I am reading Rick Miranda's Book and he states
Every algebraic curve of genus one is isomorphic to a complex torus. Now to do this let $X$ be our Riemann surface and we know that $\pi : Y\rightarrow X$, where $Y$ is the universal covering of $X$, and where $\mathbb{Z}\times \mathbb{Z}$ acts on $Y$ by two independent translations, and $Y=\mathbb{R^2}$ as a topological space, so all we need to see is that $Y\cong \mathbb{C}$ as a Riemann surface. Now for this we take a canonical divisor $K_0=div(w_0)$ in $X$ and we know that $deg(K_0)=0$ and $dimL(K_0)=1$ by Riemann-Roch and so we can take $f\in L(K_0)$ so that $w=fw_0$ will have no zeros and poles, since $deg(div(w))=0$ and $div(w)\geq 0.$ Now we can take $\pi^*(w)$ in $Y$ which will have no zeroes and will be a holomorphic $1$-form, and we fix a point $p_0$ and define $\phi:Y\rightarrow \mathbb{C}$ as $\phi(p)=\int_{\gamma_p}\pi^*w$, where $\gamma_p$ is a path from $p_0$ to $p$, and this is well-defined since the integral will only depend on the end-points.
Now my question is how do we know that is surjective or even injective. I know it depends only on the endpoints but I don't see this giving us what we want, I can't seem to be able to prove it, and the author doesn't do it or even metion this so I dont know if this is just a trivial statement or not, he says he will return to it later but I haven't gotten to that part.
Thanks in advance.
This is explained nicely in the book by Herbert Clemens, "A Scrapbook of Complex Curve Theory", Paragraph 2.9. It is not obvious at all, and involves an argument by Riemann.