Showing that $\pi(G/H, 1) = H$ under a condition

Question

Problem: Let $G$ be a simply connected (i.e., $\pi(G)=1$) topological group, and let $H$ be a discrete normal subgroup. Prove that $\pi(G/H,1) = H$.

I know that since $H$ is a discrete subgroup of $G$, we have that according to this post there exists an open set $U$ that contains $\{1\}$ s.t. the $hU$ are pairwise disjoint with $h \in H$.

Since $H$ is discrete in $G$, it seems that $\pi(H) \cong \{e\}$ since any choice of basepoint will yield only enough room for elements of $\pi(H)$ to be a trivial loop. I'm not sure if this helps us here.

Answer 1

In this answer all loops are based at the identity unless specified otherwise.

Recall that by the homotopy lifting property, since $p: G \rightarrow G/H$ is a covering map, any loop $\gamma: [0,1] \rightarrow G/H$ has a unique lift $\tilde \gamma: [0,1] \rightarrow G$ such that $\tilde \gamma(0) = 1_G$. Consider the map $f: \pi_1(G/H) \rightarrow H$ given by $f(\gamma) = \tilde\gamma(1)$. This is well-defined because the lift is unique, and by the homotopy lifting property again any homotopy $\gamma \simeq \gamma'$ lifts to a homotopy of paths in $X$; since $p(\gamma_t(1))=e$, and $H$ is discrete, we see that $\tilde\gamma'(1)=\tilde\gamma_t(1)=\tilde\gamma(1)$ for all $t$. The codomain is $H$ because $\gamma(1) = 1_{G/H}$, $\tilde \gamma(1) \in H$.

1) This is a homomorphism. For two homotopy classes of loops with representatives $\gamma, \eta$, if $f(\gamma) = h_1$ and $f(\eta) = h_2$, then we may lift $\eta$ to $\tilde\eta_{h_1}(t) = h_1\tilde \eta(t)$, whose basepoint is $h_1$ rather than $1_G$; clearly \tilde\eta_{h_1}(1)=h_1h_2. Since lifts are unique, the lift of the concatenation $\gamma*\eta$ first follows $\tilde \gamma$ and then $\tilde \eta_{h_1}$, so $\widetilde{\gamma * \eta}(1) = h_1h_2$, so it follows that $f$ is a homomorphism.

2) This map is injective. To see this, pick a loop $\gamma$ such that $\tilde \gamma(1) = 1_G$. Since $G$ is simply connected, $\tilde \gamma$ is null-homotopic (keeping the basepoint fixed, as usual). Projecting this homotopy to $G/H$ shows that $\gamma$ is null-homotopic as well. So the kernel of $f$ is trivial.

3) This map is surjective. Indeed, since $G$ is path-connected, fix $h \in H$ and pick a path $\gamma: I \rightarrow G$ with $\gamma(0) = 1_G$, $\gamma(1) = h$. This projects to a loop in $G/H$; its class $[\gamma] \in \pi_1(G/H)$ has $f([\gamma]) = h$.

So $f$ is an isomorphism, and $\pi_1(G/H) \cong H$.

Answer 2

I hope it is allowed to still add my bit. Since by your reference the action of $H$ is properly discontinuous, your problem is a special case of corollary 4.11 describing fundamental groups of orbit spaces. Just in case you are interested in some theory behind the question. Yet another way I think you can even solve it is by using the fact the $G \twoheadrightarrow G/H$ is the universal covering space so in this case we obtain the fundamental group as the deck transformation group which is obviously $H$.