# Topology homeomorphisms and quotient map question

• Apr 18th 2011, 01:20 AM
kevinlightman
Topology homeomorphisms and quotient map question
In this R denotes the real numbers, and [.] denotes subscript

Let Homeo(X) be the group of homeomorphisms f:X->X where X is a topological space.

Let G be a subgroup of Homeo(X)
~ is an equivalence relation on G where x~y iff there exists g in G s.t g(x)=y

Q1:

Let p: X -> X/~ be the quotient map.
Prove that for every U open in X, p(U) is open in X/~

Q2:
Let X=R(^n)\{0}, n>1.
Let G be the subgroup of Homeo(X) composed of g[p](x)=px
Prove that X/~ is the real projective space PR^(n-1)
Also prove that the graph of the relation ~ is closed, and that PR^(n-1) is Hausdorff

Any help would be much appreciated,

KL
• Apr 18th 2011, 11:47 AM
HappyJoe
As for question 1, you probably mean that ~ is an equivalence relation on X, right?

To show that the quotient map has the property that for all open U in X, p(U) is open in X/~ (such a map is called an open map), you need to recall what it means for a set in X/~ to be open. Indeed, a set V in X/~ is open, iff p^{-1}(V) is open. So for an arbitrary open U in X, you need to show that p^{-1}(p(U)) is open in X.

But this set is {x in X | p(x) in p(U)} = {x in X | [x] in p(U)}, where [x] is the equivalence class of x.

Continuing, we have

= {x in X | there is a y in U, such that y~x}
= {x in X | there is a y in U and a g in G, such that g(x)=y}.

So all x for which a homeomorphism g maps x to U make up that set. In other words, the set is

cup g^{-1}(U),

where the union "cup" is taken over all homeomorphisms g in G.
• Apr 18th 2011, 11:23 PM
Drexel28
Quote:

Originally Posted by kevinlightman
Q2:
Let X=R(^n)\{0}, n>1.
Let G be the subgroup of Homeo(X) composed of g[p](x)=px
Prove that X/~ is the real projective space PR^(n-1)
Also prove that the graph of the relation ~ is closed, and that PR^(n-1) is Hausdorff

Any help would be much appreciated,

KL

It's not clear what your maps are. What is g[p]
• Apr 22nd 2011, 12:24 PM
themediocrehacker
I think he means that G is the subgroup of Homeo(X) composed of g[p](x)=px
which I take to mean that we range p from X and these are the maps in the subgroup G [of Homeo(X)]
Does this help things?