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I'm self learning Algebraic topology from Rotman's Introduction to Algebraic Topology and I've come across this problem:

Prove that $CP^1 \cong S^2$

I'm really struggling with showing this.

I know that $CP^1 = \{[(z_1,z_2)] : (z_1, z_2) \in \Bbb C^2 \text{ and } (z_1, z_2) = \lambda (z_3, z_4) \text{ for some $\lambda$ in $\Bbb C - \{0\}$} $ which is the set of lines in $4$-space through the origin.

I get the feeling I'd have to somehow show a mapping from the upper hemisphere to the whole sphere, but from here I have no idea how to proceed.

Anyone have any ideas?

Oliver G
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    Hint : stereographic projection. –  Jul 31 '17 at 15:59
  • Here is an alternative way. What are elements of $S^2$ and how do you define a map from $S^2$ to $CP^1$ by charts? Then glue the chart morphism properly and show the map is homeo. – user45765 Jul 31 '17 at 16:11
  • @N.H. Could you elaborate? I'm not seeing how $S^n - {N} \cong R^n$ will help. – Oliver G Jul 31 '17 at 16:56
  • It's easy to see that $CP^1$ is the one-point compactification of $CP^1 \backslash [1:0] \cong \Bbb C$ so this implies that $CP^1 \cong S^2$ as they are both the Alexandrov compactification of $\Bbb C$. –  Jul 31 '17 at 16:57
  • What does the notation $CP^1 \backslash [1:0]$ mean? – Oliver G Aug 01 '17 at 00:11

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