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If a group $G$ acts on a topological space $X$ by homeomorphisms, why is the quotient map $X \to X/G$ necessarily an open map?

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That seems to be an immediate consequence of how the quotient topology is defined. If it is not immediate with the definitions of quotient topology and open map you use, please edit your question to include those definitions. – Henning Makholm Feb 4 '12 at 20:52
@HenningMakholm, the definition of the quotient topology I use is the second one provided on wikipedia [ ] (begins with "Equivalently, we can define them..."). I see that with the first definition there, the openness of the quotient map is, indeed, immediate. Thank you! – Rick Feb 4 '12 at 21:06
up vote 5 down vote accepted

Let $q$ be the quotient map, and let $U$ be a non-empty open set in $X$; $q[U]$ is open in $X/G$ iff $q^{-1}[q[U]]$ is open in $X$. But $$q^{-1}[q[U]]=\bigcup_{g\in G}(g\cdot U)$$ is a union of open sets, since each map $x\mapsto g\cdot x$ is a homeomorphism, and therefore open itself, as desired.

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