Quotient by an action of a group on a topological space I'm asked to prove that $(\mathbb{R}^{n}-\{0\})/G\cong S^{n-1} \times S^{1} $ where G is the group of homeomorphisms $\{\Theta^{i}, i\in \mathbb{Z}\} $ and
$$\Theta: \mathbb{R}^{n}-\{0\} \longrightarrow \mathbb{R}^{n}-\{0\}$$
$$ x \longrightarrow 2x $$
with $n\geq 2$.
So I'm having trouble even visualizing $(\mathbb{R}^{n}-\{0\})/G$. Any help would be great!
 A: To visualise this when $n = 2$, observe that the action of $G = \Bbb{Z}$ identifies each point $\mathbf{v} \in \Bbb{R}^2 - \{0\}$ with all the points $2^i\mathbf{v}$ where $i \in \Bbb{Z}$. Each point in the quotient space $(\Bbb{R}^2 - \{0\})/G$ is represented by exactly one point $\mathbf{v}$ such that $1 \le \| \mathbf{v} \| < 2$. Thus the quotient space looks like what you get from the annulus $A = \{ \mathbf{v} \in \Bbb{R}^2 \mathrel{|} 1 \le \|\mathbf{v}\| \le 2 \}$ by identifying each inner boundary point $\mathbf{v}$ ("inner" meaning $\|\mathbf{v}\| = 1$) with the corresponding outer boundary point $2\mathbf{v}$. This identification glues the edges of the annulus to form a seam as if you were making an inner tube for a bicycle tire. What it gives you is a torus: $S^1 \times S^1$.
For $n > 2$, the seam is  homeomorphic to the $n-1$-dimensional sphere $S^{n-1}$ and the resulting space is $S^{n-1} \times S^1$.
A: Hint:Remark that $S^1=R-0/<h(x)=2x>$. Consider $f:R^n-\{0\}\rightarrow S^{n-1}\times S^1$ defined by $f(x)=(x/\|x\|,[\|x\|])$ where $[\|x\|]$ is the class of $\|x\|$ in $R-0/<h>$ show that $f$ induces a homeomorphism from $R^n-0/<\Theta>\rightarrow S^{n-1}\times S^1$.
Let $(x)\in R^n-0$, remark that $f(2x)=f(x)$ thus $f$ factors by a map $F:R^n-0/<\Theta>\rightarrow S^{n-1}\times S^1$. $F$ is injective: suppose that $F([x])=F([y])$ implies that $x={{\|x\|}\over{\|y\|}}y$ and $\|x\|=2^n\|y\|$. This implies that $x=2^ny$ thus $[x]=[y]$, $F$ is obviously surjective.
