In view of the classical theory, any holomorphic map $F: \mathbb{C}/L \to \mathbb{C}/M$ such that $F([0]_L)=[0]_M$ is induced by a linear map $G: \mathbb{C} \to \mathbb{C}$ given by $z \mapsto \gamma z$, where $\gamma L \subseteq M$. Now, I have to compute the degree of $F$. So, I take $[z]_M$ and compute its inverse image $F^{-1}([z]_M)$. Clearly, it results that
$$ F^{-1}([z]_M)=\{[\gamma^{-1} (z+x_\lambda)]_L\} $$
where $x_\lambda +\gamma L$ are the elements of $M/\gamma L$. So, I expect that $deg(F)=[M: \gamma L]$. Why is this number always finite? In fact, $\gamma L$ is a subgroup of $M$ and the latter is a free group on two generators. The subgroups of $M$ are isomorphic to $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, for any $n,m \in \mathbb{Z}$. If I consider $n=1$ and $m=0$, then $[\mathbb{Z} \times \mathbb{Z}: \mathbb{Z}]$ is not finite, am I right?
If $X$ is an irreducible curve, then a morphism $f : X \to Y$ is constant or have finite fibers. Indeed, if a fiber is infinite, being closed it should be all $X$.