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.

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