$G$ topological group, $H$ discrete normal subgroup, $p$ projection, form Covering Space. Let $G$ be a topological group. Let $H$ be a discrete normal subgroup of $G$. Let $p : G \to G/H$ be the projection map. Show that $(G, p, G/H)$ form a covering space.
Here is what I have so far:
Since $G$ is a topological group, it is Hausdorff, so for any two points $x, y \in G$ there exists disjoint open sets $U, V$ such that $x \in U$ and $v \in V$.
Since $H$ is a discrete normal subgroup of $G$, $gH = Hg$ for all $g \in G$ and for any $h \in H$, $\{h\}$ is an open neighborhood of $h$.
I want to show that for any $\overline g \in G/H$ there exists an open neighborhood $U$ in $G/H$ such that $p^{-1}(U) = \displaystyle \bigcup_i S_i$ a disjoint union of open sets in $G$ such that $p \vert_{S_i} : S_i \to U$ is a homeomorphism, i.e., bijective, continuous and its inverse is continuous.
I found this problem in Rotman's Homological Algebra, in a section on sheaves. (I don't have much topo background, but this looks like an interesting problem and I haven't been able to figure it out).
What seems like a natural approach here is to construct an open neighborhood of $\overline g$ in $G/H$ and show it satisfies these properties. I am having trouble with this, any hints/sketch, etc? How would you go about approaching this problem?
 A: It's all rather straightforward once you know that $p \colon G \to G/H$ is an open map. Apart from that, we will need one special (open) neighbourhood of $1 \in G$.
Since $H$ is discrete, there is a neighbourhood $W$ of $1$ with $W \cap H = \{1\}$. Now let $V$ be an open symmetric (meaning $V^{-1} = V$) neighbourhood of $1$ with $V\cdot V \subset W$.
Then for any $g \in G$ we have $p(V\cdot g)$ an open neighbourhood of $p(g)$, and
$$p^{-1}(p(V\cdot g)) = \bigcup_{h \in H} V\cdot gh$$
is the decomposition we want.
If $v_1 g h_1 = v_2 g h_2$ with $v_1, v_2 \in V$ and $h_1, h_2 \in H$, then
$$v_1^{-1}v_2 = g h_1 h_2^{-1} g^{-1} \in V^{-1}\cdot V \cap H = \{1\},$$
so $v_1 = v_2$ and $h_1 = h_2$, hence the different $V\cdot gh$ are disjoint. This also shows that $p\lvert_{V\cdot gh}$ is injective, since if $p(v_1gh) = p(v_2gh)$, then there is a $h' \in H$ with $v_1 gh = v_2 ghh'$.
For each $h \in H$, the restriction $p\lvert_{V\cdot gh} \colon V\cdot gh \to p(V\cdot g)$ is bijective, continuous and open, hence a homeomorphism.
To see that $p$ is open, note that $U \subset G/H$ is open if and only if $p^{-1}(U)$ is open in $G$ by the definition of the quotient topology, and for $W \subset G$ open, we have $p^{-1}(p(W)) = \bigcup\limits_{h\in H} W\cdot h$ a union of open sets, hence open.
It is worth noting, in case it isn't already known, that the quotient topology makes $G/H$ a topological group, and since $H$ is discrete, it is closed, therefore $G/H$ is a $T_1$ space, and for topological groups, that implies that it is a Hausdorff space (even if $G$ isn't Hausdorff, the quotient by a closed normal subgroup is always Hausdorff).
