Bijection between $\mathbb{P}^{1}(\mathbb{C})$ and the unit sphere I have the following complex analysis problem that I am completely lost on.

Let $(z_{1},z_{2})$ be a non-zero vector in $\mathbb{C}^{2}$ and define $F:\mathbb{C}^{2}\to\mathbb{R}^{3}$ by $$F(z_{1},z_{2})=\left(\frac{z_{1}\bar{z}_{2}+\bar{z}_{1}z_{2}}{z_{1}\bar{z}_{1}+\bar{z}_{2}z_{2}},\frac{z_{1}\bar{z}_{2}-\bar{z}_{1}z_{2}}{i(z_{1}\bar{z}_{1}+\bar{z}_{2}z_{2})},\frac{z_{1}\bar{z}_{1}-\bar{z}_{2}z_{2}}{z_{1}\bar{z}_{1}+\bar{z}_{2}z_{2}}\right).$$ Show that:
(1) $F$ defines a bijection between $\mathbb{P}^{1}(\mathbb{C})$ and the unit sphere in $\mathbb{R}^{3}$;
(2) if $S$ denotes stereographic projection from $\mathbb{C}_{\infty}-\{(0,0,1)\}$ to $\mathbb{C}$ and $[z_{1}:z_{2}]\neq[1:0]$, then $S(F([z_{1}:z_{2}]))=\frac{z_{1}}{z_{2}}$.

As mentioned above, I really have no idea of where to begin. Any suggestions and comments are appreciated!
 A: As for (1), you first need to show that the domain and range are correct. The function $F$, as defined, has domain $\mathbb C^2$, not $\mathbb{PC}$. But it suffices to show that $F(z_1,z_2) = F(kz_1,kz_2)$ for any $k\neq0$ . Then $F$ applied to an element of $\mathbb{PC}$ can be defined as $F$ applied to any representative in $\mathbb C^2$, because all representatives have the same value of $F$. (Remember the definition of a function: any input must have exactly one output.)
The codomain (a superset of the range) given is $\mathbb R^3$. (Check that all components of $F$ are actually Real numbers.) You need to show that the function maps to the unit sphere; $\lVert F(z_1,z_2)\rVert = 1$ .
A bijection is an injection and a surjection. $F$ being an injection means that it's one-to-one, that different inputs cannot give the same output. $F$ being a surjection, also called "onto", means that the range is the entire unit sphere, not just a part of it. Another way to think of bijectivity is that $F^{-1}$ is uniquely defined everywhere on the unit sphere.

(2) is slightly confusing. I haven't seen a definition of $\mathbb C_\infty$ , but the fact that $S$ is applied to the output of $F$ implies that $\mathbb C_\infty$ is simply the unit sphere in $\mathbb R^3$. (I think this is bad notation, similar to saying that $\mathbb R^2 = \mathbb C$.)
Stereographic projection is invertible, so (2) is equivalent to $F(z_1,z_2) = S^{-1}(\frac{z_1}{z_2})$. I'm familiar with the Real version of stereographic projection as
$$(x,y,z) = S^{-1}(X,Y) = \left(\frac{2X}{X^2+Y^2+1}, \frac{2Y}{X^2+Y^2+1}, \frac{X^2+Y^2-1}{X^2+Y^2+1}\right)$$
(Is there a specific definition that you have to work with?)
Compare this with the given components of $F$, which can be rewritten as
$$F(z_1,z_2) = \left(\frac{2\,\text{Re}(z_1\overline z_2)}{|z_1|^2+|z_2|^2}, \frac{2\,\text{Im}(z_1\overline z_2)}{|z_1|^2+|z_2|^2}, \frac{|z_1|^2-|z_2|^2}{|z_1|^2+|z_2|^2}\right)$$
or, noting that $\frac{z_1}{z_2} = \frac{z_1\overline z_2}{|z_2|^2}$ if $z_2 \neq 0$ ,
$$F(z_1,z_2) = \left(\frac{2\,\text{Re}(z_1/z_2)}{|z_1/z_2|^2+1}, \frac{2\,\text{Im}(z_1/z_2)}{|z_1/z_2|^2+1}, \frac{|z_1/z_2|^2-1}{|z_1/z_2|^2+1}\right)$$
I think you can see the connection, if $z_1/z_2 = X+iY$.
