map from $\mathbb{C}$ to $\mathbb{C}/L$ is open map?

Question

Let $w_1,w_2\in\mathbb{C}$ be linearly independent vectors and let$$L=\{m_1w_1+m_2w_2:m_1,m_2\in\mathbb{Z}.\}$$

How does one show that the projection map $\pi:\mathbb{C}\rightarrow\mathbb{C}/L$ is open map? Well, we define $U \subset \mathbb{C}/L$ is open iff $\pi^{-1}(U)$ is open in $\mathbb{C}$ and let hence $\pi$ is continuous, let $V$ be open in $\mathbb{C}$. Then to show $\pi(V)$ is open, enough to show $\pi^{-1}(\pi(V))$ is open in $\mathbb{C}$, but I am not able to visualize the situation or fact here, shall be happy if some one formally and informally write how to prove this. Thank you.

Answer 1

Here are two "general" explanations for why $\pi: \mathbb{C} \rightarrow \mathbb{C}/\Lambda$ is an open map.

1) $\pi$ is a covering map, hence a local homeomorphism, hence an open map.

I think this is quite intuitive and easy to see: about any point on the torus $\mathbb{C}/\Lambda$, the preimage of a sufficiently small disk-shaped neighborhood will be a "lattice" of small disks in $\mathbb{C}$. In fact $\pi$ is a regular covering map, hence the quotient by an action of the group $\Lambda$.

This leads to a more general answer.

2) For a group $G$ acting on a topological space $X$, the quotient map $\pi: X \rightarrow X/G$ is open.

Proof: By definition of the quotient topology on $X/G$, we must show that if $U \subset X$ is open, then $\pi^{-1} \pi U$ is open. But $\pi^{-1} \pi U = \bigcup_{g \in G} g U$. Since each $g \bullet$ is a homeomorphism (that is part of the definition of a group action on a topological space) and $U$ is open, $g U$ is open, so $\pi^{-1} \pi U$ is a union of open sets, hence open.

Note that Paul Garrett remarks that most reasonable quotient maps are open. While I certainly agree with this as a principle, in any particular reasonable situation one still needs to summon a proof. Studying in general the problem of openness of quotient maps seems (to me, of course) to be unrewarding and technical -- c.f. Bourbaki's General Topology, which does entirely too much of this for my taste -- so it is worthwhile to collect "easy" explanations like those above. In fact 2) above was taken directly from lecture notes for a course on modular curves I taught recently: and it is from page 1 of those notes!

Answer 2

Hint: show that $$\pi^{-1}(\pi(V)) = \bigcup_{m,n}(V+m\omega_1+n\omega_2).$$

Why is this open?