Polish by compact is Polish? Let $X,Y$ be separable and metrizable, with $Y$ Polish, and suppose there is a topological quotient map $f:X\to Y$ with compact fibers. Is $X$ Polish?
I have a specific space in mind, so if the answer is no and some extra criterion comes to mind ($Y$ is not locally compact, and $f$ is not a fiber bundle), I would appreciate it. 
 A: I think the answer is no. Consider the following subsets of the Euclidean plane. 
Let $X_0 = \{(p,q)\in (\mathbb R\setminus \mathbb Q)\times \mathbb R: 
 0\le p \le 1, 0 \le q \le 1 \}$, $X_1 = \{(p,q)\in (  \mathbb Q)\times \mathbb R: 
 0\le p \le 1, -1  \le q \le 0 \}$, 
Let  $X=X_0\cup X_1$, $Y=[0,1]$, and let $f$ be the projection.  
(ADDED: $X$ is not Polish, as $X$ contains a closed subset homeomorphic to $\mathbb Q$.  Closed subsets of Polish spaces are Polish, but $\mathbb Q$ is not.) 
A: $X$ need not be Polish; here's a counterexample:
Let $Y$ be the real line with its usual topology, and let $A$ be an arbitrary subset of $Y$. (I'll specialize $A$ later, but for now, let it be arbitrary.) Let
$$ 
X=\{(a,b)\in Y\times[0,1]:b=0\text{ or }a\in A\}.
$$
Then the projection $(a,b)\mapsto a$ is a topological quotient map.  Its fibers are copies of $[0,1]$ (over the points in $A$) and singletons (over the other points of $Y$), so they are compact. $X$ is a separable metric space, a subspace of $Y\times[0,1]$. 
What remains is to see whether $X$ is Polish. Since it's a subspace of $Y\times[0,1]$, it will be metrizable by a complete metric (and therefore Polish) if and only if it is a $G_\delta$ subset of $Y\times[0,1]$. There are only $\mathfrak c$ (the cardinality of the continuum $G_\delta$ subsets of $Y\times[0,1]$, and there are $2^{\mathfrak c}$ different possibilities for $A$, all yielding different sets $X$.  So, for most choices of $A$, the resulting $X$ sill not be Polish.
