Proof:

We shall show that if $g\cdot y = y$ for some $y \in Y$ and $g\in G$ then $g = 1$. If $g \neq 1$, choose a neighborhood $U$ of $y$ as in definition (20.1) which in particular implies $g\cdot U\cap U = \emptyset$. But $y \in g\cdot U\cap U$ and we get a contradiction.

The set of all orbits of the action with its quotient topology is denoted by $Y/G$ and the following theorem expresses the covering properties of the quotient map

$$\eta : Y \longrightarrow Y/G.$$

Note that for each $g\in G$, the map $y \mapsto g\cdot y$ is a bijective map. If each of these maps is a homeomorphism of $Y$ onto itself, we say that $G$ acts as a group of homeomorphisms on $Y$.