I just read on wikipedia that the the complex projective line is homeomorphic to the riemann sphere. How do I prove this? But, before that I have an extremely silly doubt that has been eating me. In the complex projective line, antipodal points are identified, but in the Riemann sphere they are clearly distinct. Then how can the two be homeomorphic.

I found out the following map from the complex projective line to $S^2$. But how do I prove this is a homeomorphism? Finding the inverse of this functions seems complicated, and then how do I prove that the inverse map of a open subset is a open subset, and vice versa?

$f:\mathbb{C}\mathbb{P}^1 \to S^2 = [z:w] \to \frac{(2Re(w\bar{z}), 2Im(w\bar{z}),|w|^2-|z|^2)}{(|w|^2+|z|^2)} $


Sorry I'm late to the party; I would comment on Berci's post but I'm not allowed to comment just yet.

While I'm sure you can actually prove that's a homeomorphism by computing inverses and showing that everything's continuous/smooth, I think most people would balk at actually carrying out such computations, and would simply accept that it can be done without actually doing it out.

But if you want, there is another way to see this fact using CW-complexes: on pages 6-7 of Hatcher's book he describes the standard CW-structure for $\mathbb{CP}^n$. In this structure you get one $2k$-cell for each $k$ between $0$ and $n$, and no other cells.

This helps here, because it means that the CW-structure of $\mathbb{CP}^1$ is just a $2$-cell glued onto a $0$-cell ($=$ a point). But there is only one way to do that, and that's to map all of the boundary of the $2$-cell to the point, which gives you the $2$-sphere.


The real projective line is basically a real line plus one point in the infinity.

The complex projective line is the complex plane plus one point in the infinity.

Have you heard about the stereographic projection?

  • 1
    $\begingroup$ Berci, thanks for the answer. Yes, I know about the stereographic projection. But how do I show the homeomorphism between the complex projective line and the Riemann sphere directly, using the map mentioned in my question. $\endgroup$ – user23238 Mar 7 '13 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy