Showing the quotient map is open

Question

If I have a topological space $X$ and a subgroup $G$ of $\operatorname{Homeo}(X)$. Then defining an equivalence relation $x \sim y$ iff there is a $g\in G$ s.t. $g(x) = y$. I'm trying to show that the quotient map $q: X \to X/R$ is open.

I can just about see that, if $U$ is an open set in X, then $p^{-1}(p(U)) = \cup_{g \in G} g(U)$ - reason being that this will give all the elements that will map into the equivalence classes of $U$ under $q$.

Now I'm struggling to see why this means that $p^{-1}(p(U))$ is open. Just because we know that $U$ is open, how do we know that $g(U)$ is open. (Which would then give a union of open sets).

Thanks

Answer 1

We have $$p^{-1}(p(U))=\{gu\mid g\in G, u\in U\}=\bigcup_{g\in G}g(U)$$ But each $g(U)$ is open since $g$ is a homeomorphism. So the union is open too.

Answer 2

This is because a homeomorphism is an open map (equivalently, its inverse is continuous). Thus, for any $g\in G$ and any open subset $U$ of $X,$ we have $g(U)$ open in $X,$ too.