I have a question about an answer written here. For completeness, here is the question and a sketch of the solution:
Let $G$ be a topological group acting on a space $X$. Show that if both $G$ and $X/G$ are connected, then $X$ is connected.
Solution sketch: suppose on the contrary $X=U\cup V$ for disjoint, nonempty, open $U,V\subseteq X$. If $\pi:X\to X/G$ denotes the canonical quotient map, then we have $\pi(U)\cup\pi(V)=X/G$. The goal is to show that this disconnects $X/G$. One uses the connectedness of $G$ to show that $\pi(U)\cap\pi(V)=\varnothing$, and then, as the answer linked states, "by definition of the quotient topology, both $\pi(U)$ and $\pi(V)$ are open in $X/G$", hence $X/G$ is disconnected, a contradiction.
My question: I'm pretty sure I'm over thinking this, but why can we say $\pi(U)$ and $\pi(V)$ are open in $X/G$? Quotient maps need not be open maps in general. Is this a fact about quotients by group actions?
Quotients maps by group actions are open maps because the saturation of an open set $$\pi^{-1} (\pi(U)) = \bigcup_{g \in G} g\cdot U$$ is open, so $\pi(U)$ is open.