There does not exist a holomorphic map between torus and Riemann sphere So the question is as follows. Prove that there is no meromorphic function $f$ such that at every $z\in \mathbb{C}$ we have $f(z)=f(z+1)$ and $f(z)=f(z+i)$ with only simple poles at the points $m+ni, m,n \in \mathbb{Z}$. 
So, this looks like a holomorphic map from the torus to the Riemann sphere. Since the map is continuous, the image of the torus is compact, hence closed. On the other hand, holomorphic maps are open, hence the image is also open, so the map is surjective. 
Now, if I knew that $f'(z) \neq 0$, then I would have a covering, and there are no coverings from the torus to the sphere. However, I can't make that statement.  
I'm wondering, is there a proof that uses only arguments from complex analysis?
Thanks for any help!

Assume for a contradiction that such a meromorphic function $f\colon\mathbb{C}\rightarrow\mathbb{C}$ existed.
Let $\Gamma\colon= \mathbb{Z}+\mathbb{Z}i$ denote the Gaussian integers.
Consider the holomorphic mapping $F\colon\mathbb{C}/\Gamma\rightarrow \mathbb{P}^1$ associated to $f$. Here, $\mathbb{C}/\Gamma$ and $\mathbb{P}^1$ denote the complex torus with respect to $\Gamma$ and the Riemann sphere, respectively.
Since any complex torus is compact, the mapping $F$ is proper. Thus, there exists a natural number $\operatorname{deg}(F)$, the degree of $F$, such that $F$ takes every value in $\mathbb{P}^1,$ counting multiplicities, $\operatorname{deg}(F)$ times. Since $f$ has poles at the lattice points $\Gamma$, and only there, the mapping $F$ has precisely one pole. This pole is simple by assumption. Thus, $\operatorname{deg}(F)=1$ and $F$ is biholomorphic. This contradicts the fact that the torus $\mathbb{C}/\Gamma\cong S^1\times S^1$ and the Riemann sphere $\mathbb{P}^1\cong S^1$ are not homeomorphic.
