The map $f : \mathbb{R}^{n+1} \setminus \{0\} \to S^n$ defined by $f(x) = \frac{x}{\|x\|}$ is an open map

Question

I want to show that $f(x) = \frac{x}{\|x\|}$ is an open map from $\mathbb{R}^{n+1} \setminus \{0\} \to S^n$ where $S^n = \{ x \in \mathbb{R}^{n+1} \mid \|x\| = 1\}$ is the $n$-sphere. That is, I want to show that for any set $U$ open in $\mathbb{R}^{n+1}$, $f(U)$ is open in $S^n$. This is easy to visualize, but I'm having trouble proving it. I have a feeling it's easier than I think and that I'm missing something obvious.

Answer 1

Let $U\subseteq\mathbb{R}^{n+1}\setminus\{0\}$ be open. Let $\hat{U}:=\{rx\mid r\in\mathbb{R}^+,x\in U\}$. Intuitively, $\hat{U}$ contains all the rays from the origin passing $U$. For $r\in\mathbb{R}^+,x\in U$, we have $f(rx)=f(x)$, so clearly $f(U)=f(\hat{U})$. Furthermore, $\hat{U}$ is open, since $\hat{U}=\bigcup_{r\in\mathbb{R}^+}rU$. Lastly, convince yourself that $f(\hat{U})=\hat{U}\cap S^n$, so that $f(U)=\hat{U}\cap S^n$ is open in $S^n$ by the definition of the subspace topology.

Answer 2

A basis for the open sets in $\Bbb R^{n+1}\setminus\{0\}$ is the open balls $B(x,r)$ centered at points $x$.

But the image under $f$ of such sets are sets of the form $B(x/\|x\|,r')\cap S^n$, which are open in $S^n$, by the definition of the subspace topology.

Since $f$ is open on a basis, it's an open map.