# Openness in quotient space

• Mar 8th 2009, 03:01 PM
HenryB
Openness in quotient space
Let X be a topological space. I can prove that the set Homeo(X) of homeomorphisms $\displaystyle f:X \rightarrow X$ 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 ~$\displaystyle _{G}$ y iff there exists $\displaystyle g \in G$ such that $\displaystyle g(x) = y$" is an equivalence relation.

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

Let $\displaystyle X = {R}^n \ {0}, n\geq 2$, and let G be the subgroup of Homeo(X) composed of the maps $\displaystyle g(x) = cx$ where c is a constant. Prove that $\displaystyle x /$ ~$\displaystyle _{G}$ is the real projective space $\displaystyle P {R}^{n-1}$.
• Mar 10th 2009, 12:53 AM
clic-clac
Hi

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