topological group

Question

I read a theorem in book of topological group, but I face to several problem:

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 $G/H$.

my questions are:

1:why is there an open neighborhood $ V$ of$ e$ in $G $ such that $ \overline{V \cap H}$ is compact?

2:why is there $ U_{0}$ such that $ U_{0}^{3} \subset W $?

3: why can we say$ \overline{U_{0}}^{3} \subset \overline{U_{0}^{3}} $?

4: what is the meaning of perfect mapping in "the restriction of $ \pi $ to $ \overline{U_{0}} $ is a perfect mapping of $ \overline{U_{0}} $ onto the subspace $ \pi ( \overline{U_{0}} )$" ?

Answer 1

1. Let $W$ be a compact neighborhood of $e$ in $H$. Then there is an open subset $A$ of $G$ containing $e$ such that $A\cap H\subset W$. So, $\overline{A\cap H}\subset\overline W=W$ , which is compact.
2. Because the function from $G^3$ to $G$ defined by $\varphi(x,y,z)=xyz$ is continuous. Therefore $\varphi^{-1}(W)$ is a neighborhood of $(e,e,e)$ and you take a neighborhood $U_0$ of $e$ such that $U_0\times U_0\times U_0\subset\varphi^{-1}(W)$.
3. It's a general fact for continuous functions $f$ that $f\bigl(\overline A\bigr)\subset\overline{f(A)}$.
4. It means that it's a homeomorphism.

Answer 2

1 is easy: $H$ is locally compact, so it has a precompact neighbourhood $W$ of $e$. By definition of the subspace topology there is an open $V$ in $G$ such that $V \cap H = W^\circ$. Hence $V \cap H \subseteq W$ and is thus precompact as well.

2: Multiplication is continuous, so the only hard part is figuring out how to guarantee that $U_0$ is symmetric. Here is a simple argument for a related fact, namely that every neighbourhood $U$ of the identity contains a symmetric neighbourhood $V$ such that $V^2 \subseteq U$: Since multiplication is continuous there is a neighbourhood $W$ of $e$ such that $x,y \in W \implies xy \in U$. Now take $V = W \cap W^{-1}$ and the result follows.

3: The multiplication map $(x,y) \mapsto m(x,y) = xy$ is continuous, so $m(\bar A) \subseteq \overline{m(A)}$ just like any other continuous function.

4: Look up "perfect map" on wikipedia or whatever topology book/notes you are working from. The definition is crystal clear and easy to check.