complex torus has topological genus one could any one give me a hint how to show a complex torus has topological genus one by constructing an explicit homeomorphism to $S^1\times S^1$?
Complex Torus: $\mathbb{C}/L$, where $L=\{\mathbb{Z}\omega_1+\mathbb{Z}\omega_2\}$.
Thank you.
 A: For convenience, I will identify $S^1$ with the topological group $\mathbb{R} / \mathbb{Z}$. Let $\Lambda$ be the lattice $\mathbb{Z} \omega_1 + \mathbb{Z} \omega_2$. Since $\omega_1$ and $\omega_2$ are not parallel, there exists a choice of (real!) coordinates $(x, y)$ on $\mathbb{C}$ such that $\omega_1 = (1, 0)$ and $\omega_2 = (0, 1)$. It is clear that there is a continuous (indeed, smooth) group homomorphism $\mathbb{C} \to S^1 \times S^1$ given by the formula
$$(x, y) \mapsto (x + \mathbb{Z}, y + \mathbb{Z})$$
and $\Lambda$ lies in the kernel of this homomorphism, so it factors as a homomorphism $\mathbb{C} / \Lambda \to S^1 \times S^1$. It is not hard to check that this is bijective. On the other hand, $\mathbb{C} / \Lambda$ is compact and $S^1 \times S^1$ is Hausdorff, so this is not just a continuous bijection but a homeomorphism. 
There is a much more exciting answer involving Weierstrass $\wp$-functions, but for that one should take a different definition of "genus one" and/or "complex torus"...
A: From a topological viewpoint $\Bbb C/L$ is indistinguishable from $\Bbb R^2/\Bbb Z^2$.
But then it should be clear that
$$
\frac{\Bbb R^2}{\Bbb Z^2}\simeq\frac{\Bbb R}{\Bbb Z}\times\frac{\Bbb R}{\Bbb Z}
\simeq S^1\times S^1.
$$
