easiest way to show the map to circle is open

Question

What is the quickest way to show that $F:\mathbb{R}^{n+1}\setminus\{0\}\to S^n$, $x\mapsto \frac{x}{||x||}$ is an open map?

Answer 1

Given $A$ open $$F(A)=\bigcup_{r>0} rA\cap \mathbb{S^{n}}$$ is a union of open sets in $\mathbb S^n$

Answer 2

$\mathbb R^{n+1}\setminus\{0\}$ is homeomorphic to $S^n\times \mathbb R_{>0}$, via: $$ \phi\colon x\mapsto \left(\frac{x}{\|x\|},\|x\|\right) $$ and its inverse $$ \psi\colon(z,r)\mapsto rz $$ Then your map $F$ may be written as the composition $$ \mathbb R^{n+1}\xrightarrow{\phi}S^n\times\mathbb R_{>0}\xrightarrow{\text{pr}_1}S^n $$ where $\text{pr}_1$ is the projection map. Homeomorphisms and projections maps are open, and the composition of open maps is open, so $F$ is open.

Answer 3

$S^n$ is the orbit space of the action of $\mathbb{R}^+$ on $\mathbb{R}^{n+1}$ (i.e., it is the quotient space obtained by letting the multiplicative group of positive reals act on $\mathbb{R}^{n+1}$ by scalar multiplication.) Since your map is naturally identified with the orbit map (i.e., the quotient map arising from the group action), and since orbit maps are always open, the result follows.