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!
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 $z_1/z_2 = X+iY$.