Show that $\Phi$ is a covering map Let $G$ be a locally connected topological group, $H < G$ a closed locally connected subgroup and $H_0$ the $H$ identity connected component.
How can i show that $\Phi : G/H_0 \to G/H$ is a covering map where
$\Phi (g H_0) = gH$
I'm assuming the neighborhood i want to show the homeomorphism is the same i get from $G$ being locally connected, but i'm stuck at the calculation.
 A: Here is a proof in the case when $G$ is metrizable with a metric $d$. (More generally, the same proof works if $G/\bar{e}$ is 1st countable, where $\bar{e}$ is the closure of $e$ in $G$.)
WLOG (Birkhoff–Kakutani theorem), $d$ is right-invariant. Since $H$ is locally connected, there exists $\delta$ such that $B_\delta(e)\cap H= B_\delta(e)\cap H_0$. Now, take $\epsilon=\delta/2$. Then for each $h\in H- H_0$, $B_\epsilon(e)\cap B_\epsilon(h)=\emptyset$.
We have the projections $q: G\to G/H_0$ and $p: G/H_0\to G/H$. Then for each $h\in H$, $U:=p\circ q(B_\epsilon(h))$ is open in $G/H$,
$$
p^{-1}(U)= \coprod_{s\in S} q(B_\epsilon(s))
$$
where $S$ is the set of representatives of the cosets $hH_0$, and $p$ restricted to each  $q(B_\epsilon(s))$ is a homeomorphism to $U$.
A: I ended up finding a proof of this result on Knapp's book, but we need to ask G to be connected rather than loc connected, the proof is as it follows:
We have $H/H_0 \subset G/H_0$, where $H/H_0$ has discrete topology, so there exists an open set $U \subset G/H_0$ s.t.
$$
    U\cap\Phi(H) = \Phi(H_0)
$$
This implies that
$$
 \underbrace{\Phi^{-1}(U)}_{W} \cap H = H_0
$$
We may take an smaller open set, $V$, with $V^2 \subset W$ and $V = V^{-1}$.
It is clear that
$$
\Phi^{-1}(VH) = \bigcup_{g \in H} V g H_0 
$$
As $VgH_0$ are open connected sets, we need to show only that
i $Vg_1 H_0 \cap Vg_2H_0 = \emptyset $, if $g_1 \neq g_2$.
ii $\left.\Phi\right|_{VgH_0}$ is a homeomorphism.
The former follows from:
\begin{align*}
   & Vg_1H_0\cap Vg_2H_0 \neq \emptyset \\
   \implies  & VH_0g_1\cap VH_0g_2 \neq \emptyset \quad \text{as $H_0$ is normal in $H$.} \\
   \implies & V^{-1}VH_0 \cap H_0 g_2g_1^{-1} \neq \emptyset \\
   \implies & V^{-1}V \cap H_0 g_2g_1^{-1} \neq \emptyset,
\end{align*}
and since we have $V^{-1}V\cap H \subset H_0$, then $g_2^{-1}g_1 \in H_0$ and therefore $Vg_1H_0 =  Vg_2H_0$. Thus, $i.$ follows.
As for $ii.$, $\Phi$ is a projection, therefore, open, continuous and surjective, so we must see that $\Phi$ is injective from $VgH_0$ to $VH$. If
$$
\Phi(v_1gH_0) = \Phi(v_2gH_0)
$$
so $v_1g = v_2g g'$ for some $g' \in H$, thus
$$
V^{-1}V \ni v_2^{-1}v_1 = g g'g^{-1} \in H 
$$
Hence $v_1 = v_2h$ for some $h \in H_0$, as $H_0$ is normal we also have
$$
v_1gH_0 = v_2hgH_0 = v_2 g H_0
$$
Concluding that $\Phi$ is injective in the restriction.
Note: i couldn't fix the itemize in the latex so i left it like this.
