Explicit homeomorphism between $\mathbb{S^2}$ and $\mathbb{P^1(C)}$ I know that $\mathbb{P^1(C)} \cong \mathbb{P^1(C)} \cup \{N\} $, where $N$ is the north-pole of the sphere, is homeomorphic to the sphere $S^2$ thanks to the stereographic projection, but I am not sure if the explicit projection could be the following:
$$f: S^2 \longrightarrow \mathbb{P^1(C)} \\(x,y,z) \mapsto \left( \frac{x+iy}{1-z},\frac{x-iy}{1+z} \right) .$$
Is it the right approach?
 A: The map you provide will not work. Even restricted to the sphere without the north pole, it is not even injective. Indeed, $(1,0,0)$ and $(-1,0,0)$ go to the same point, since 
$$f(1,0,0)=[1:1]=[-1:-1]=f(-1,0,0).$$
To answer the question, let's try going step by step: 
The stereographic projection $\mathrm{Steo}:S^2\backslash\{N\} \to \mathbb{R}^2$ is given by 
$$(x,y,z) \mapsto \left(\frac{x}{1-z},\frac{y}{1-z} \right).$$
When you identify $\mathbb{R}^2 \simeq \mathbb{C}$, you have a formula which appears in your attempt (I'll still call the stereographic projection by the same name):
\begin{align*}
\mathrm{Steo}:S^2\backslash\{N\} &\to \mathbb{C} \\
(x,y,z) &\mapsto \frac{x+iy}{1-z}.
\end{align*}
Now, $\mathbb{C}$ embeds naturally in $\mathbb{C}P^1$ via 
\begin{align*}
g:\mathbb{C} &\to \mathbb{C}P^1\\
z &\mapsto [z:1].
\end{align*}
This strategy is alluded to in the comments by Max. Now, we have  that this $g$ misses a single point: $[1:0]$. This is due to the fact that if $b \neq 0$, then $[a:b]=[ab^{-1}:1]$ (and if $b=0$, $[a:0]=[aa^{-1}:0]=[1:0]$, where we recall that $a$ can't be zero if $b$ is zero).
So we have the explicit map 
\begin{align*}
g \circ \mathrm{Steo}:S^2 \backslash\{N\} &\to \mathbb{C}P^1 \backslash \{[1:0]\}\\
(x,y,z) &\mapsto \left[\frac{x+iy}{1-z}:1\right],
\end{align*}
which is an homeomorphism. There is the problematic missing north pole, and missing $[1:0]$. However, they are not a problem at all. Indeed, we can extend $g$ to have domain $S^2$ and codomain $\mathbb{C}P^1$ by sending $(0,0,1) \mapsto [1:0]$, which is essentially the uniqueness of the one-point compactification.
So, the final mapping becomes:
\begin{align*}
g \circ \mathrm{Steo}:S^2  &\to \mathbb{C}P^1\\
(x,y,z) &\mapsto \left[\frac{x+iy}{1-z}:1\right], \quad z \neq 1 \\
(0,0,1) &\mapsto [1:0], \quad z=1.
\end{align*}
Since you only want an homeomorphism, this escaping via general topology is enough. If you want to check for differentiability etc, you will need to follow through Pedro's suggestion in the comments, which is essentially considering the other map analogous to $g$ which helps covering the $[1:0]$ problematic case via a chart.
