Last time we saw that if $G$ acts on a tree, $T$ then $G/T$ has the structure of a graph of groups (remeber stabilizers). Such a $\mathcal{G}=G/\Gamma$ is called developable.

Exercise 18: Show that $SL_2\mathbb{Z}\cong(\mathbb{Z}/4\mathbb{Z})\ast_{\mathbb{Z}/2\mathbb{Z}}(\mathbb{Z}/6\mathbb{Z})$.

Theorem 13 (Scott-Wall): Let $\mathscr{X}$ be a graph of spaces. The universal cover of $X_{\mathscr{X}}$ is itself a graph of spaces. Also, each vertex space (resp. edge space) is the universal cover of a vertex space of $\mathscr{X}$ (.edge space).

Sketch of Proof: For any $v\in V(\Xi)$, let $L_v=X_v\cup(\cup_{e\in E}X_e\times [\pm 1,0])$, where $E$ is the set of edges that are the edges incident to $v$. It should be noted that $X_v$ is a deformation retract of $L_v$. Also, recall that the edge maps are $\pi_1$ injective. From covering space theory we know that given a map $f:Y\to X$ and a covering space $\hat X$ of $X$ that $f$ lifts to a map $\hat f:Y\to \hat X$ if and only if $f_\ast (\pi_1(Y))\subset \pi_1(\hat X)$. It therefore follows that $\tilde X_v \hookrightarrow \tilde L_v$. So $\tilde L_v$ is built from $\tilde X_v$ by attaching covering spaces of edge spaces. Because $\partial_{\pm}$ is $\pi_1$ injective, these covering spaces of edge spaces really are universal covers. By iteratively gluing together copies of the $\tilde L_v$ we can construct a simply connected cover of $X_{\mathscr{X}}$.

Given $X_{\mathscr{X}}$ we have constructed the universal cover $\tilde X_{\mathscr{X}}$. The underlying graph, $T$ of $\tilde X_{\mathscr{X}}$ is a tree because $\pi_1(\tilde X_{\mathscr{X}})\to\pi_1(T)$ is a surjection. We now need to check that the action of $G=\pi_1(X_{\mathscr{X}})$ on $T$ is interesting.

Let $\mathcal{G}$ be a graph of groups and let $G=\pi_1(\mathcal{G})$. Fix a base point $\ast\in X_v \subset X_{\mathcal{G}}$  and a choice of lift $\tilde\ast\in \tilde X_{\tilde v}\subset \tilde X_{\mathcal{G}}$. Let $g\in G_v=\pi_1(X_v)$.  The space $\tilde X_{\tilde v}$ is a universal cover of $X_v$ by Theorem 13, and so the lift of $g$ to the universal cover of $X_{\mathcal{G}}$ at $\tilde\ast$ is contained in $\tilde X_{\tilde v}$. Therefore the preimages of $\ast$ that are contained in $\tilde X_{\tilde v}$ correspond to the elements of $G_v$.

Now consider $g\in G\smallsetminus G_v$.  If $g$ is lifted at $\tilde \ast$ then the terminus of this lift is not in $\tilde X_{\tilde v}$, but in some other component of the preimage of $X_v$.  Call the component where the lift terminates $\tilde X_{\tilde v_1}$. If $g,h\in G$ are such that both have lifts that terminate in $\tilde X_{\tilde v_1}$ then $h^{-1}g\in G_v$. We have just proved the following lemma.

Lemma 16: Let $\mathcal{G}$ be a graph of groups and let $T$ be the underlying graph of the universal cover $\tilde X_{\mathcal{G}}$. For any vertex $v\in V(\Gamma)$ the set of vertices of $T$ lying above $v$ is in bijection with $G/G_v$ and $G$ acts by left translation.

We can also prove the following two lemmas in a similar fashion.

Lemma 17: For any $e\in E(\Gamma)$ the set of edges of $T$ that lie above $e$ is in bijection with $G/G_e$ and $G$ acts by left translation.

Lemma 18: If $e\in E(\Gamma)$ adjoins a vertex $v\in V(\Gamma)$ then for any $\tilde v\in V(T)$ lying above $v$ the set of edges of $T$ adjoining $\tilde v$ lying above $e$ is in bijection wiht $G_v/G_e$ and $G_v$ acts by left translation.