I'm reading a book 'Modular forms', Toshitsune Miyake. In p.6, the author proves the following theorem.
Assume that a topological group $G$ acts transitively on a topological space $X$. If $G$ is a locally compact group with a countable basis, and $X$ is a locally compact Hausdorff space, then for each element $x \in X$, the space of the right cosets $G/G_x$ is homoemorphic to $X$ by the correspondence $gG_x \to gx$.
It is obvious that this map is bijective and continuous. To show that the correspondence map is open, let $U$ be an open subset of $G$. Let $gx$ be any point of $Ux$. The author takes a compact neighborhood $V$ of the unit element of $G$ so that $V^{-1}=V$ and $gV^2 \subset U$.
My question is, how can we take such compact neighborhood?
Best Answer
We just need one basic fact, namely that $hX$ is open for any $h \in G$ and open $X \subseteq G$, which is proved here.
So, $g^{-1}U$ is an open neighborhood of the identity $e$. By continuity of the group operation there exist open $W_1,W_2$ such that $(e,e) \in W_1 \times W_2$ and $W_1W_2 \subseteq g^{-1}U$. Then $W_1 \cap W_2$ is an open neighborhood of $e$, so contains a compact neighborhood $Z$ of $e$. Set $V := Z \cap Z^{-1}$. Then