Let $G$ be a connected topological group. And $H$ be a discrete subgroup.

Theorem : $\pi_1(g/H)=H$

This is the content in the book Algebraic Topology -Greenberg and Harper

I want to know the proof. In this book, take an open set $U$ around $1$ such that $H\cap U = \emptyset $. And take open set $V$ in $U$ such that $gh^{-1} \in U$ for all $g$, $h \in V$

I cannot understand why this argument is helpful to prove the theorem. 
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    $\begingroup$ I think you need that $G$ is simply connected. $\endgroup$ – Jason DeVito Nov 13 '12 at 2:26
  • $\begingroup$ Yes. Here $G$ is simply connected and $H$ is normal. $\endgroup$ – HK Lee Nov 13 '12 at 2:56

There are two steps to the proof:

  1. Show that the projection $p\colon G\to G/H$ is the universal cover of $G/H$.
  2. Show that the group of deck transformations of $p$ is isomorphic to $H$. Since $\pi_1(G/H)$ is isomorphic to the group of deck transformations of $p$, this will prove the theorem.

Assume for the moment we know (1) to be true. Then statement (2) is not so hard to prove. For each $h\in H$, let $L_h\colon G\to G$ be the map $L_h(g) =hg$. Note that $L_{h_1}\circ L_{h_2} = L_{h_1h_2}$, so the set $\{L_h: h\in H\}$ forms a group isomorphic to $H$ under composition. Every $L_h$ is a deck transformation of $p$. On the other hand, suppose $\varphi$ is a deck transformation of $p$. Then $\varphi(1)\in H$, so $L_{\varphi(1)^{-1}}\circ \varphi$ is a deck transformation of $p$ that fixes $1\in G$. The only deck transformation that fixes a point is the identity, so $L_{\varphi(1)^{-1}}\circ \varphi = Id$, and hence $\varphi = L_{\varphi(1)}$. This proves that the group of deck transformations of $p$ is exactly $\{L_h : h\in H\}\cong H$.

Now we have to show (1), i.e., that $p$ is a covering map. Let $U$ and $V$ be as you stated in your question.

Claim: For all $h\in H$ such that $h\neq 1$, the sets $V$ and $hV$ are disjoint.

Indeed, if this were not the case, then there would exist $v,w\in V$ and $1\neq h\in H$ such that $v = hw$. Now we use your construction of $V$ to conclude that $h = vw^{-1}\in U$. Thus $h\in U\cap H = \{1\}$, a contradiction of the assumption that $h\neq 1$. This proves the claim.

Fix $g\in G$, and let $W = p(Vg)$. Note that $W$ contains $p(g)$, and is an open set in $G/H$, since $p^{-1}(W) = \bigsqcup_{h\in H} hVg$ is open (this is the definition of the quotient topology). Moreover, the restrictions $p|_{hVg}\colon hVg\to W$ are homeomorphisms for each $h\in H$, again by the definition of the quotient topology. This is exactly what needs to happen for $p$ to be a covering map.


A handy way to see this is to consider the long exact sequence $...\to\pi_1(G)\to\pi_1(G/H)\to\pi_0(H)\to\pi_0(G)\to...$ you find $\pi_1(G)=0$ and $\pi_0(G)=0$ because $G$ is simply connected and $\pi_0(H)=H$ because $H$ is discrete. This just give $\pi_1(G/H)=H$.

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    $\begingroup$ Thanks for the answer. I edited the post to format it nicely. Check out math.stackexchange.com/help/notation for a quick reference on mathjax formatting (essentially the same as LATEX) in the future. $\endgroup$ – TravisJ Apr 21 '15 at 0:46

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