Discrete Closed Subgroup H of a Simply Connected Topological Group G isomorphic to fundamental group of G / H. A problem in Rotman's Algebraic Topology is as follows:
Given a simply connected topological group G with a closed discete normal subgroup H, show that $\pi_1(G / H) \cong H$.
I believe I have this more or less worked out (briefly: in that the quotient map is shown to evenly cover G / H by using the discreteness of H and the fact that we can shrink the neighborhood of 1 so that it is closed under inverses / multiplication, and that the path multiplication of the image in the quotient of path representatives from 1 to elements in H corresponds isomorphically to multiplication of elements in H by lifting), but I don't see why closedness of H is a necessary condition. (We are not given that G is $T_0$, or Hausdorff, or anything that would imply that $H$ is automatically closed.)
Can someone point me in the right direction?
Thank you.
Edit: Maybe the fact that H closed iff $G / H$ is Hausdorff is relevant? I'm willing to be that if I dug through Rotmans proof of $\pi_1 (S^1) = \mathbb{Z}$ I would find that he uses the closedness of $S^1$ somewhere. But I'm still not sure.
 A: The assumption that $H$ is closed is in fact unnecessary, and as far as I can tell your argument works fine without it.
In fact, if you know the result is true for $H$ closed, it follows immediately that it also holds without assuming $H$ is closed.  To prove this, let $K=\overline{\{1\}}\subseteq G$.  Then the quotient $G'=G/K$ is $T_0$, the quotient map $p:G\to G'$ is a homotopy equivalence, and $H'=p(H)$ is a discrete normal subgroup of $G'$.  Any discrete subgroup of a $T_0$ topological group is closed (proof sketch: if elements of $H$ accumulate at some $g\in G\setminus H$, then elements of the form $h_1^{-1}h_2\in H$ where $h_1,h_2$ approach $g$ would approach $1$, contadicting discreteness).  Thus $H'$ is closed in $G'$, and the result in the closed case gives $\pi_1(G'/H')\cong H'$.  But $G'/H'$ is just the $T_0$ quotient of $G/H$ and so the natural map $G/H\to G'/H'$ is a homotopy equivalence, and $p$ is a group isomorphism from $H$ to $H'$.  Thus $\pi_1(G/H)\cong H$.
(Here I am using the general fact that if $X$ is a topological space and $Y$ is the $T_0$ quotient of $X$, then the quotient map $p:X\to Y$ is a homotopy equivalence.  To prove this, note that any section $s:Y\to X$ is automatically continuous.  To prove $sp\simeq 1_X$, just take the homotopy $H(x,0)=s(p(x))$ and $H(x,t)=x$ for all $t>0$.)
