The Branch Schema of a Subgroup of the Modular Group Let $\Gamma$ be a subgroup of the modular group $PSL(2, \mathbb{Z})$.
What is the best and easiest way to grasp the notion of the Branch Schema of the subgroup $\Gamma$.
Why do we have only four cases for it when $\Gamma$ is a normal subgroup of finite index of the modular group $PSL(2, \mathbb{Z})$ (that are $(1,1,n)$, $(1,3,n)$, $(2,1,n)$ and $(2,3,n)$ where $n$ is a natural number)?
Why, in the first three cases $(1,1,n)$, $(1,3,n)$ and $(2,1,n)$, is the space $\Gamma\backslash\mathbb{H}$ simply connected (where $\mathbb{H}$ is the upper half-plane)?
R. C. Gunning. Lectures on Modular Forms. Pages 12 and 13


 A: This is about the branched covering $\Gamma\backslash \Bbb{H}^*\to SL_2(\Bbb{Z})\backslash \Bbb{H}^*$ (where $\Bbb{H}^*=\Bbb{H}\cup\Bbb{Q}\cup i\infty$)
eg. the holomorphic map $$f:\Gamma z\to SL_2(\Bbb{Z})z$$
It is locally biholomorphic (unramified) everywhere except maybe at $f^{-1}(SL_2(\Bbb{Z}) i\infty), f^{-1}(SL_2(\Bbb{Z}) i),f^{-1}( SL_2(\Bbb{Z}) e^{2i\pi /3})$.
Proof: making a special case for $i\infty$, then this is the points where the branched covering $\Bbb{H}\to SL_2(\Bbb{Z})\backslash \Bbb{H}$ is ramified.
Then apply Riemann Hurwitz formula to find the genus of $\Gamma\backslash \Bbb{H}^*$.
The numbers you need to know are $$n(\Gamma z)=|f^{-1}(SL_2(\Bbb{Z}z)|, \qquad N=[SL_2(\Bbb{Z}):\Gamma]$$
Since $\Gamma$ is normal then each element of $\gamma\in \Gamma\backslash SL_2(\Bbb{Z})$ is an automorphism $\gamma(\Gamma z)=\Gamma \gamma z$ of $\Gamma\backslash \Bbb{H}^*$ such that $f= f\circ \gamma$ and $\frac{N}{n(\Gamma z)}$ is the number of automorphisms leaving $\Gamma z$ fixed. Note that $n(\Gamma z)=n(\Gamma \gamma z)$.
$\frac{N}{n(\Gamma i\infty)}$ can be an integer, $\frac{N}{n(\Gamma i)}$ is either $1$ or $2$, and $\frac{N}{n(\Gamma e^{2i\pi/3})}$ is either $1$ or $3$.
Proof: again look at $\Bbb{H}\to SL_2(\Bbb{Z})\backslash \Bbb{H}$.
