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I want to map the set $\{(x,y)\in \mathbb{R}^2 | 0\leq x\leq 1, 0\leq y\leq x^2\}$ to the unit sphere $S^{2}\subset \mathbb{R}^3$ in a way so that the image of the point $(0,0)$ is the north pole and the cusp of the 2d set is still a cusp on the sphere. I wasn't capable to find the correct function term of the projection. Can someone help?

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  • $\begingroup$ Why do you write $S^{3 - 1}$? Also note that $S^2 \subset \Bbb R^3$ usually, so maybe you mean $S^1$? $\endgroup$ – Ruben Jul 17 at 13:49
  • $\begingroup$ I want to project it to $S^2 \subset R^3$. That's why I wrote $S^{3-1}$; I wanted to emphasize this fact. $\endgroup$ – user406143 Jul 17 at 13:50
  • $\begingroup$ Ah. I think there are more appropriate terms than project, as projection usually means going down in dimension. So you want to map the set you describe to $S^2 \subset \Bbb R^3$. $\endgroup$ – Ruben Jul 17 at 13:54
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    $\begingroup$ Ok, thanks, I edited the question. $\endgroup$ – user406143 Jul 17 at 14:00
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One can just use the map $(x,y) \mapsto (x,y,1)$ and then normalize the resulting 3d element. This should do the trick.

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