Theorem: suppose that $G$ is a topological group, $H$ is a locally compact subgroup of $G$, and $P$ is a closed symmetric subset of $G$ such that $P$ contains an open neighborhood of the neutral element $e$ in $G$, and that $\overline{P^{3} } \cap H$ is compact. Let $ \pi : G \longrightarrow G / H $ be the natural quotient mapping of $G$ onto the quotient $G/H$. then the restriction of $f$ of $\pi$ to $P$ is a perfect mapping of $P$ onto the subspace $ \pi( P ) $ of $ G /H $.
proof:
clearly $f$ is continuous . first we show that $ f^{-1}f (a) $is compact, for any $ a \in P $ indeed from the definition of $ f $we have $ f^{-1}f (a) = aH \cap P $ . the subspace$ aH \cap P $ and $ H \cap a^{-1}P $ are obviously homeomorphic and closed in$ G $ . since$ a^{-1} \in P^{-1} =P $ , we have$ H \cap a^{-1}P \subset H \cap P^{2} \subset \overline{P^{3}} \cap H$ . hence $H \cap a^{-1}P $is compact and so is the set $ f^{-1}f (a) $.
closed: let us fix any subset $ M $ of $ P $ , and let$ a $ be a point of $ P $ so that$ f(a) \in \overline{f(M)} $ . we show $ f(a) \in f(M) $. assume the country. then $ (aH \cap \overline{P^{2}} ) \cap M = \emptyset $. note that $aH \cap \overline{P^{2}} $is compact, since $aH \cap \overline{P^{2}} $ is homeomorphism to$H \cap a^{-1} \overline{P^{2}} $ , which is a closed subset of the compact space $H \cap \overline{P^{3}} $ .
since $aH \cap \overline{P^{2}} $ is compact and $ M $ is closed and disjoint from $aH \cap \overline{P^{2}} $, there exists an open,symmetric neighborhood $ W $ of$ e$ in $ G $so that $ ( W( aH \cap \overline{P^{2}}) \cap M = \emptyset $ and$ W \subset P $ .
since the quotient mapping $ \pi $ is open and $ Wa $ is open neighborhood of $ a $ , the set$ \pi ( Wa) $ is an open neighbourhood of$ \pi (a) $ in $ G / H $ . therfore, the set$ \pi ( Wa) \cap \pi (M) $ is not empty, and we can fix $ m \in M $ and $ y in W $so that $ \pi(m )= \pi(ya) $ , that is $ mH = yaH $ . then $ m =yah $, for some $ h \in H $ . however, $ ah = y^{-1}m \in \overline{P^{2}}$, since $ y^{-1} \in W^{-1} =W \subset P$and $ M \subset P $ . beside $ ah \in aH $. hence,$ ah \in (aH \cap {P^{2}}$ and $ m = yah \in W ( aH \cap \overline{P^{2}}) $,since $ y \in W $. thus $ m \in M \cap ( aH \cap \overline{P^{2}}) $, which contradicts the choice of$ W $ . hence$ f(a ) \in f(M) $ , and $f(M) $is closed in$ (P) $ .
my questions are:
1:To show that a perfect mapping, must show that the mapping is closed and continuous mapping? why$ is f^{-1}f (a) = aH \cap P $?(in the first part)
2:why is $ (aH \cap \overline{P^{2}}) \cap M = \emptyset $?( in second part)
3:why is $ ( W( aH \cap \overline{P^{2}}) \cap M = \emptyset $? ( in third part)
4:why is not the set$ \pi ( Wa) \cap \pi (M) $ empty?( in forth part)
thanks for your pay attention.
Ad 1. It follows from this simple lemma (which I leave as an exercise):
Ad 2. He assumes that $f(a)\not\in f(M)$. This means that $aH\cap M=\emptyset$ because otherwise if $m\in aH\cap M$ then $f(a)=f(m)$ (because $m\in aH$) and thus $f(a)\in f(M)$.
Ad 3. Let's make it into a lemma:
Proof. Put $V:=G-M$ (the complement of $M$). Obviously $V$ is an open neighbourhood of $C$ such that $V\cap M=\emptyset$.
Now consider the multiplication $\circ:G\times G\to G$. Since $\circ$ is continous then $\circ^{-1}(V)$ is open in $G\times G$. By the definition of product topology for any $c\in C$ there are open subsets $U_c, W_c\subseteq G$ such that $W_c\times U_c\subseteq \circ^{-1}(V)$, $e\in W_c$ and $c\in U_c$ (we just take open neighbourhoods around $(e, c)$). By compactness of $C$ we have a finite set of indexes $c_1,\ldots, c_m$ such that $\{U_{c_1},\ldots,U_{c_m}\}$ covers $C$. Put $$W:=W_{c_1}\cap\cdots\cap W_{c_m}$$ $$U:=U_{c_1}\cup\cdots\cup U_{c_m}$$ It follows that $C\subseteq U$ and $WU\subseteq V$. Since $V\cap M=\emptyset$ then $WC\cap M=\emptyset$.
All in all $W$ is open and contains $e\in W$ and $WC\cap M=\emptyset$. To make $W$ symmetric simply refine it: $W':=W\cap W^{-1}$. Obviously $W'$ is open, $e\in W'$, $W'$ is symmetric and $W'\subseteq W$. Thus $W'C\cap M=\emptyset$. $\Box$
Ad 4. Since $f(a)\in\overline{f(M)}$ then by the definition any open neighbourhood $U$ of $f(a)$ has to intersect nonempty with $f(M)$. Note that $M\subseteq P$ so $f(M)=\pi(M)$ and obviously $f(a)=\pi(a)$.