Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb Z)$. The quotient $\pi: \mathscr H \rightarrow Y(\Gamma) = \Gamma \backslash \mathscr H$ is given a one-dimensional complex manifold structure as follows: let $\pi(\tau) = x \in \Gamma \backslash \mathscr H$ for $\tau \in \mathscr H$. There exists an open neighborhood $U \subseteq \mathscr H$ of $\tau$ such that:
(i) if $\gamma \in \Gamma$ with $\gamma(U) \cap U \neq \emptyset$ then $\gamma(\tau) = \tau$ and
(ii): $U$ has no elliptic points except for possibly $\tau$.
Let $\delta(z) = \frac{z - \tau}{z - \bar{\tau}}$, which sends $\tau$ to zero and $\bar{\tau}$ to $\infty$. Let $h$ be the order of the image of the stabilizer of $\tau$ as a group of automorphisms of $\mathscr H \cup \{\infty\}$, and let $\rho(z) = z^h$. Note that the image is cyclic of order $h$. Let $\psi = \rho \circ \delta$. Then one shows that for $\tau_1, \tau_2 \in U$, we have $\pi(\tau_1) = \pi(\tau_2)$ if and only if $\psi(\tau_1) = \psi(\tau_2)$. Setting $V = \psi(U)$,
$$\phi: \pi(U) \rightarrow V, \pi(\tau_1) \mapsto \psi(\tau_1)$$
is a well defined homeomorphism which defines a chart $(\pi(U),\phi)$.
Let $(\pi(U_i), \phi_i), i = 1, 2$ be two charts defined for points $x_1 = \pi(\tau_1), x_2 = \pi(\tau_2)$ and neighborhoods $U_1, U_2$ as above. Let $h_1, h_2$ be the orders of the images of the stabilizers of $\tau_1, \tau_2$. I'm trying to understand why these charts are compatible. It would be a mess to write down my attempt at doing so, so I would appreciate any hint or insight into how this is done.