With the notation $\overline{\mathbb C}= \mathbb C\cup \{\infty\}$, I want to show that every meromorphic bijection $f:\overline{\mathbb C}\rightarrow \overline{\mathbb C}$ is a Möbius transformation. A Mobius transformation is any
$$ \frac{az+b}{cz+d} , \quad a,b,c,d\in\mathbb C$$
with $ad-bc\ne 0$. I have that the meromorphic functions in $\overline {\mathbb C}$ are the rational functions. I can see that, because $f$ is rational then it has a partial fraction decomposition. But so does $z^2$ and that's a rational function … so I can guess that somehow the assumption of being bijective comes in right about now. And somehow it should tell us that the maximum degree of the numerator and denominator is 1.
I'm also guessing that somehow I should first inspect the poles and argue that the largest order of any pole is 1, and then make some kind of argument about $f(1/z)$.
So ok, to make an attempt on some of these guesses: $f$ has the form
$$ f(z) = {P(z)\over Q(z)} $$
If $\deg P \geq \deg Q$ then we can perform long division to separate this into
$$ {P(z)\over Q(z)} = R(z)+{S(z)\over Q(z)} $$
Where $R$ and $S$ are polynomials and $\deg S < \deg Q$. I'm not sure why the degree of $R$ should have to be 0 but it seems like it must, because I'm guessing we'll turn the ${S(z)\over Q(z)}$ into a sum of polynomials in $1/(z-z_k)$ for the $z_k$ roots of $Q$. Then with each of these being a Mobius transformation, we want to argue that their sum is a Möbius transformation …
Well, I'm actually now unsure that their sum is a Möbius transformation now that I think of it. Maybe this whole path was a giant dead-end. Should I be trying to do something entirely different from partial fractions?
Best Answer
You can compose any such bijection $\varphi:\overline{\mathbb C}\to\overline{\mathbb C}$ with a Möbius transformation $\mu$ to get a meromorphic bijection mapping $\infty\mapsto\infty$. Restricting to $\mathbb C$, this is an entire bijection $\mathbb C\to\mathbb C$, with a pole at $\infty$. The pole at $\infty$ makes it a polynomial, and bijective polynomials are exactly the linear ones, which are Möbius transformations. So $\varphi\circ\mu$ is a Möbius transformation. And since Möbius transformations are a group under composition, $\varphi$ is one, too.