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Let $X$ a topological space and $G$ a compact topological group, why is the quotient map $$\pi \colon X \to X/G$$ closed?

For every closed subset $C$ of $X$ $$\pi^{-1}(\pi(C)) = \bigcup_{g \in G}g\cdot C,$$ so we have to show that this subset is closed, but it is not evident to me the reason why this eventually infinite union should be closed.

Context: I’m trying to prove that $\pi$ is a proper map.

Edit: the suggested question is about the multiplication map $\mu \colon G \times X \to X$. This map is closed, unfortunately I can’t see why this should help.

donovan
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1 Answers1

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Let $\mu \colon G \times X \to X$ the action map, this map is closed because $G$ is compact.

Let $C$ be a closed subset of $X$, then $G \times C$ is a closed subset of $G \times X$ so is $\mu(G \times C) \subset X$. We have: $$\mu(G \times C)=G \cdot C=\bigcup_{g\in G}g\cdot C=\pi^{-1}(\pi(C)),$$

so $\pi(C)$ is a closed subset of $X/G$.

donovan
  • 400