I have a group $G$ acting on a set $X$.
Let $U$ be any subset of $X$.
Let $G(x)=\{gx\mid g \in G\}$
This is basically orbit of $x$ under the action
Now I define $q$ : $X \rightarrow \frac{X}{G}$ as
$q(x) = G(x)$
Then how can I show that $q^{-1}(q(U))$ is union of open sets $gU$ where $g \in G$?
I am proceeding like this
$$q^{-1}(q(U)) =\{x \in X:q(x) \in q(U)\}$$
$$q^{-1}(q(U))=\{x \in X:G(x) = G(y) \text { for some y in U}\}$$
From here, how can I claim the desired result?
Two orbits $G(x),G(y)$ are equal, if and only if, $x=gy,\, \text{for some } g\in G$. Hence, you have $$\{x\in X: G(x)=G(y), \text{ some } y\in U\}=\{x\in X: x=gy, \text{ some }g\in G,\, y\in U\}=\bigcup_{g\in G}gU.$$