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Let X be a topological space that is not too bad (let's say "not too bad" = "compactly generated Hausdorff"), and let ∼ be an equivalence relation such that X /∼ is compact Hausdorff.

Does there exist a compact subspace A⊂X that meets every equivalence class of ∼?
(This would then imply that A /∼ is homeomorphic to X /∼).

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I found a non-separable counterexample, so let me add "separable" to "not too bad". –  André Henriques Aug 3 '12 at 14:11
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4 Answers 4

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The answer is positive if $X$ is second countable and locally compact, and $X/\sim$ is first countable (in addition to being compact Hausdorff). Proof: we claim that any point in $X/\sim$ has a neighborhood which is contained in the image of a compact subset of $X$. Given this, the rest is easy: use compactness of $X/\sim$ to find a covering of it by finitely many such neighborhoods and take the union of the compact sets whose images contain them. (This is Agol's technique.)

To prove the claim, suppose it fails and let $x \in X/\sim$ be a falsifying point. Fix a countable base $(U_n)$ of $X$ (second countability) and wlog assume each $U_n$ is precompact (local compactness). By first countability of $X/\sim$, we can now find a sequence $(x_n)$ in $X/\sim$ that converges to $x$ and such that $x_n$ is not in the image of $U_1 \cup \cdots \cup U_n$. Since $x$ is contained in the image of some $U_n$, eventually $x_n \neq x$, so wlog we can assume $x_n \neq x$ for all $n$. Now let $C$ be the set of points in $X/\sim$ whose image is one of the $x_n$. This set cannot be closed, for then its complement would be open and $(x_n)$ could not converge to $x$. Therefore it must not contain some boundary point $\bar{x}$, and this point must map onto $x$. Finally, by local compactness of $X$ some $U_n$ must contain $\bar{x}$, which contradicts the choice of the sequence $(x_n)$. We conclude that the claim must hold.

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You don't need to assume first countability of the quotient space, since it is automatically second countable. –  BS. Aug 5 '12 at 15:01
    
@BS: a quotient of a second countable space need not be second countable, or even first countable. Let $X$ be the disjoint union of countably many copies of the one point compactification of ${\bf N}$, and let $\sim$ be the equivalence relation which identifies all the copies of $\infty$ ... but does assuming $X/\sim$ is compact help? –  Nik Weaver Aug 5 '12 at 15:29
    
You're right. On the other hand, since $X$ is first countable, its quotient is sequential (the topology is determined by convergent sequences), and this suffices for your argument. This implies in turn that the quotient is a quotient of a compact sec. count., hence metrisable space (by your result). Finally this implies the quotient is a subspace of the Hausdorff compact metrisable space of compacts of $X$ (the equivalences classes), hence is metrisable, i.e. second countable. Thus compactness of the quotient helps, eventually. –  BS. Aug 5 '12 at 17:22
    
I don't quite see how being sequential suffices for my argument ... –  Nik Weaver Aug 5 '12 at 18:00
    
You're again right. Too bad... –  BS. Aug 5 '12 at 18:28
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Here's a non-separable counterexample:

Let $D^2:=\{(\theta,r)|\theta\in S^1, r\in[0,1]\}/\sim$ be the unit disc in $\mathbb R^2$, parametrized in polar coordinates.
Let's call a subset $K\subset D^2$ thin if $0\in K$, it is compact, and for every convergent sequence $(\theta_n,r_n)\to 0$ of elements of $K$, the limit $\lim\theta_n\in S^1$ exists.

Let $X$ be disjoint union of all thin subspaces of $D^2$. Then $X$ maps surjectively onto $D^2$, and this is a quotient map.

However, no compact subspace of $X$ maps surjectively onto $D^2$. A compact subspace of that infinite disjoint union is necessarily contained in a finite disjoint union, and the union of finitely many thin subspaces cannot be the whole of $D^2$.

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Nice! I'm deleting my faulty answer. –  Nik Weaver Aug 3 '12 at 15:11
    
To get a separable counterexample, can't you just let $X$ be the disjoint union of a countable family of thin subspaces that cover $D^2$, for example, $K_\alpha$ for rational angles $\alpha$ where $K_\alpha = \{(\theta, r) : |\theta - \alpha| \le r\}$? –  Trevor Wilson Aug 3 '12 at 16:11
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@Trevor: the hard part is getting the natural map onto $D^2$ to be a quotient map. Since there are countably many $K_\alpha$, you can find a sequence of values $(\theta_n, r_n)$ that converges to $0$ and eventually avoids every $K_\alpha$. Then $D^2$ minus this set (but including $0$) will be open in the quotient topology, so you get the wrong topology. –  Nik Weaver Aug 3 '12 at 18:51
    
@Nik: Oh, I see. Thanks for the explanation. –  Trevor Wilson Aug 3 '12 at 19:11
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Here's a general version of the $D^2$ idea. Replace $D^2$ with any first countable compact Hausdorff space $Y$ at least one point of which has no countable neighborhood, and let $X$ be the disjoint union of all subsets of $X$ of the form: a convergent sequence together with its limit. –  Nik Weaver Aug 6 '12 at 1:53
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Here is a counterexample, inspired by Henrik Rüping's answer, but somewhat simpler:

Take $X$ to be the disjoint union of
$\quad X_1:=([0,1]\times [0,1])\setminus \{(0,0)\}$ and
$\quad X_2:=(]0,1]\times [0,1])\cup ([-1,0]\times\{0\})$.
Both $X_1$ and $X_2$ are given the subspace topology from $\mathbb R^2$.

The quotient space is $[0,1]^2$, with quotient map $\pi:X\to [0,1]^2$.
The map $\pi|_{X_1}$ is the inclusion $X_1\hookrightarrow [0,1]^2$.
The map $\pi|_{X_2}$ is a continuous bijection $X_2 \to [0,1]^2$ given by $\pi(-a,0)= (0,a)$ for $(-a,0)\in [-1,0]\times\{0\}$.

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If $X$ is locally compact, and the map $X \to X/\sim$ is an open mapping, then I think this holds. Consider a collection of open subsets $\cup_i U_i=X$, such that $\overline{U_i}$ is compact. Then the image of $U_i$ in $X/\sim$ is open by hypothesis. Choose finitely many $U_i$'s whose images cover $X/\sim$, then the union of their closures gives a compact subset mapping onto $X/\sim$.

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Quotient maps are always open, aren't they? So I think you've shown there is no locally compact counterexample. –  Nik Weaver Aug 1 '12 at 23:23
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The quotient map $f: [0,1]\to S^1$ which glues $0$ to $1$ is not open: $f^{-1}\left(f\left([0,\delta)\right)\right)$ contains $1$ but does not contain $(1-\epsilon,1]$ for any $\epsilon$. –  Noah Stein Aug 2 '12 at 2:22
    
Oh, you're right. So let me think about whether my counterexample can be modified to be locally compact. –  Nik Weaver Aug 2 '12 at 3:58
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