compact quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:17:57Z http://mathoverflow.net/feeds/question/103728 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103728/compact-quotient compact quotient André Henriques 2012-08-01T21:18:19Z 2012-08-06T18:27:18Z <p>Let <b>X</b> be <b>a topological space</b> that is not too bad (let's say "not too bad" = "compactly generated Hausdorff"), and let ∼ be an equivalence relation such that <b>X /∼ is compact Hausdorff</b>.</p> <p>Does there exist a <b>compact subspace A⊂X</b> that meets every equivalence class of ∼?<br> (This would then imply that <b>A /∼</b> is <b>homeomorphic to X /∼</b>).</p> http://mathoverflow.net/questions/103728/compact-quotient/103738#103738 Answer by Agol for compact quotient Agol 2012-08-01T23:00:37Z 2012-08-01T23:00:37Z <p>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 $X/\sim$. </p> http://mathoverflow.net/questions/103728/compact-quotient/103866#103866 Answer by André Henriques for compact quotient André Henriques 2012-08-03T14:30:00Z 2012-08-03T14:30:00Z <p>Here's a non-separable counterexample:</p> <p>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.<br> Let's call a subset $K\subset D^2$ <i>thin</i> 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.</p> <p>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. </p> <p>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 $D^2$.</p> http://mathoverflow.net/questions/103728/compact-quotient/104014#104014 Answer by Nik Weaver for compact quotient Nik Weaver 2012-08-05T11:56:43Z 2012-08-06T18:27:18Z <p>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.)</p> <p>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.</p> http://mathoverflow.net/questions/103728/compact-quotient/104022#104022 Answer by André Henriques for compact quotient André Henriques 2012-08-05T14:00:00Z 2012-08-05T14:00:00Z <p>Here is a counterexample, inspired by Henrik Rüping's answer, but somewhat simpler:</p> <p>Take $X$ to be the disjoint union of<br> $\quad X_1:=([0,1]\times [0,1])\setminus \{(0,0)\}$ and<br> $\quad X_2:=(]0,1]\times [0,1])\cup ([-1,0]\times\{0\})$.<br> Both $X_1$ and $X_2$ are given the subspace topology from $\mathbb R^2$.</p> <p>The quotient space is $[0,1]^2$, with quotient map $\pi:X\to [0,1]^2$.<br> The map $\pi|_{X_1}$ is the inclusion $X_1\hookrightarrow [0,1]^2$.<br> The map $\pi|_{X_2}$ is a continuous bijection $X_2 \to [0,1]^2$ given by $\pi(-a,0)= (0,a)$ for $(-a,0)\in [-1,0]\times\{0\}$.</p>