Universal covering space for non-semilocally simply connected spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:11:28Z http://mathoverflow.net/feeds/question/111310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces Universal covering space for non-semilocally simply connected spaces Glen M Wilson 2012-11-02T21:34:44Z 2012-11-03T17:02:35Z <p>Consider a topological space $X$. Let us consider a universal covering space to be a covering $p : \tilde{X} \rightarrow X$ which is a covering of all other covering spaces. (Perhaps I should call this an initial covering space). That is, for any other covering space $q : Y \rightarrow X$, there is a map $f : \tilde{X} \rightarrow Y$ such that $p = q \circ f$. </p> <p><strong>Question:</strong> Does there exist a space $X$ which is not semilocally simply connected with a universal covering space in this sense? Such a covering space is necessarily not simply connected. </p> <p>I've seen some papers by Brazas, Biss, Cannon and Conner, but I couldn't see how to especially use their results in finding such an example. I'd expect such an example to be in the literature somewhere. Could somebody give me a reference in this case? </p> <p>One idea is to find a non semilocally simply connected space $X$ with finite fundamental group, e.g. $\mathbb{Z}_2$. In this case, it might be possible that $X$ itself acts as a universal covering space. </p> http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces/111322#111322 Answer by Tom Goodwillie for Universal covering space for non-semilocally simply connected spaces Tom Goodwillie 2012-11-03T00:15:31Z 2012-11-03T17:02:35Z <p>You probably mean to assume that $X$ is connected, and to allow only covering spaces $E\to X$ such that $E$ is connected.</p> <p>The connected covering spaces of $X$ are the objects of a category, where a morphism from $E\to X$ to $E'\to X$ means a continuous map $E\to E'$ that respects the projection to $X$. In general such a map $E\to E'$ is not itself a covering space, although this is true in the good case, i.e. when $X$ is connected and locally path-connected and semi-locally simply connected.</p> <p>In the good case, that category of connected covering spaces is equivalent to the category of sets with transitive action of $G=\pi_1(X,x)$. Note that the universal covering space is not initial; it corresponds to a set with a free transitive $G$-action; it has a map to every object, but not a unique one.</p> <p>Certainly there are non-slsc examples in which the category of connected covering spaces has a "universal"object. For example, the "topologist's sine curve", which is connected but not path-connected, has no covering spaces except the one-sheeted one. </p> http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces/111388#111388 Answer by Jeremy Brazas for Universal covering space for non-semilocally simply connected spaces Jeremy Brazas 2012-11-03T16:15:34Z 2012-11-03T16:15:34Z <p>My answer to <a href="http://mathoverflow.net/questions/58341/is-a-space-with-no-covering-spaces-simply-connected" rel="nofollow">this question</a> gives an example of a locally path connected (but non-semilocally simply connected) space $HA\subset\mathbb{R}^3$ called the Harmonic archipelago: draw the Hawaiian earring on a disk and between each hoop push the surface straight up to make a hill of height 1. </p> <p><img src="http://www.freeimagehosting.net/newuploads/f8n4t.png" alt="Harmonic archipelago"></p> <p>The fundamental group is uncountable but the only covering of $HA$ is itself so it is the unique object in its category of coverings. This weird covering behavior happens because you can deform any loop as close as you want to the basepoint (over finitely many hills) but cannot actually contract it all the way <strong>to</strong> the basepoint because the homotopy would have to take the loop over infinitely many of hills (contradicting compactness).</p> <p>Topologies on the fundamental group detect this behavior. For instance the quotient topology from the loop space (the <em>quasitopological fundamental group</em>, $\pi_{1}^{qtop}(X)$) satisfies: If $p:Y\to X$ is a covering map, then $p_{\ast}:\pi_{1}^{qtop}(X)\to \pi_{1}^{qtop}(Y)$ is an open embedding of quasitopological groups. So if $\pi_{1}^{qtop}(X)$ is an indiscrete group there are no proper open subgroups so the only covering of $X$ is itself.</p> <p>This works for all spaces including non-locally path connected ones. Really, I'd say it is easier to produce non-locally path connected examples. Here is a compact space $X$ with $\pi_{1}^{qtop}(X)\cong \mathbb{Z}_n$ indiscrete.</p> <p>Let $T\subset \mathbb{R}^2$ be the <a href="http://en.wikipedia.org/wiki/Topologist%27s_sine_curve" rel="nofollow">closed topologists sine curve</a>, $a\in T$ be the endpoint in the open path component and $b$ be a point in the other path component - say the origin. Now let $$Y=\frac{T\times S^1}{T\times \{1\}\cup \{a\}\times S^1}$$</p> <p>In other words, if $a$ is the basepoint of $T$, $Y$ is the reduced suspension $\Sigma T$.</p> <p>You have $\pi_1(Y)\cong \mathbb{Z}$ where the loop $L:S^1\to Y$, $t\mapsto (b,t)$ represents a generator, but every neighborhood of $L$ contains a trivial loop so the quotient topology on $\pi_1(Y)$ is the indiscrete topology. If you attach a 2-cell to $Y$ using $L^n$ as an attaching map, you get a space $X$ where $\pi_{1}^{qtop}(X)\cong \mathbb{Z}_n$ is indiscrete. Both $Y$ and $X$ spaces have no nontrivial coverings.</p> <p>Constructing locally path connected examples with countable indiscrete fundamental groups is much harder. In fact, I think it is an open question whether or not there exists a Peano continuum with finite, indiscrete fundamental group. For such spaces it is more efficient to talking about coverings via the "shape topology" and Spanier groups (the last section of <a href="http://arxiv.org/abs/1207.1310" rel="nofollow">this paper</a> shows the shape topology consists precisely of the data of the category of coverings). Applied to your question: locally path connected $X$ has a categorical universal covering iff there is a minimal open subgroup in $\pi_1(X)$ with the shape topology. If the minimal open subgroup is the trivial subgroup you get back a classical universal covering and a discrete group.</p>