Let $X$ be a compact Riemann surface (or more generally a compact manifold?) and let $\mathbb{C}^a/\Lambda$ be a complex torus. Suppose we have a holomorphic map $$g: X\to\mathbb{C}^a/\Lambda.$$
By more general theory (universal property of the albanese variety), it seems that for any meromorphic $f: X\to\mathbb{C}$, we have $$\hat g(\text{div}(f))=0\in\mathbb{C}^a/\Lambda,$$ where $\hat g: \text{Div}(X)\to \mathbb{C}^a/\Lambda$ maps $\sum a_ix_i$ to $\sum a_i g(x_i)$.
After thinking about it for a while, I can't see how I could get this equality in the torus at all. I am probably missing something important...
Suppose $X$ is obtained by gluing the sides of a $4\times$genus-gon $D$ in the usual way. Then $g$ lifts to a map $\tilde g:D\to\mathbb C^a$, as $D$ is 1-connected. Let us consider the integral $$I=\frac{1}{2\pi i}\int_{\partial D}\tilde g\frac{df}{f}\in\mathbb C^a.$$ Modulo $\Lambda$ we have (by looking at the residues) $I=\hat g(\operatorname{div}(f))$. On the other hand, if $A$ and $A'$ are two sides of $D$ that get glued in $X$, say $A$ with positive and $A'$ with negative orientation, then $\tilde g$ differs by a constant $\lambda\in\Lambda$ on $A$ and $A'$, hence the part of $I$ coming from $A$ and $A'$ is $$\frac{\lambda}{2\pi i}\int_{A}\frac{df}{f}.$$ Since $\int_{A}\frac{df}{f}\in 2\pi i\mathbb Z$ ($df/f=d\log f$ and $A$ is closed on $X$), we get $I\in\Lambda$, i.e. $I$ is $0$ modulo $\Lambda$.