This question stems from exercise 3.24 of Rotman's Algebraic Topology.
Let $G$ be topological group and let $H$ be a discrete normal subgroup of $G$. Now it is easy to see that the canonical map $\pi$ from $G$ to $G/H$ (which is in this case an identification) is actually a covering map. In details:
- We may choose some open neighborhood $U$ of the identity (in $G$) that intersect $H$ at the identity only, by the discreteness of $H$.
- We may also choose an open neighborgood $V$ of the identity that such that $$V \subset U, VV \subset U, V=V^{-1}$$ using the topological group structure of $G$.
- Now the map $\pi\restriction_{V}:V\rightarrow\pi(V)$ is a homeomorphism, and the cosets $hV$ are disjoint, so we actually have a map $\pi\restriction_{HV}:HV\rightarrow\pi(V)$ which evenly covers $\pi(V)$.
- Finally, homogeneity of $G$(and $G/H$) implies that $\pi$ is a covering map.
Exercise 3.24 states that when $G$ is simply connected, we have in fact that $\pi_{1}(G/H)$ isomorphic to $H$. Now this can be proved by mimicking the standard proof (using 'winding numbers') of the fact that the fundamental groups of the circle is $\mathbb{Z}$, which usese the path lifting, homotopy lifting properties of covering maps. We just have to do the same thing using the map $\pi$ this time, constructing a group isomorphism from $\pi_{1}(G/H)$ to its 'winding numbers', that is, $H$.
My question is, Rotman assumes in this exercise that H be a discrete CLOSED normal subgroup of $G$, and also says at the end: Remark. If $G$ is $T_{0}$, then every discrete subgroup of $G$ is necessarily closed.
It is well known that every topological group must be regular, so I'm perfectly fine with that last 'remark' since regularity and $T_{0}$-ness forces the topological group to be $T_{3}$. However, I've been going through the details of the proof that $\pi_{1}(G/H)$ is isomorphic to $H$, but nowhere have I found a place the closedness of $H$ is necessary. It seems like everything works just fine without the assumption that $H$ is closed.
Is closedness really unnecessary here? or am I missing something important?
