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.