Square root function on complex plane and on $S^2$

Square root function on complex plane and on $S^2$

If we let $f : \mathbb{C} \to \mathbb{C}$ be given by $f(z)=z^2$, then
writing $z=re^{i\theta}$ where $\theta = \arg(z)$ and $r=|z|$, we have
that $f(z)=r^2 e^{i(2\theta)}$. In that case, let $a \in \mathbb{C}$, then
we can also write $a$ in polar form as $a = R(\cos \phi + i \sin \phi)$.
Now, if we think, which $z$ gives $f(z)=a$, it's easy to come out with
$z=r^{1/2}e^{i\theta/2}$. The only problem is that if we also pick $z =
r^{1/2}e^{i\theta/2+\pi}$ then it'll also do the job since $\cos$ and
$\sin$ have period $2\pi$. In other words, $|f^{-1}(a)|> 1$ and so, we
have problem on defining the inverse function.
Now, I've heard that using the sphere $S^2$ we can solve the problem. My
idea was: if we construct $\mathbb{C}$ simply as $\mathbb{R}^2$ with a
product, then if we forget the product, $\mathbb{C}$ is just
$\mathbb{R}^2$. In that case, let $N \in S^2$ be the north pole and let $U
= S^2 \setminus \{N\}$, then we know we can construct a chart for $S^2$
using stereographic projection, mapping $U$ to $\mathbb{R}^2$. Since
$\mathbb{C}$ is essentially $\mathbb{R}^2$, my idea was to consider the
same chart, $(x,U)$ but with $x: U \to \mathbb{C}$.
In that case my idea was "now we transfer the $f(z)=z^2$ function to
$U\subset S^2$ and show that the version of this function there satisfies
$|f^{-1}(a)|=1$ for every $a \in U$ and we'll be able to define the square
root", what I did was to construct $\xi : U \to U$ by $\xi = x^{-1} \circ
f \circ x$, so it takes a point on the sphere, maps to $\mathbb{C}$, takes
the square, and maps back to the sphere. Now, a mathematician I know said
this won't work, and $\xi$ will still have the same problem, and that
indeed $S^2$ cannot solve the problem.
Isn't this the way we use $S^2$ to solve the problem to define the square
root? How can we do so then? Is it true that $S^2$ don't solve this
problem?
Thanks very much in advance!