Homeomorphism between $\mathbb{C}/L$ and $S^1\times S^1$ We have the lattice $L = \{m_1w_1 + m_2w_2 \mid  m_1, m_2 \in \mathbb{Z}, w_1, w_2 \in \mathbb{C}\}$. We want to construct a homeomorphism between $\mathbb{C}/L$ and $S^1\times S^1$. I've read that the function $f(z)= (e^{2\pi  ai},e^{2\pi bi})$ with $z=aw_1 +bw_2$ could be an option. Can someone explain to me in more detail why this function gives us the homeomorphism we want?
Thank you.

Consider the map $f: \mathbb{C}/L\to S^1\times S^1$ given by 
$f(a\omega_{1}+b\omega_{2}+L)=(e^{2\pi a i}, e^{2\pi b i})$.
First, $f$ is well-defined. Indeed, if $a\omega_{1}+b\omega_{2} + L = c\omega_{1}+d\omega_{2}+L$, then $c=a+m_{1}$ and $d=b+m_{2}$ for some integers $m_{1}$ and $m_{2}$. Using $e^{2\pi n i}=1$ for any integer $n$, we easily get
$f(a\omega_{1}+b\omega_{2}+L)=f(c\omega_{1}+d\omega_{2}+L)$.
Next, $f$ is injective because if $f(a\omega_{1}+b\omega_{2}+L)=f(c\omega_{1}+d\omega_{2}+L)$, then $(e^{2\pi a i}, e^{2\pi b i}) = (e^{2\pi c i}, e^{2\pi d i})$. From here, it follows that $a-c$ and $b-d$ are integers, so that $a\omega_{1}+b\omega_{2}+L=c\omega_{1}+d\omega_{2}+L$.
Now, $f$ is clearly surjective, and thus bijective. Since exponential function is continuos, $f$ is a continuous bijection. A continuous bijection from a compact space $\mathbb{C}/L$ to a Hausdorff space $S^1\times S^1$ is homeomorphism, so we are done.
