Openness in quotient space

**HenryB** · March 8th 2009, 03:01 PM

Let X be a topological space. I can prove that the set Homeo(X) of homeomorphisms $f:X \rightarrow X$ becomes a group when endowed with the binary operation of composition.

I can also show that if G is a subgroup of Homeo(X), then the relation "x ~ $_{G}$ y iff there exists $g \in G$ such that $g(x) = y$ " is an equivalence relation.

Now the question: Let G and ~ $_{G}$ be as above, and let $p: X \rightarrow X /$ ~ $_{G}$ be the quotient map. Prove that for every U open in X, P(U) is open in X / ~ $_{G}$ .

Let $X = {R}^n \ {0}, n\geq 2$ , and let G be the subgroup of Homeo(X) composed of the maps $g(x) = cx$ where c is a constant. Prove that $x /$ ~ $_{G}$ is the real projective space $P {R}^{n-1}$ .

**clic-clac** · March 10th 2009, 12:53 AM

Hi

Let $H$ be a group, and consider a group action of $H$ on a topological space $Y.$ Then the quotient map $q:Y\rightarrow Y/H$ is open: indeed, since translations (given a $h\in H,$ maps in $Y^Y$ like $y\mapsto h.y$ ) are homeomorphisms, given an open set $U$ of $Y, q^{-1}(q(U))=$ $\bigcup\limits_{h\in H}$ $h.U$ is a union of open sets (the image of an open set under a homeomorphism is an open set) thus $q^{-1}(q(U))$ is open and by quotient topology definition, $q(U)$ is open.

Ok, that's a more general case, but it can give you an idea to solve your problem, the chosen group being particular.

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