I'm asked to show that the Weierstrass's elliptic function $\wp: \mathbb{C}/\Gamma \rightarrow \mathbb{C}P^1$ has exactly 4 branch points. My problem is that I don't see why there are 4 branch points and not just 3.
I looked at the zeroes of the derivative $\wp '$. Since $\wp '$ is doubly periodic and odd this implies that $\frac{w_1}{2}$, $\frac{w_2}{2}$ and $\frac{w_1+w_2}{2}$ are zeros of $\wp '$, where $w_1,w_2$ span $\Gamma$. But I know that an elliptic function has the same number of poles as it has zeros (where the order of the poles / zeroes matters). Since $\wp '$ has only one pole (of order 3) I know that the three zeros are all of order 1 and in particular there can't be a fourth zero.
Where is my mistake?
The final branch point is at $z = 0$, and the reason for this is a bit subtle.
The image of $z = 0$ under $\wp$ is the "point at infinity" on $\mathbb {CP}^1$. In fact, $z = 0$ is the unique point on $\mathbb C / \Gamma$ whose image under $\wp$ is the point at infinity on $\mathbb {CP}^1$. Since the map $\mathbb C/\Gamma \to \mathbb {CP}^1$ is of degree two, this observation alone is enough for us to conclude that $z = 0$ is a branch point.
For an alternative perspective, recall that our map $\mathbb C / \Gamma \to \mathbb {CP}^1$ is given by $z \mapsto [1 : \wp(z)] $ for $z \neq 0$. But to define the map at $z = 0$, we need to rewrite the map as $z \mapsto [\wp(z)^{-1} : 1]$. Now $\wp(z)$ has a double pole at $z = 0$, so $\wp(z)^{-1}$ has a double zero at $z = 0$, hence $z = 0$ is a branch point of our covering with ramification index equal to two.