# Weierstrass $\wp$-function defines a map from the torus to an elliptic curve. Why is it injective?

For $$L$$ a lattice in $$\mathbb C$$, the Weierstrass $$\wp$$-function is the meromorphic function

$$\wp(z) = \frac{1}{z^2} + \sum\limits_{0 \neq \lambda \in L}\frac{1}{(z-\lambda)^2} - \frac{1}{\lambda^2}$$ It can be shown to satisfy the differential equation $$\wp'(z) = 4\wp(z)^3 - g_2\wp(z) - g_3$$, where

$$g_2 = 60 \sum\limits_{0 \neq \lambda \in L} \frac{1}{\lambda^4}$$

$$g_3 = 120 \sum\limits_{0 \neq \lambda \in L} \frac{1}{\lambda^6}$$ If $$E$$ is the elliptic curve in $$\mathbb P^2$$ defined by the homogeneous polynomial $$y^2z = 4x^3 - g_2xz^2-g_3z^3$$, then

$$F(z) = \begin{cases} (\wp(z);\wp'(z);1) & \textrm{if }z\not\in L \\ (0;1;0) & \textrm{if } z \in L \end{cases}$$

can be shown to define a holomorphic function $$\mathbb C \rightarrow E$$. Since $$\mathscr P$$ and $$\mathscr P'$$ are well defined on $$\mathbb C/L$$, so is $$F$$, and $$F$$ induces a holomorphic function

$$\bar{F}: \mathbb C /L \rightarrow E$$ which is automatically surjective, because $$F$$ is an open map (being holomorphic and nonconstant), and $$\mathbb C/L$$ and $$E$$ are compact. I want to say that $$\bar{F}$$ is a biholomorphism, which is equivalent to saying $$\bar{F}$$ is injective.

How do we know that $$\bar{F}$$ is injective?

[I will treat $$\wp$$ and related functions always as functions with domain $$\mathbb{C}/L$$, rather than $$\mathbb{C}$$.]
First, note that $$\wp$$ has degree $$2$$: its only pole is a double pole at $$0$$, so it takes every value in $$\mathbb{C}\cup\{\infty\}$$ twice, with multiplicity. Also, $$\wp$$ is even. So, if $$z\neq -z$$, then $$z$$ and $$-z$$ are two preimages of $$\wp(z)$$ and thus must be the only such preimages (and both must have multiplicity $$1$$). If $$z=-z$$, we now see that $$\wp$$ is a $$2$$-to-$$1$$ mapping in a deleted neighborhood of $$z$$, so $$z$$ is a preimage of $$\wp(z)$$ of multiplicity $$2$$ and is the only such preimage. In particular, we see that for any $$z,w\in\mathbb{C}/L$$, $$\wp(z)=\wp(w)$$ iff $$z=\pm w$$
So, it suffices to show that $$\wp'(z)\neq \wp'(-z)$$ for any $$z$$ such that $$z\neq -z$$. Now $$\wp'$$ is odd, so $$\wp'(z)=\wp'(-z)$$ implies $$\wp'(z)=0$$. But $$\wp'(z)=0$$ means that $$z$$ is a preimage of $$\wp(z)$$ of multiplicity $$2$$. By the previous paragraph, this implies $$z=-z$$, as desired.