In my general topology class, we're discussing the basics of fundamental groups of topological spaces. When we discussed how $S^1$ had a fundamental group isomorphic to $\mathbb{Z}$, it occurred to me that $S^1$ arose naturally as a quotient space induced by a group action of $\mathbb{Z}$ on $\mathbb{R}$, which (when we'd only discussed the one space) struck me as an interesting coincidence, interesting enough that it likely held in other situations. What I came up with was the following, about which I had two questions: (1) is it sound, and (2) does this result have a name attached to it? I find it strange that this result seems so simple and powerful and yet wouldn't have a name.

Claim: If $X$ is a simply connected topological space (i.e. $X$ is path-connected, and any two loops about the same point are path-homotopic), and the group $G$ acts on $X$ continuously in such a way that the map $p : X \to X / G$, where $p(x) = \{ g . x : g \in G \}$ is a covering map, then $\pi_1 (X / G) \cong G$.

Proof: Let $x_0 \in X$. Recall first the following fact: If $q: X \to Y$ is a covering map, and $\sigma$ is a loop in $Y$ about $q(x_0)$, i.e. where $\sigma(0) = \sigma(1) = q(x_0)$, then there exists a unique path $\overline{\sigma}$ in $X$ such that $\overline{\sigma}(0) = x_0$ and $q \circ \overline{\sigma} = \sigma$; furthermore, if $\sigma, \tau$ are loops in $(Y, p(x_0))$, and $\sigma$ and $\tau$ are path-homotopic, then so are their liftings (i.e. $\sigma \simeq_P \tau \Rightarrow \overline{\sigma} \simeq_P \overline{\tau}$).

Now, we establish a map $\phi : G \to \pi_1 (X / G)$, and confirm that it's a group isomorphism. Since $X$ is path-connected, we know that for any $g \in G$ there exists a path $\gamma_g : [0, 1] \to X$ such that $\gamma_g (0) = x_0, \gamma_g (1) = g . x_0$. Set $\phi(g) = [p \circ \gamma_g]$, which is a loop about $p(x_0)$. To see that the map is well-defined, i.e. that it's independent of which path from $x_0$ to $g . x_0$ we chose, suppose that $\alpha, \beta$ are paths in $X$ from $x_0$ to $g . x_0$. Then \begin{align*} \alpha & = \alpha \beta^{-1} \beta \\ & = (\alpha \beta^{-1}) \beta \\ & \simeq_P \beta , \end{align*} since $\alpha \beta^{-1}$ is a loop at $x_0$, and thus path-homotopic to the trivial loop, so \begin{align*} p \circ \alpha & \simeq_P p \circ \beta \\ \Rightarrow \phi (\alpha) & = \phi (\beta) . \end{align*}

Now we show that $\phi$ is a group homomorphism. Let $g, h \in G$, and let $\alpha$ denote a path from $x_0$ to $g . x_0$, and $\beta$ a path from $x_0$ to $h . x_0$. Furthermore, set $\gamma = g . \beta$, which constitutes a path from $g . x_0$ to $(gh) . x_0$. Thus $\alpha \gamma$ is a path from $x_0$ to $(gh) . x_0$, so \begin{align*} p \circ (\alpha \gamma) & \simeq_P (p \circ \alpha) (p \circ \gamma) \\ & = (p \circ \alpha)(p \circ (g. \beta)) \\ & = (p \circ \alpha)(p \circ \beta) \\ \Rightarrow [p \circ (\alpha \gamma)] & = [p \circ \alpha] [p \circ \beta] \\ \Rightarrow \phi(gh) & = \phi(g) \phi(h) . \end{align*} Thus $\phi$ is a group homomorphism.

It remains only to show that $\phi$ is both injective and surjective. To see it's onto, we refer to the first part of the fact stated earlier. If $\sigma$ is a loop in $(Y, p(x_0))$, then there exists a path $\overline{\sigma}$ with beginning point $x_0$ such that $\sigma = p \circ \overline{\sigma}$. But we also note that $p(\overline{\sigma}(1)) = p(x_0)$, so $\overline{\sigma}(1) \in p^{-1} \{ p (x_0) \} = G. x_0$. Thus there exists some $g \in G$ such that $\overline{\sigma}(1) = g . x_0$. Thus $[\sigma] = [p \circ \overline{\sigma}] = \phi(g)$.

To see it's injective, let $[p \circ \alpha] = [p \circ \beta]$, i.e. $p \circ \alpha \simeq_P p \circ \beta$, where $\alpha, \beta$ are paths from $x_0$, and $\alpha(1) = g. x_0, \beta(1) = h . x_0$. Then $p \circ \alpha \simeq_P p \circ \beta$, so therefore $\alpha \simeq_P \beta$, and therefore $\alpha(1) = \beta(1)$, so $g = h$. Thus $\phi(g) = \phi(h) \Rightarrow g = h$, i.e. $\phi$ is an injection.

  • $\begingroup$ This is a well known theorem and result in topology. I do not think there is a name attached to it (it is theorem 7.2 in Topology and Geometry from Bredon), but it is linked with the study of regular cover. This is a wonderful subject, have fun studying it. $\endgroup$ – Maxime Scott Feb 24 '17 at 18:28
  • $\begingroup$ @MaximeScott I took for granted that it'd be well-known. Another question I didn't think to ask before posting: Is there a "converse", i.e. is there a canonical way to take a space $X$ and induce a simply connected space $Y$ such that $\pi_1 (X)$ acts on $Y$ such that $y \mapsto \pi_1 (X) . y$ is a covering map? $\endgroup$ – AJY Feb 24 '17 at 18:34
  • 1
    $\begingroup$ The object that you seems to be interested in is called the universal cover of a space. You actually need a little more hypothesis on your space for it to admit a canonical universal cover (a simply connected space that covers it). It has to be locally simply connected. The usual example of a space that is not locally simply connected is called the Hawaiian earring (here's a link to the wikipedia page en.wikipedia.org/wiki/Hawaiian_earring ) $\endgroup$ – Maxime Scott Feb 24 '17 at 18:41
  • $\begingroup$ There is a nice book called Algebraic Topology by Allen Hatcher where he study among many other thing covering space theory. The book is available online at this address : math.cornell.edu/~hatcher/AT/ATpage.html . The first chapter talks about what you are interested in and continues with other fundamental notions of algebraic topology. $\endgroup$ – Maxime Scott Feb 24 '17 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.