Prove that $\varphi^{-1}$ is the inverse of $\varphi$ This is exercise 2.27 of Lee's introduction to topological manifolds.
I proved (geometrically) that $$\varphi(x,y,z)=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$
and that $$\varphi^{-1}(x,y,z)=\frac{(x,y,z)}{\max\{|x|,|y|,|z|\}}$$
How can I prove directly that $\varphi\circ\varphi^{-1}=id$ and $\varphi^{-1}\circ\varphi=id$ ?

@AndrewD.Hwang $$\varphi^{-1}(\varphi(x,y,z))=\frac{\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}}{\max{\left(\frac{|x|}{\sqrt{x^2+y^2+z^2}},\frac{|y|}{\sqrt{x^2+y^2+z^2}},,\frac{|z|}{\sqrt{x^2+y^2+z^2}}\right)}}=\frac{\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}}{\frac{\max{(|x,|y|,|z|})}{\sqrt{x^2+y^2+z^2}}}=\varphi^{-1}(x,y,z)$$ And the same problem for $\varphi\circ\varphi^{-1}$.
 A: I'll take it as granted that $$\phi\quad (x,y,z)\mapsto(u,v,w):={(x,y,z)\over\sqrt{x^2+y^2+z^2}}$$ maps $C$ bijectively onto $S^2$. Consider now the map
$$\psi:\quad (u,v,w)\mapsto {(u,v,w)\over\max\{|u|,|v|,|w|\}}\qquad\bigl((u,v,w)\in S^2\bigr)\ .$$
Then $\psi$ is continuous on $S^2$: From $u^2+v^2+w^2=1$ it follows that $\max\{u^2,v^2,w^2\}\geq{1\over3}$, whence $\max\{|u|,|v|,|w|\}\geq{1\over\sqrt{3}}$, and $\max$ is a continuous function of its arguments. Furthermore one easily checks that
$$\psi(\lambda u,\lambda v,\lambda w)=\psi(u,v,w)\qquad(\lambda>0)\ .$$
We therefore have
$$\psi\circ\phi(x,y,z)=\psi\left({(x,y,z)\over\sqrt{x^2+y^2+z^2}}\right)=\psi(x,y,z)={(x,y,z)\over\max\{|x|,|y|,|z|\}}=(x,y,z)$$
for all $(x,y,z)\in C$.
A: Let us denote $\| (x,y,z) \|_2 = \sqrt{x^2+y^2+z^2} $ and $\| (x,y,z) \|_\infty = \max\{|x|,|y|,|z|\} $. One can easily check that these are norms in $\mathbb R^3$ and thus are absolutely homogeneous (i.e. $\|\alpha x\| = |\alpha|\|x\|$ for any scalar $\alpha$).
Furthermore, let $\mathbb S^2 = \{ x\in\mathbb R^3\,:\, \|x\|_2 = 1\}$ and $C = \{ x\in\mathbb R^3\,:\, \|x\|_\infty = 1\}$. Define functions $\varphi\colon C\to\mathbb S^2,\ \psi \colon \mathbb S^2\to C$ with formulas $\varphi(x) = \frac{x}{\|x\|_2}$ and $\psi(y) = \frac{y}{\|y\|_\infty}$. Now, let $x\in C$ and $y\in \mathbb S^2$, i.e. $\|x\|_\infty = 1$ and $\|y\|_2 = 1$. Then we have
$$ \displaystyle\varphi(\psi(y)) = \frac{\frac{y}{\|y\|_\infty}}{\left\|\frac{y}{\|y\|_\infty}\right\|_2} = \frac{\frac{y}{\|y\|_\infty}}{\frac{\|y\|_2}{\|y\|_\infty}} = \frac y{\|y\|_2} = y$$ and similarly $$ \displaystyle\psi(\varphi(x)) = \frac{\frac{x}{\|x\|_2}}{\left\|\frac{x}{\|x\|_2}\right\|_\infty} = \frac{\frac{x}{\|x\|_2}}{\frac{\|x\|_\infty}{\|x\|_2}} = \frac x{\|x\|_\infty} = x$$ This works in $\mathbb R^n$ for any $n\in\mathbb N$.
