Let $X$ be a locally compact, second countable Hausdorff topological space and let $Y$ be a Hausdorff quotient of $X$. Let $q:X\to Y$ denote the quotient map. Then for $y\in Y$, $q^{-1}(y)$ is a closed subset of $X$ but not necessarily compact. 


Is it possible to find another locally compact, second countable Hausdorff space $X\_1$ such that $Y$ is quotient of $X\_1$ and such that for all $y\in Y$, $q\_1^{-1}(y)$ is compact, where $q\_1:X\_1\to Y$ is the quotient map? Is it possible even to arrange that each $q\_1^{-1}(y)$ is finite?



Hausdorff quotient space with compact or finite inverse images
I think the answer is no. I'll basically point to two exercises in Engelking.


A map $q:X\to Y$ is called *hereditary quotient*, if for any $B\subset Y$, the restriction of $q$ on $q^{-1}(B)$ is a quotient map (see Exercise 2.4.F for some characterizations). In particular, it is said there that any quotient map onto a Hausdorff Frechet-Urysohn space is hereditary quotient.


On the other hand Exercise 3.7.D says that if $q$ is hereditary quotient with compact preimages of points, then $w(Y)\le w(X)$ and if $X$ is locally compact, then $Y$ is locally compact (the point is that hereditary quotient + compact preimages + $X$ is locally compact implies that the map is perfect, and in particular closed).


Now let $X=\mathbb{R}$ and let $q$ be the map that sends all integer into a point, and so you get a countable wedge of circles. Note that it is not locally compact at the point where all the circles meet. However, it is a metrizable space and in particular, it is Frechet-Urysohn. 


Hence, if $X\_1$ is locally compact and $q\_1:X\_1\to Y$ is quotient, then it is hereditary quotient, and so if we further assume that preimages under $q\_1$ are compact, then $Y$ is locally compact, which leads to a contradiction.


