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Define on $E = (0,\infty) \times (0,\pi) \times (0,2\pi)$ the spherical coordinates map: $$ \psi: E \to \mathbb{R}^3, \ \ \psi(r,\theta,\phi) = (r \sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \theta). $$ I am trying to show that this map is injective. Intuitively, this is quite clear. We know that the vector $(x,y,z)$ coordinate in the 3-d space is uniquely determined by its length, $r$, its angle on the plane, $\theta$, and its angle in the $z$-direction, $\phi$. However, I don't know what is the best way to prove this formally. Basically we would like to show that if $\psi(r,\theta,\phi) = \psi(r',\theta',\phi')$, that is, $$ \begin{cases} r \sin \theta \cos \phi = r' \sin \theta' \cos \phi' \\ r \sin \theta \sin \phi = r' \sin \theta' \sin \phi' \\ r \cos \theta = r' \cos \theta', \end{cases} $$ then $(r,\theta,\phi) = (r',\theta',\phi')$. I have tried to decompose as two different functions, one on the unit sphere, and one to scale the length of the vector, but this does not work since the composition does not work out properly.

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Hint: Use the quotient of the first two equations to show $\phi = \phi'$. Then use the quotient of the second two equations to get $\theta = \theta'$. Then use the third equation for the last step.

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  • $\begingroup$ If I'm correct this leads to a step where $\sin(\phi - \phi') = \sin(\phi' - \phi)$. Does it then follows that $\phi - \phi' = \phi' - \phi$? I don't see how this follows since $\phi - \phi', \phi' - \phi \in (-2\pi,2\pi)$. $\endgroup$
    – Sigurd
    Jun 16 '18 at 23:30
  • $\begingroup$ I didn't get that. Could you explain how you got there? $\endgroup$
    – theyaoster
    Jun 16 '18 at 23:35
  • $\begingroup$ Sure. But I probably did something wrong since I'm not good with trigonometry. By quotient of two equations, what exactly does that mean? $\endgroup$
    – Sigurd
    Jun 17 '18 at 0:00
  • $\begingroup$ What I meant by quotient is: given equations $a = b$ and $c = d$, divide them so you conclude that $a/c = b/d$. $\endgroup$
    – theyaoster
    Jun 17 '18 at 0:02
  • $\begingroup$ I see. Is it correct that I then get $\cos \phi \sin \phi' = \cos \phi' \sin \phi$? How to proceed to show that $\phi = \phi'$? $\endgroup$
    – Sigurd
    Jun 17 '18 at 12:04

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