Orbit Space of Action of $SO(n)$ on $\mathbb{E}^n$ I'm trying to solve problem 28, chapter 4 of M. A. Armstrong's Basic Topology :

Describe the orbits of the natural action of $SO(n)$ on $\mathbb{E}^n$ as a group of linear
transformations, and identify the orbit space.

I take the norm function as an identification map and try to use the fact that $\mathbb{E}^n$ with this identification is homeomorphic to $[0,1)$.
I know each orbit of action $SO(n)$ on $\mathbb{E}^n$ is the same as an equivalence class of identification space but I need to show that the identification space and orbit space of acting $SO(n)$ on $\mathbb{E}^n$ are homeomorphic to claim the orbit space is homeomorphic to $[0,1)$. Can anyone help me to show that?
 A: Perhaps if I unwind the notation of this problem for you, it will be more evident what you have to do.
Let me introduce some notation for the norm function, say,
$$N(\vec p) = \|\vec p\| = \sqrt{p_1^2 + \cdots + p_b^2}
$$
What you have to prove is that the set of orbits of the action of $SO(n)$ on $\mathbb E^n$ is equal to the set of subsets of the form
$$S_r = \{\vec p \in \mathbb E^n \mid N(\vec p) = r\}, \quad r \in [0,\infty)
$$
Of course, if $r>0$ then $S_r$ is the sphere of radius $r$ centered on the origin, and $S_0 = \{\vec 0\}$.
The next main step is to prove is that the orbits of the action of $SO(n)$ on $\mathbb E^n$ are precisely the spheres $S_r$ of radius $r>0$ and the set $S_0$, and I'll assume that you know how to do that.
The final step involves applications of concepts of quotient maps and quotient topologies. One must check that:

*

*The norm function $N : \mathbb R^n \to [0,\infty)$ is a quotient map.

Once that has been checked, one must then apply

*

*The universal property of quotient maps.
It follows that

*

*$N$ induces a homeomorphism from the quotient topology on the orbit space of the action to the space $[0,\infty)$.

If these things are unfamiliar to you, you'll have to learn about quotient maps and quotient topologies. I suggest a good topology book, such as Munkres "Topology".
