# Orbit space projection is open

Lemma 21.1. For any continuous action of a topological group $G$ on a topological space $M$, the quotient map $\pi \colon M \to M/G$ is an open map.

Proof. For any $g \in G$ and any subset $U \subseteq M$, we define a set $g \cdot U \subseteq M$ by $$g \cdot U = \{ g \cdot x : x \in U \}.$$ If $U \subseteq M$ is open, then $\pi^{-1}(\pi(U))$ is equal to the union of all sets of the form $g \cdot U$ as $g$ ranges over $G$. Since $p \mapsto g \cdot p$ is a homeomorphism, each such set is open, and therefore $\pi^{-1}(\pi(U))$ is open in $M$. Becaues $\pi$ is a quotient map, this implies that $\pi(U)$ is open in $M/G$, and therefore $\pi$ is an open map.

(Original image here.)

I’m having a hard time understanding why $\pi^{-1}\pi(U)$ is the union of $gU$. Isn’t $M/G = \{gU_n\}$ the set of orbits? When you are pulling back to $M$, shouldn’t it be just the point $x$ such that $x \to gx$?

$\pi^{-1}(\pi(U))$ is the set of $a\in M$ with $\pi(a)\in\pi(U)$. That is it consists of the $a\in M$ such that $\pi(a)=\pi(b)$ for some $b\in U$. But $\pi(a)=\pi(b)\iff G\cdot a=G\cdot b\iff a\in G\cdot b\iff a=g\cdot b$ for some $g\in G$. Therefore $a\in\pi^{-1}(\pi (U))$ iff there is $g\in G$ and $b\in U$ with $a=g\cdot b$, that is $\pi^{-1}(\pi(U))=\bigcup_{g\in G} g\cdot U$.