Bbb N}$ with the following properties:

1. $\bigcup\{V_s\mid s\in\Bbb N^1\}=\bigcup\{W_s\mid s\in\Bbb N^1\}$.
2. $V_s\subseteq W_s$ for all $s\in\Bbb N^{<\Bbb N}$.
3. For all $k\in\Bbb N$, $s\in\Bbb N^k$, $$\bigcup\{V_t\mid t\in\Bbb N^{k+1},t\supset s\}=V_s\cap\bigcup\{W_t\mid t\in\Bbb N^{k+1},t\supset s\}.$$
4. For all $y\in\widehat{Y}$, $k\in\Bbb N$ and $s\in\Bbb N^k$, there are only finitelymany $t\supset s$ with $t\in\Bbb N^{k+1}$ with $y\in V_t$, and there are only finitely many $t\in\Bbb N^1$ with $y\in V_t$.

$4$ is the only difficult condition, and it is where we use that open sets are $F_\sigma$. Namely write each $W_t$ as $\bigcup_n C_{n,t}$ for closed sets $C_{n,t}$ with $C_{n,t}\subseteq C_{n+1,t}$ for all $n\in\Bbb N$. Then we can demand that for $t_0,t_1\in\Bbb N^{k+1}$,$s\in\Bbb N^k$ with $s\subset t_0,t_1$, if $$t_0(k)<t_1(k),$$ then $$W_{t_1}\cap C_{n,t_0}=\varnothing$$ for $n<t_1(k)$. As a consequence for every $y\in\widehat{Y}$ and $k\in\Bbb N$ there are only finitely many $s\in\Bbb N^k$ with $y\in V_s$. After this is all done it's clear that $$Y\subseteq\bigcap_{k\in\Bbb N}\bigcup_{s\in\Bbb N^k}V_s,$$ so showing the revere inclusion is enough to conclude that $Y$ is $G_\delta$ in $\widehat{Y}$. To show the reverse inclusion fix $y$ in the RHS and consider $$T_y=\{s\in\Bbb N^{<\Bbb N}\mid \text{ there exist infinitely many $t\supset s$ with $y\in V_t$}\},$$ which is a tree in the sense that $s\subset t\in T_y\implies s\in T_y$. By 3 if $s\in T_y$, then $y\in V_s$ and by 4 for each $s\in T_y$ there is a proper extension $s'\supset s$ also in $T_y$, thus we have infinite branches in the tree and we can find $f\in\Bbb N^{\Bbb N}$ such that

1. $f\upharpoonright k\in T_y$ for all $k\in\Bbb N$.
2. $y\in V_{f\upharpoonright k}$, for all $k\in\Bbb N$.

And since $X$ is complete we can also find an $x\in\bigcap_{k\in\Bbb N}N_{f\upharpoonright k}$. Since $\{N_{f\upharpoonright k}\mid k\in\Bbb N\}$ is a neighbourhood basis for $x$ and $\pi$ is open and continuous, it follows that $y$ is contained in every open set containing $\pi(x)$, thus $y=\pi(x)$ and we obtain the reverse inclusion, which concludes the proof.

Answer 2

This may not be the answer you want, since I am only providing further reading, but the following is a result of Hausdorff (1934).

Theorem A. If $X$ is completely metrizable, and there is a continuous open map $f:X\to Y$ from $X$ onto a metrizable space $Y$, then $Y$ is completely metrizable.

I now quote from A Note on Completely Metrizable Spaces by E. Michael, where a different proof of this result is discussed.

The simplicity of this argument contrasts strikingly with Hausdorff's complicated and lengthy proof in [H2].

So you may want to check out the details in Michael's paper. In particular, the proof discussed there goes through the notion of a "complete open sieve". It is shown (in earlier works) that a metrizable space is completely metrizable if and only if it has a complete open sieve. So, in Theorem A, $X$ has a complete open sieve, which is mapped to a complete open sieve in $Y$ under $f$, hence $Y$ is completely metrizable. I did not look further into these details, but perhaps in the setting that $X$ is Polish it is easier to describe the complete open sieve in question.

As a final remark, a generalization of Theorem A to the case that $Y$ is paracompact is given as Exercise 5.5.8(d) in General Topology by Engelking.

Answer 3

If $X$ is Polish, it embeds as a $G_\delta$ in the Hilbert cube $[0,1]^\omega$. But the open map doesn't help us there, because it concerns sets around $X$. So we need an internal characterisation in terms of open sets preferably.

Well, a metrisable space is completely metrisable iff it's Čech-complete and that has the following characterisation:

$X$ has a family of open covers $\mathcal{U}_n, n \in \omega$ such that every family of closed sets $\mathcal{F}$ that has the finite intersection property and such that $$\forall n \in \omega: \exists F \in \mathcal{F}: \exists U \in \mathcal{U}_n: F \subseteq U$$ then $\bigcap \mathcal{F} \neq \emptyset$.

This compact-like property is equivalent to Čech-completeness in Tikhonov spaces and so equivalent to complete metrisability if $X$ is already known to be metrisable. It was a tempting idea to just to take the images under $g$ of these open covers for the completely metrisable $X$ to get open covers of $Y$. But $F \subseteq g[U]$ does not always imply $g^{-1}[F] \subseteq U$, so that idea didn't work either (I couldn't show the crucial property for the covers in $Y$ from that property in $X$, but maybe someone else sees an easy way to "fix" this using paracompactness or suitable refinements.

But then I found in Engelking (to find the example of non-Čech-complete space which is the open image of a Čech-complete space) the following theorem

[Pasynkov 1967]: If $f:X \to Y$ is open continuous onto $Y$ and $X$ is locally Čech-complete and $Y$ is paracompact, then $Y$ is Čech-complete.

From this your desired result follows quite easily. But further on in Exercise 5.5.8 in Engelking was exactly what we need, due to Michael and earlier Hausdorff and Sierpiński (1934 resp 1930 for more restrictive classes):

Michael's 1959 version: If $f:X \to Y$ is open continuous and onto, and $X$ is completely metrisable and $Y$ is paracompact (Hausdorff) then $Y$ is completely metrisable.

The paper is: A theorem on semi-continuous set valued functions, Duke Math. J. 26 (1959), 647-651 (part of a nice series of papers on selection theorems, real classics, BTW)