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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?

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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.

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  • $\begingroup$ So it would seem that the simplest case of interest provides a counter-example. I should have done my homework better! But many thanks. $\endgroup$
    – Douglas Somerset
    Commented Jun 1, 2018 at 7:39
  • $\begingroup$ I am mystified. The two exercises from Engelking cited in the answer would seem to show that if $X$ is locally compact, second countable and Hausdorff, and preimages are compact, then Y is locally compact. But there is a standard example $Y=S_{\omega}$ due to Arhangelski and Franklin in which $X$ has all those properties, and preimages are pairs of points, yet $Y$ is nowhere locally compact. Can Engelking be wrong? His exercise (actually 3.7.E in my copy) cites a 1963 paper from Tong which I have not been able to locate. $\endgroup$
    – Douglas Somerset
    Commented Jun 1, 2018 at 18:22
  • $\begingroup$ Is $Y$ a Frechet-Urysohn space? and what is $X$? I understand that $S_\omega$ is sequential, but this is not enough. $\endgroup$
    – erz
    Commented Jun 1, 2018 at 19:16
  • $\begingroup$ $X$ is a countable union of convergent sequences; and basically one pairs off the countable set of limit points in $X$ with the countable set of non-limit points, so that in $Y$ every point is the limit of a sequence. The quotient is Hausdorff, and is a Frechet-Urysohn space to my understanding. $\endgroup$
    – Douglas Somerset
    Commented Jun 1, 2018 at 19:22
  • $\begingroup$ I presume that the explanation is that $S_{\omega}$ is sequential but is not a Frechet-Urysohn space. I was not distinguishing the two notions. $\endgroup$
    – Douglas Somerset
    Commented Jun 1, 2018 at 20:07

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