Topological Groups and Covering Spaces

Question

So the question is suppose $G$ is a topological group and $H$ is a closed, discrete subgroup of $G$, we have to show that the quotient map $p: G\to \frac GH$ is a covering projection. The way I'm looking at it is to consider the action of $H$ on $G$ by left multiplication, such that the orbits are the right cosets. If you can prove that about every $g \in G$ there is a neighbourhood $V_g$ such that $hV_g \cap V_g$ is disjoint for all non identity $h \in H$ then we are done. How do you do that?

Answer 1

$H$ is discrete means that for every $h\in H$ there exists an open subset $U_h$ of $G$ such that $U_h\cap H =\{h\}$.

Take $h=1\in H$ the identity. Let $m:G\times G\rightarrow G$ be the multiplication. $m^{-1}(U_1)\cap (U_1\times U_1)$ is an open subset which contains $(1,1)$. There exists open neighborhoods of $1$, $U'_1,U"_1$ such that $U'_1\times U"_1\subset m^{-1}(U_1)\cap (U_1\times U_1)$. Write $W'_1=U'_1\cap U"_1$. Let $i:G\rightarrow G$ be the inversion defined by $i(g)=g^{-1}$. Define $W_1= i(W'_1)\cap W'_1$. If $u,v\in W_1$, $u,v^{-1}\in W'_1$ thus $u\in U'_1, v^{-1}\in U"_1$ and $uv^{-1}\in U_1$.

For every $g\in G$, consider the neighborhood $V_g= W_1g$. For every $h\in H$, $x\in hV_g\cap V_g$ is equivalent to saying that $x=hy, y\in V_g$ and $x=z, z\in V_g$. You can write $y=y'g, y'\in W_1$ and $z=z'g, z'\in W_1$. We deduce that $hy'g=z'g$, thus $hy'=z'$ and $h= z'{y'}^{-1}\in U_1$. This implies that $h=1$.