Every degree $2$ holomorphic endomorphism of $\mathbb{CP}^1$ with two ramification points such that $g(a_1) = 0$ and $g(a_2) = \infty$ can be written…

Question

I’d like to show that if $g:\mathbb{CP}^1 \rightarrow \mathbb{CP}^1$ is a holomorphic map of degree 2 with two ramification points $a_1 \neq a_2$ such that $g(a_1) = 0$ and $g(a_2) = \infty$, then $g$ can be written in the form $g(z) = (\varphi(z))^2$, where $\varphi \in \operatorname{Aut}(\mathbb{CP}^1)$.

I’ve thought about it but I don’t really know how to attack this problem, so any hints would be very appreciated.

Some thoughts: I know that $g$ should locally look like $z \mapsto z^2$ around the ramification points (in local coordinates). I also was wondering if $g$ could generally be written in the form $a((z-a_1)/(z-a_2))^2$, where $0 \neq a \in \mathbb{C}$.

Answer 1

A few additional thoughts:

• It might make our lives easier if we assume that $a_1 = 0$ and $a_2 = \infty$. (If this isn't the case to begin with, then we can make this the case by composing $g$ with a suitable automorphism of $\mathbb {CP}^1$.)
• With $a_1 = 0$ and $a_2 = \infty$, you're right that $g$ "looks like" $z \mapsto z^2$ locally around $0$. But we can make your statement even more precise. By considering the Taylor series expansion of $g$ around $0$, we should be able to argue that $g(z) = z^2 h(z)$, where $h$ is a non-vanishing entire function.
• Using Cauchy's theorem, we should able to argue that $h$ has a holomorphic logarithm, and therefore, $h$ has a holomorphic square root.

I hope this is enough for you to get going with. If not, do leave a comment.

---

Edit: Actually, I realise there is another way to think about this problem. Are you aware that every meromorphic function on $\mathbb {CP}^1$ is a rational function? If you try to construct a rational function with a double zero at $a_1$ and a double pole at $a_2$ and no other zeroes or poles, you'll quickly realise that you don't have that many options. I think you're spot on to conjecture that $g$ can be written as $a((z-a_1)/(z-a_2))^2$ (for some $a \in \mathbb{C} \setminus \{ 0 \}$).