Suppose a compact Lie group $G$ acts on a manifold $M$ and let $\pi : M \rightarrow M/G$ be the orbit map. Can I say that $\pi$ is closed map?
If $C \subseteq M$ is a closed set in $M$ then I only need to show that $\pi^{-1}(\pi(C)) = \displaystyle\bigcup_{g\in G} g \cdot C$ is a closed set. But since I don't necessarily have that $G$ is finite I can't say for sure that this union is closed. However, I know that it will be closed if the family $\{ g \cdot C\}_{g \in G}$ is locally finite. So my question is -
When $G$ is compact will the above mentioned family be locally finite?
Thanks!
The family $\{gC\}_{g \in G}$ may not be locally finite, as you can see the case of $G = SO(2,\mathbb{R})$ acting on $\mathbb{R}^2$ by rotations, and $C = \{p\}$, where $ (0,0)\ne p \in \mathbb{R}^2$.
However, the map $M \to M/G$ is indeed closed. Like you said, you have to check that the saturation $G\cdot C$ of a closed subset $C$ of $M$ is again closed. You can see that using sequences. Consider $g_n c_n \to d$. Then a subsequence of $g_n$ will be convergent. Passing to such a subsequence, we may assume that $g_n\to g$. Therefore, $c_n = g_n^{-1} \cdot (g_n c_n) \to g^{-1} d$. Hence $c_n$ is also convergent, clearly to a point in $c \in C$, and so $g_n c_n \to g c \in G\cdot C$.