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How to show the torus has no self intersection? ie how do you show that for any two distinct points $(\theta_1,\phi_1) \ne (\theta_2,\phi_2)$ we always have $F(\theta_1,\phi_1) \ne F(\theta_2,\phi_2)$? where

$$F : (\theta, \phi) \rightarrow ((a+b\sin(\theta))\cos(\phi), (a+b\sin(\theta))\sin(\phi),b\cos(\theta))$$

$$\theta , \phi \in [0,2 \pi), a>b>0$$

I have assumed there does exist $\theta_1 , \theta_2 , \phi_1 , \phi_2 \in [0,2 \pi)$ such that $F(\theta_1 , \phi_1)=F(\theta_2 , \phi_2)$ and that $(\theta_1 , \phi_1) \ne (\theta_2 , \phi_2)$ (in hope of reaching a contradiction), leading to:

$$ (a+b\sin \theta_1)\cos \phi_1 = (a +b\sin \theta_2)\cos \phi_2 $$

$$(a+b\sin \theta_1)\sin \phi_1 = (a+b\sin \theta_2)\sin \phi_2$$

$$b\cos \theta_1 = b\cos \theta_2$$

From the last line, we get $\theta_1 = 2 \pi - \theta_2$ when $\theta_2 \ne 0$ and $\theta_1=\theta_2$ when $\theta_2=0$

I have started considering when $\theta_2=0$:

when $\theta_2=0, \theta_1=0 \Rightarrow \phi_1 \ne \phi_2$

$\Rightarrow$ $$\cos \phi_1 = \cos \phi_2$$

$$\sin \phi_1 = \sin \phi_2$$

the only thing I can think to do is divide the first equation by the second but I don't see how this helps, thanks for your time.

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  • $\begingroup$ thanks I edited $\endgroup$ – StevenCr99 Feb 3 '18 at 18:41
  • $\begingroup$ at this point you must show that if $\cos\phi_1=\cos\phi_2$ then $\sin\phi_1\neq\sin\phi_2$ for $\phi_1\neq\phi_2$ and $\phi_1,\phi_2\in[0,2\pi)$, and viceversa $\endgroup$ – Masacroso Feb 3 '18 at 19:05
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The image of the point $(\phi,\theta)$ is $(x,y,z)=(c\cos\phi,c\sin\phi,b\cos\theta)$ where $c=a+b\sin\theta>0$ as $a>b$. Thus $\phi$ is uniquely determined in $[0,2\pi)$ as the argument of the complex number $x+yi$. Also the absolute value of $x+yi$ is $|x+yi|=c=a+b\sin\theta$. Thus $(x,y,z)$ determines $\sin\theta$ and $\cos\theta$ and they determine $\theta$ uniquely within $[0,2\pi)$.

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  • $\begingroup$ thanks, what are you saying about theta being uniquely determined? $\endgroup$ – StevenCr99 Feb 3 '18 at 20:01

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