I recently started to study conguence subgroups and quotients of the upper half-plane by their action. I found various proofs of the existence of the fundamental region for congruence subgroups that use the hyperbolic metric. I accept those but I would prefer proofs that only use transformations of a fundamental domain of $SL_2(\mathbb{Z})$.
What i mean is this: let $\Gamma$ be a congruence subgroup and $\{\gamma_i\}\in SL_2(\mathbb{Z})$ such that $SL_2(\mathbb{Z})=\bigcup_i \gamma_i \Gamma$; let now $\mathcal{D}$ be a fundamental domain for $SL_2(\mathbb{Z})$; the set $\mathcal{F}=\bigcup_i\gamma_i\mathcal{D}$ has some properties of a fundamental domain such as having no conjugate points in the interior and having a representative for each equivalence class in $\mathbb{H}$. This set however is not necessarily connected.
The problem is that I can't quite show that there exist a choice of coset representatives such that the union is connected as it should. The only ideas that come up to mind are having a list of "adjacent" to the current region triangles that are not conjugate to one another and choosing a transformation as a representative by iterating. However I can't find a way to show that once the list is finished the set of tranformations obtained is in fact a choiche of representatives. Are there any proofs that work that way?
Let me answer a more general question: given any subgroup $\Gamma < SL_2(\mathbb Z)$, how do you use a $\mathcal D$ for $SL_2(\mathbb Z)$ to construct a fundamental domain for $\Gamma$?
Let me specialize to the finite index case. Let's enumerate the left cosets of $\Gamma$ as $C_1,...,C_K$, with $C_1=\Gamma$.
Denote $\mathcal D_0 = \mathcal D$. The polygon $\mathcal D_1$ has four sides (two meeting at a straight angle), incident to four translates of $\mathcal D_1$ by four group elements $g_1,g_2,g_3,g_2$. These four group elements generate the entire group $SL_2(\mathbb Z)$. Therefore, assuming $K \ge 2$, at least one of these elements, say $g_{j_1}$, is NOT in $\Gamma=C_1$, and so there exists $k_1 \in \{2,...,K\}$ such that $g_{j_1} \in C_{k_1}$. Change the enumeration so that $k_1=2$, $g_{j_1} \in C_2$.
Define $\mathcal D_2 = \mathcal D_1 \cup \left(g_{j_1} \cdot \mathcal D_1\right)$. This is a polygon having $6$ sides, incident to six translates of $\mathcal D_1$ by six group elements $g_5,...,g_{10}$. These six group elements, together with $g_1$, generate the entire group. They are therefore not all contained in $C_1 \cup C_2 = C_1 \cup g_1 C_1$, and so, assuming $K \ge 3$, one of them, say $g_{j_2}$ is contained in $C_{k_2}$ for $3 \le k_2 \le K$. Change the enumeration so that $k_2=3$.
Define $\mathcal D_3 = \mathcal D_2 \cup g_{j_2} \cdot \mathcal D_1$. Now perhaps you can see how the induction continues...