Appell Humbert Theorem and characters $\alpha(u+v)=\alpha(u)\alpha(v) \ \forall u,v \in \Gamma$.

Let $V$ be a complex vector space of dimension $g$, let $\Gamma$ be a lattice in $V$, and $X=V/\Gamma$ a complex torus. Let's call $G$ the group of characters, whose elements are maps $\alpha : \Gamma \to U(1)$, where $U(1)$ is the complex unit circle, such that $\alpha(u+v)=\alpha(u)\alpha(v) \ \forall u,v \in \Gamma$.

Part of the statement of Appell Humbert Theorem says that

$$\frac{H^1(X, \mathcal{O}_X)}{H^1(X,\mathbb Z)} \simeq G.$$

I can't figure out why this is actually true.

I know that $H^1(X, \mathcal{O}_X) \simeq \overline V^*$ ($\mathbb C$-antilinear maps on $V$, or equivalently $(0,1)$-forms on $V$). And $H^1(X,\mathbb Z) \simeq \Gamma^*$ (linear applications $\Gamma \to \mathbb Z$). Thus we must show

$$\frac{\overline V^*}{\Gamma^*} \simeq G$$

where the inclusion $\Gamma^* \subset \overline V^*$ is probably intended as $\mathbb C$-antilinear extention.

It is said that a representative for an element of such a quotient is a $(0,1)$-form $\omega$ that is invariant under $\Gamma$-translation, but I don't know why, and that it's easy to see that it goes in $Pic(X)$ in the class of $L(0,\alpha)$ where

$$\alpha(u)=e^{2i\pi\int_0^{u}\omega +\overline \omega}.$$

It is said that we must use Dolbeault Isomorphism, but I can't see how. This is of course a character thanks to the invariance by $\Gamma$-translation.

Thanks!