Möbius transforms on the projective line Let's suppose that I have a mapping
$$
\phi:\mathbb{P}^1\rightarrow\mathbb{P}^1
$$
of degree two. How can I use Möbius transforms  to write this map as
$$
\phi([x:y])=\frac{y^2}{x^2}?
$$
I already know, from Hurwitz's formula, that $\phi$ has two ramification points, both of multiplicity two.
Thanks for any help!
 A: (Answer has been re-written. It is essentially the argument given by Theo Johnson-Freyd in a comment, augmented by an explicit example.)
Suppose $\Phi : X \to Y$ is a degree-two morphism, where $X \cong Y \cong \mathbb{P}^1$.  Then, as the OP says, an argument based on Hurwitz's theorem tells us that $\Phi$ has two ramification points, each of multiplicity two.
If we let $z$ be an affine coordinate on $X$, and take for granted the well-known fact that Möbius transformations are two-transitive${}^1$ on $\mathbb{P}^1$, we can map the two ramification points to $z = \pm\infty$.  Similarly, if $w$ is an affine coordinate on $Y$, we can map their images to $w = \pm\infty$.
The morphism $\Phi$, restricted to the affine patch, becomes a map $\phi : \mathbb{C}\to\mathbb{C}$, $w = \phi(z)$, which is generically two-to-one, but ramified at zero.  This can only be $w = \alpha z^2$ for some $\alpha \in \mathbb{C}^*$.  Rescaling $w$ (which is a Möbius transformation fixing $0$ and $\infty$), we get $w = z^2$, which is what we wanted.

An example:
Let the morphism be given, as a map from homogeneous coordinates $[x:y]$ on $X$ to an affine coordinate $w$ on $Y$, as
$$
w = \frac{x^2 + y^2}{xy} ~.
$$
Introducing the affine coordinate $z = x/y$, this becomes
$$
w = z + \frac 1z~.
$$
Setting $dw/dz = 0$, we find the ramification points are $z = \pm 1$.  So the first thing to do is find a transformation which sends these to $0$ and $\infty$. Define
$$
z' = \frac{z+1}{z-1} ~.
$$
Now we find
$$
w = 2\left(1 + \frac{2}{z'^2 - 1}\right)~,
$$
hence
$$
\frac{w + 2}{w-2} = z'^2 ~.
$$
So the branch points are at $w = \pm 2$ (indeed, these points corresponds to the images of $z = \pm 1$; this is a simple consistency check).  The left-hand side of the above is precisely the Möbius transformation we need to send these points to $0$ and $\infty$; setting $w'$ equal to this expression, we have $w' = z'^2$, and are done.
${}^1$ In fact they are three-transitive, but we don't need this.
