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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 $\varphi^{-1}\circ\varphi=id$ ?

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@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 $\varphi\circ\varphi^{-1}$.

user5402
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  • Do you know what $\varphi$ represents? (Because it's fairly clear that it's not invertible in the first place - take two parallel vectors and see what happens under $\varphi$) – Milo Brandt Aug 06 '15 at 21:09
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    Please include the complete statement of the exercise. In particular, what are the domain and codomain of $\varphi$? You might also explain where you have run into issues in your solution. – vociferous_rutabaga Aug 06 '15 at 21:13
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    I agree with Milo Brandt, $\varphi$ is not invertible. Unless your domain was the surface of a sphere, but since you didn't give that detail I will assume that the domain of $\varphi$ is all of 3-space. – Tucker Aug 06 '15 at 21:21
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    Perhaps $\varphi$ maps the cube (side 2 center at the origin) bijectively onto the sphere (radius 1 center at the origin). And $\varphi^{-1}$ is the other way around. – GEdgar Aug 06 '15 at 21:30
  • @MiloBrandt I know I forgot to include the domain of the function. I assumed that you have a copy of the book. My fault but no need to turn this site into another stackoverflow where any small detail missing puts the question on hold or close it. I saw your comment now and I included some more details. As simple as that. – user5402 Aug 07 '15 at 12:16
  • @whatever: In this case, the domain and codomain aren't exactly small details. :) But could you please address MPO's question: Where have you run into trouble? Otherwise there's not much to say (once the question is re-opened) beyond "Check that if $(x, y, z) \in C$, then $\varphi^{-1} \circ \phi(x, y, z) = (x, y, z)$, and if (x, y, z) \in S^{2}$, then $\varphi \circ \varphi^{-1}(x, y, z) = (x, y, z)$." – Andrew D. Hwang Aug 07 '15 at 12:50
  • @AndrewD.Hwang I added my work in the question. – user5402 Aug 07 '15 at 14:49

2 Answers2

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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 $(x,y,z)\in C$.

Christian Blatter
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  • My problem was with the last equality. I forgot that $\max{(|x|,|y|,|z|)}=1$ on the cube. What a stupid mistake but that's what happens when you don't read any book for years. Thank you. – user5402 Aug 07 '15 at 16:12
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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 $n\in\mathbb N$.

Ennar
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