I read a few days ago on topological groups that a user asked. I recently studied topological groups and this is an interesting case, but I have a question in this theorem.
Theorem: suppose that $G$ is a topological group, $H$ is a locally compact subgroup of $G$, and $\pi\colon G \to G / H$ is the natural quotient mapping of $G$ onto the quotient $G/H$, then there exists an open neighborhood $U$ of the neutral element $e$ such that $\pi(\overline{U})$ is closed in $G/H$ and the restriction of $\pi$ to $\overline{U}$ is a perfect mapping of $\overline{U}$ onto the subspace $\pi(\overline{U})$, (thus, $\pi$ is an open locally perfect mapping of $G$ onto $G/H$).
proof: $H$ is closed in $G$. Since $H$ is locally compact, there exists an open neighborhood $V$ of $e$ in $G$ such that $\overline{V \cap H}$ is compact. Since the space $G$ is regular, we can select an open neighborhood $W$ of $e$ such that $\overline{W}\subset V$. Then, $\overline{W} \cap H$ is compact, since $\overline{W} \cap H$ is a closed subset of the compact subspace $\overline{V \cap H}$. let $U_{0}$ be any symmetric open neighborhood of $e$ such that $ U_{0}^{3} \subset W$. Since $\overline{U_{0}}^{3} \subset \overline{U_{0}^{3}} $, the restriction of $\pi $ to $\overline{U_{0}}$ is a perfect mapping of $ \overline{U_{0}}$ onto the subspace $\pi[\overline{U_{0}}]$ . Since $\pi$ is an open mapping , the set $\pi[U_{0}]$ is open in $G/H$ . Since the space $G/H$ is regular, we can take an open neighborhood $V_{0}$ of $\pi(e)$ in $G / H$ such that $\overline{V_{0}} \subset \pi[U_{0}]$. Then $U = \pi^{-1}[V_{0} \cap U_{0}]$ is an open neighborhood of $e$ contained in $ \overline{U_{0}} $, and the restriction of $f$ to the closure of $U$ is a perfect mapping of $\overline{U}$ onto the subspace $ \pi[\overline{U}]$. However, $\pi[\overline{U}]$ is closed in $\pi[\overline{U_{0}}]$,and $ \pi (\overline{U}) \subset \overline{V_{0}} \subset \pi (U_{0}) \subset \pi [\overline{U_{0}}]$, therefore , $\pi[\overline{U}]$ is closed in the closed set $\overline{V_{0}}$ which implies that $\pi[\overline{U}]$ is closed in $G/H$.
1: why is $U = \pi^{-1}[V_{0} \cap U_{0}]$ an open neighborhood of $e$ contained in $ \overline{U_{0}} $, and the restriction of $f$ to the closure of $U$ is a perfect mapping of $\overline{U}$ onto the subspace $ \pi[\overline{U}]$?
2: why $\pi[\overline{U}]$ is closed in $\pi[\overline{U_{0}}]$?
3: why is $\pi[\overline{U}]$ closed in the closed set $\overline{V_{0}}$?
4:why is $\pi[\overline{U}]$ closed in $G/H$?
I guess $U$ should be defined as $\pi^{-1}[V_0]\cap U_0$, because $U_0\subset G$. whereas $V_0\subset G/H$. Then $U$ is an open neighborhood of $e$ contained in $ \overline{U_{0}} $, and the restriction of $\pi$ to the closure of $U$ is a perfect mapping of $\overline{U}$ onto the subspace $ \pi[\overline{U}]$.
Since the restriction of $\pi $ to $\overline{U_{0}}$ is a perfect (in particular, a closed) mapping of $ \overline{U_{0}}$ onto the subspace $\pi[\overline{U_{0}}]$ and $\overline{U}$ is closed in $\overline{U_0}$.
This is explained in the last sentence of the proof.
Because $\pi[\overline{U}]$ is closed in the set $\overline{V_{0}}$, which is closed in $G/H$.