# Projection to the sphere

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?

• Why do you write $S^{3 - 1}$? Also note that $S^2 \subset \Bbb R^3$ usually, so maybe you mean $S^1$? – Ruben Jul 17 at 13:49
• 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. – user406143 Jul 17 at 13:50
• 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 $S^2 \subset \Bbb R^3$. – Ruben Jul 17 at 13:54
• Ok, thanks, I edited the question. – user406143 Jul 17 at 14:00

One can just use the map $$(x,y) \mapsto (x,y,1)$$ and then normalize the resulting 3d element. This should do the trick.