Group Acts by Homeomorphisms Implies Projection on Quotient Space is Open

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $G$ be a group acting by homeomorphisms on a topological space $X$.


Then the projection map $\pi: X \to X / G$ is open.


Proof

Let $U \subset X$ be open.

We have to show that $\pi \sqbrk U$ is open.

By definition of quotient topology, this is the case if and only if $\pi^{-1} \sqbrk {\pi \sqbrk U}$ is open.

By definition of saturation under group action:

$\ds \pi^{-1} \sqbrk {\pi \sqbrk U} = \bigcup_{g \mathop \in G} g U$

Because $G$ acts by homeomorphisms, $\pi^{-1} \sqbrk {\pi \sqbrk U}$ is open.

$\blacksquare$


Also see