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I am considering the natural surjection $\pi : \mathcal{H} \to Y(\Gamma)$ where $\mathcal{H}$ is the complex upper half plane and $Y(\Gamma)$ the modular curve of the congruence subgroup $\Gamma$. $Y(\Gamma)$ inherits the quotient topology, meaning a subset of $Y(\Gamma)$ is open if its inverse image under $\pi$ in $\mathcal{H}$ is open. \pi is continuous. Then the problem is the following:

This makes $\pi$ an open mapping.

I want to show the above. I have had two ideas.

1) Simply show that some open set $U \subset \mathcal{H}$ is mapped to an open set $A \subset Y(\Gamma)$. Let $A \subset Y(\Gamma)$ be an open set. Then $\pi^{-1}(A)$ is open by the above. Hence $\pi(\pi^{-1}(A))$ is open and $\pi(\pi^{-1}(A)) \subset A$. So if I can show that $\pi(\pi^{-1}(A)) \supset A$ we would in fact have equality and $A$ would be open. But this does not hold right, since it's only $\pi$ that is surjective?

2) Show that $\pi$ is non-constant and holomorphic and use the Open Mapping Theorem. I tried to find the limit $$\lim_{z \to z_0} \frac{\pi(z)-\pi(z_0)}{z-z_0}$$ but to no avail.

Can you help?

Numbersandsoon
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    Approach $1)$ unfortunately doesn't work, what you want to prove is that $\pi^{-1} (\pi (U))$ is open for anu $U$ open, and not the other way around. And $2)$ is using too much structure, you actually don't need to consider the complex structure on $Y(\Gamma)$. For the answer you could check this previous question: http://math.stackexchange.com/questions/61173/when-quotient-map-is-open . – Giovanni De Gaetano Sep 01 '14 at 07:44
  • Got it, thanks! – Numbersandsoon Sep 01 '14 at 08:00

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