2

I am reading Milnor's Topology from the Differentiable Viewpoint, in particular page 8-9 about applying regular values to prove the fundamental theorem of algebra. So he defines stereographic projection as $h_{+}: S^{2}-\{(0,0,1)\}\to \mathbb{R}^{2}\times \{0\}$ and also he defines it from the south pole, namely, $h_{-}:S^{2}-\{(0,0,-1)\}\to \mathbb{R}^{2}\times \{0\}.$ Although he says its elementary geometry, I don't understand why the composition $h_{+}h^{-1}_{-}(z)=\frac{z}{|z|^{2}}$. Any help would be greatly appreciated thank you very much.

Rene Cabrera
  • 342
  • 1
  • 9
  • What did you come up with for your maps $h_+$ and $h_-$? – T.J. Gaffney Mar 10 '15 at 11:21
  • I used the stereographic projection to define $h_{+}(x,y,z)=\Big(\frac{x}{1-z},\frac{y}{1-z},0\Big)$ and $h_{-}(x,y,0)=\Big(\frac{2x}{|z|^{2}+1},\frac{2y}{|z|^{2}+1},\frac{1-|z|^{2}}{1+|z|^{2}}\Big),$ but I think Milnor avoids this and uses something more elementary in geometry. Any help is greatly appreciated. Thanks. – Rene Cabrera Mar 12 '15 at 05:09
  • Your definition of $h_-$ can not be correct, the 3rd coordinate has to be $0$. Or do you mean $h_-^{-1}$? – user39082 Mar 13 '15 at 10:23
  • Probably you should take polar coordinates in $R^2$. Then it is kind of obvious that $h_+h_-^{-1}$ preserves the $\phi$-coordinate and you have to argue that the circle of radius $r$ in one chart corresponds to the circle of radius $\frac{1}{r}$ in the other. – user39082 Mar 13 '15 at 10:29
  • He's an arse for calling that "elementary geometry"... – goblin GONE May 12 '17 at 13:28

0 Answers0