CP1 and S^2 are smooth manifolds and diffeomorphic (proof)

Proposition: $\mathbb{C}P^1$ is a smooth manifold.

Proof:
Define $U_1=\{[z^1, z^2]\mid z^1\neq 0\}$ and $U_2=\{[z^1, z^2]\mid z^2\neq 0\}$. Also define $g_i: U_i\to\mathbb{C}$ by $g_1([z^1, z^2])=\frac{z^2}{z^1}$ and $g_2([z^1, z^2])=\frac{\overline{z^1}}{\overline{z^2}}$.

Let $f:\mathbb{C}\to\mathbb{R}^2$ be the homeomorphism from $\mathbb{C}$ to $\mathbb{R}^2$ defined by $f(x+iy)=(x,y)$ and define $\phi_i: U_i\to\mathbb{R}^2$ by $\phi_i=f\circ g_i$.

Note that $\{U_1, U_2\}$ is an open cover of $\mathbb{C}P^1$, and $\phi_i$ are well-defined homeomorphisms (from $U_i$ onto an open set in $\mathbb{R}^2$). Then $\{(U_1,\phi_1), (U_2,\phi_2)\}$ is an atlas of $\mathbb{C}P^1$.

The transition function $\displaystyle \phi_2\circ\phi_1^{-1}: \phi_1(U_1\cap U_2)\to\mathbb{R}^2,$
\begin{aligned} \phi_2\phi_1^{-1}(x,y)&=\phi_2 g_1^{-1}(x+iy)\\ &=\phi_2([1,x+iy])\\ &=f(\frac{1}{x-iy})\\ &=f(\frac{x+iy}{x^2+y^2})\\ &=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) \end{aligned}
is differentiable of class $C^\infty$. Similarly, $\phi_1\circ\phi_2^{-1}:\phi_2(U_1\cap U_2)\to\mathbb{R}^2$ is of class $C^\infty$. Hence $\mathbb{C}P^1$ is a smooth manifold.

Proposition:
$S^2$ is a smooth manifold.

Proof:
Define $V_1=S^2\setminus\{(0,0,1)\}$ and $V_2=S^2\setminus\{(0,0,-1)\}$. Then $\{V_1, V_2\}$ is an open cover of $S^2$.

Define $\psi_1: V_1\to\mathbb{R}^2$ by $\psi_1(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z})$ and $\psi_2: V_2\to\mathbb{R}^2$ by $\psi_2(x,y,z)=(\frac{x}{1+z},\frac{y}{1+z})$.

We can check that $\psi_1^{-1}(x,y)=(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2})$. Hence $\psi_1$ is a homeomorphism from $V_1$ onto an open set in $\mathbb{R}^2$. Similarly, $\psi_2$ is a homeomorphism from $V_2$ onto an open set in $\mathbb{R}^2$. Thus $\{V_1,V_2\}$ is an atlas for $S^2$.

The composite $\psi_2\circ\psi_1^{-1}: \psi_1(V_1\cap V_2)\to\mathbb{R}^2$ is differentiable of class $C^\infty$ since both $\psi_2$, $\psi_1^{-1}$ are of class $C^\infty$. Similarly, $\psi_1\circ\psi_2^{-1}$ is of class $C^\infty$. Thus $S^2$ is a smooth manifold.

We can also compute the transition function explicitly:
$\displaystyle \psi_2\psi_1^{-1}(x,y)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}).$
Note that $\psi_2\psi_1^{-1}=\phi_2\phi_1^{-1}$.

Define $h:\mathbb{C}P^1\to S^2$ by $h(\phi_1^{-1}(x,y))=\psi_1^{-1}(x,y)$ and $h(\phi_2^{-1}(x,y))=\psi_2^{-1}(x,y)$.

We see that $h$ is well-defined since if $\phi_1^{-1}(x,y)=\phi_2^{-1}(u,v)$ then $\displaystyle (u,v)=\phi_2\phi_1^{-1}(x,y)=\psi_2\psi_1^{-1}(x,y)$ so that $\psi_2^{-1}(u,v)=\psi_1^{-1}(x,y)$.

Similarly, we have a well-defined inverse $h^{-1}: S^2\to\mathbb{C}P^1$ defined by $h^{-1}(\psi_1^{-1}(x,y))=\phi_1^{-1}(x,y)$ and $h^{-1}(\psi_2^{-1}(x,y))=\phi_2^{-1}(x,y)$.

We check that (from our previous workings)
\begin{aligned} \psi_1 h\phi_1^{-1}(x,y)&=(x,y)\\ \psi_2 h\phi_1^{-1}(x,y)&=\psi_2\psi_1^{-1}(x,y)\\ \psi_1 h\phi_2^{-1}(x,y)&=\psi_1\psi_2^{-1}(x,y)\\ \psi_2 h\phi_2^{-1}(x,y)&=(x,y) \end{aligned}
are of class $C^\infty$. So $h$ is a smooth map. Similarly, $h^{-1}$ is smooth. Hence $h$ is a diffeomorphism.