Im having some trouble with the following:
Let n be a positive integer, and let $||\cdot||$ denote the usual Euclidean norm on $\mathbb{R}^n$, that is, $||x||=(x^2_1+…+x^2_n)^{1/2}$ where $x=(x_1,…,x_n) \in \mathbb{R}^n$. We consider $\mathbb{R}^n$ as a topological space equipped with the standard topology, and define an equivalence relation $\sim$ on $\mathbb{R}^n$ by $x \sim y \iff ||x||=||y||$.
Show that $\mathbb{R}^n/\sim$ (equipped with the quotient topology) is homeomorphic to [0,1) (equipped with the standard topology).
I know that we say that X and Y (two topological spaces) are homeomorphic if there exist a homeomorphism, a function $f:X \rightarrow Y$ which is bijective, continuous and $f^{-1}:Y \rightarrow X$ is continuous aswell.
I know I have to find a function/homeomorphism (a quotient map), but I am unsure of how to do so.
I was thinking something like:
$v \mapsto f(||v||) \cdot v$
But I honestly dont know how to approach such a problem…
Best Answer
Proof. $F$ is of course well defined. It is easy to see that it is a bijection as well. Furthermore it is continuous because if $\pi:X\to X/\sim$ is the standard projection then $\pi^{-1}(F^{-1}(U))=f^{-1}(U)$ and thus $F^{-1}(U)$ is open.
Finally if $F^{-1}(U)$ is open, then by the above so is $f^{-1}(U)$ and thus if $f$ is a quotient map, then $U$ is open. Meaning $F$ is a quotient map as well. But a bijective quotient map is a homeomorphism. That's because for bijections $V=F^{-1}(F(V))$, and since $F$ is a quotient map, then this implies it is open. Or equivalently its inverse is continuous. $\Box$
With that consider
$$f:\mathbb{R}^n\to [0,\infty)$$ $$f(v)=\lVert v\rVert$$
This function is of course continuous.
Furthermore it is a quotient map. Indeed, assume that $f^{-1}(U)$ is open while $U\subseteq [0,\infty)$ is not. This means that there is $r\in U$ and a sequence $(x_n)\subseteq [0,\infty)\backslash U$ convergent to $r$. Now choose a vector $v$ in $\mathbb{R}^n$ with $\lVert v\rVert=r$. Of course $v\in f^{-1}(U)$. Next take $v_n=\frac{x_n}{r}v$ and note that $\lVert v_n\rVert = x_n$. Finally $v_n\to v$. However $v_n\not\in f^{-1}(U)$ which cannot happen since the last set is open. Note that the special case when $r=0$ has to be treated separately, which I leave as an exercise. Also note that the same works for any dimension (even infinite) and any norm.
Thus, by our lemma, $f$ induces a homeomorphism between $\mathbb{R}^n/\sim$ and $[0,\infty)$. As the final step pick your favourite homeomorphism $[0,\infty)\to [0,1)$, e.g. $\text{arctan}$ or something.