$\mathbb{C}P^1$ diffeomorphic to $S^2$

Question

I am trying to show that the complex projective line is diffeomorphic to the 2-sphere. I'm using the $C^{\infty}$ structure on $\mathbb{C}P^1$ given by $U_1 = \{ [z_1 : 1], z_1 \in \mathbb{C} \}$, $\phi_1 : z_1 \mapsto (\operatorname{Re}(z_1), \operatorname{Im}(z_1)) \in \mathbb{R}^2$ and $U_1 = \{ [1 : z_2], z_2 \in \mathbb{C} \}$, $\phi_2 : z_2 \mapsto (\operatorname{Re}(z_2), \operatorname{Im}(z_2)) \in \mathbb{R}^2$ and stereographic projections on $S^2$. I've defined two maps from $S^2$ to $\mathbb{C}P^1$ by $$ f_+ (x_0, x_1, x_2) = [1+x_0 : x_1 + ix_2] $$ on $\{x_0 \neq -1\}$ and $$ f_- (x_0, x_1, x_2) = [x_1 - ix_2 : 1-x_0] $$ on $\{x_0 \neq 1\}$. I've shown these are equal on the intersection of their domains. I seem to have shown smoothness too; the plan is to use the inverse function theorem to show the inverse is smooth, but when trying to compute the Jacobian I get into a great big mess and can't show that the determinant is non zero. Am I even going in the right direction?

Answer 1

Here is a different approach.

Defining $[U_0] = \{(1: t) ; t \in \mathbb C\}$ and $[U_{\infty}] = \{(s: 1) ; s \in \mathbb C\}$, then $\mathbb C P^1 = [U_0] \cup [U_{\infty}]$. Now define the atlas $$\mathcal A = \{([U_i], \varphi_i)_{i \in \{0, \infty\}} ; \varphi_0 (1:t) = t\,\,\,\text{and} \,\,\, \varphi_{\infty}(s: 1 ) = s \}$$

Then $$\begin{align}\varphi _{\infty} \, \circ \, \varphi_0^{-1}(t) &= \varphi_{\infty} (1:t)\\ & = \varphi_{\infty} \left(\frac{1}{t}: 1\right)\\&= \frac{1}{t}\end{align}$$ whenever $t \neq 0$. Further, $\varphi_i ([U_i]) = \mathbb C$.

On the other hand, $\psi _{N} \, \circ \, \overline\psi_S^{-1}(t) = \frac{1}{t}$ and $\psi_N(\mathbb S^2 - \{N\}) = \mathbb C = \psi_S(\mathbb S^2- \{S\})$.

Thus

i) $\varphi_0([U_0]) = \psi_N (\mathbb S^2 - \{N\})$ and $\varphi_{\infty}([U_{\infty}]) = \psi_S (\mathbb S^2 - \{S\})$ ;

ii) $\varphi_{\infty \, 0} = \psi_{N \, S}$.

We conclude that the $1$ - cocycles are equivalent, therefore $S^2$ and $\mathbb C P^1$ are diffeomorphic.

Answer 2

Consider a point $(z:w)$ with $w\neq 0$. Then send this point to $z/w\in\Bbb C$. Use now the stereographic projection $\Bbb C\to S^2-N$, and finally send the point $(1:0)$ to the north pole of $S^2$. You need to check that this is a diffeomorphism $\Bbb CP^1\to \Bbb C^\ast\to S^2$

The inverse of the first map sends a point $h\in \Bbb C$ to $\left(\frac{h}{\sqrt{1+|h|^2}},\frac{1}{\sqrt{1+|h|^2}}\right)$ and infinity to $(1,0)$, so you can obtain the inverse $S^2\to\Bbb CP^1$ explicitly.