Proving $\Bbb C P^1 \cong S^2$ From Rotman's Algebraic Topology:

Prove that $\Bbb C P^1 \cong S^2$, where $\Bbb C P^1  = (\Bbb C^2 - \{0,0\})/{\sim}$ and $\sim$ is the equivalence relation defined by:  $x \sim y$ iff $x = \lambda y$ for some $\lambda \in \Bbb C - \{0\}$, and $S^2$ is defined as $ \{x \in \Bbb R^3 : |x| = 1\}$.

I can see that $S^2 \cong \Bbb {\hat C}$, where $\hat {\Bbb C}$ is the one-point compactification of $\Bbb C$.  And the function $[(x,y)] \mapsto x/y$ from $\Bbb C P^1$ to $\hat {\Bbb C}$ looks like it may be useful but I can't figure how to show this function is continuous.
Anyone have any ideas?
 A: Consider the map
$h: \mathbb{C}^2 \setminus \{\mathbf{0}\} \to S^2$ defined by:
$$h(u,v) = \frac{1}{|u|^2+|v|^2}(i(\bar{u}v-\bar{v}u), \bar{u}v + u\bar{v}, |u|^2 - |v|^2).$$
Note that if $(u,v)$ is scaled by a nonzero complex factor $\lambda$, $h(u,v)$ remains the same. In other words:
$$ h(\lambda u, \lambda v) = h(u,v),$$
for all nonzero $\lambda \in \mathbb{C}$. This map $h$, which is (essentially) the Hopf map, induces a diffeomorphism from $P^1(\mathbb{C})$ onto $S^2$. Of course, some details need to be checked, but I hope you can continue from here.
A: Consider a decomposition of $\mathbb{CP}^n$ in terms of its decomposition of its affine charts $U_1,\ldots,U_{n+1}$, where $U_i$ consists of classes where the $i$-th entry in the tuple is nonzero (this property is invariant of class representative). Generally, $U_i \cong \mathbb{C}^n$; in this case, e.g., $U_1 \cong \mathbb C$. In general, $\mathbb{CP}^n \setminus U_i$ can be identified with $\mathbb{CP}^{n-1}$; can you see why? And in this particular case, what does e.g., $\mathbb{CP}^1\setminus U_1$ look like?
Once you have these pieces, the process assembling a diffeomorphism with the one-point compactification of the complex plane should be clearer.
A: It seems that your real question is why the map  $[(x,y)] \mapsto x/y$ is continuous. So let us consider
$$f : \mathbb C^2 \setminus \{ (0,0) \} \to \hat{\mathbb C}, f(x,y) = x/y .$$
Here we understand $x/0 = \infty$ (note that the  expression $0/0$ cannot occur because $(0,0)$ is not in the domain of $f$). This map is surjective since $f(z,1) = z$ and $f(1,0) = \infty$.
Obviously $f$ is continuous in all points $(\zeta,\omega)$ with $\omega \ne 0$. Now consider $(\zeta,0)$ with $\zeta \ne 0$. Let $((z_n,w_n))$ be a sequence in $\mathbb C^2 \setminus \{ (0,0) \}$ converging to $(\zeta,0)$. This means $z_n \to \zeta$ and $w_n \to 0$. Since $\zeta \ne 0$, we have $\lvert z_n  \rvert > \lvert \zeta \rvert/2$ for $n \ge n_0$. Moreover, given $R > 0$, we have $\lvert w_n  \rvert < 2/R\lvert \zeta \rvert$ for $n \ge n_R$. Thus for $n  \ge \max(n_0,n_r)$
$$\lvert f(z_n,w_n)  \rvert = \begin{cases} \infty > R & w_n = 0 \\ \lvert z_n/w_n \rvert=  \lvert z_n \rvert/ \lvert w_n\rvert > \lvert \zeta \rvert/2\lvert w_n\rvert > R & w_n \ne 0 \end{cases}$$
This shows that $f(z_n,w_n) \to \infty = f(\zeta,0)$. Therefore $f$ is continuous in $(\zeta,0)$.
It is easy to verify that $f(z,w) = f(z',w')$ if and only if $(z,w) \sim (z',w')$. Hence $f$ induces a continuous bijection
$$F : \mathbb{CP}^1 \to  \hat{\mathbb C}. $$
Since $ \mathbb{CP}^1$ is compact and $\hat{\mathbb C}$ is Hausdorff, $F$ is a homeomorphism.
Remark:
The same proof shows that $[(x,y)] \mapsto x/y$ is a homeomorphism  $\mathbb{KP}^1 \to  \hat{\mathbb K}$ for $\mathbb K = \mathbb R, \mathbb C, \mathbb H$ = field of quaternions. We have $\hat{\mathbb R} \approx  S^1,\hat{\mathbb C} \approx  S^2, \hat{\mathbb H} \approx  S^4$.
