Recall that our goal is to determine, for a given map of graphs of spaces such as the one shown below, whether the map $\Phi$ can be extended to a covering map $\widehat{\Phi}$.

Figure 1

Let $\mathcal{X}, \mathcal{X}'$ be graphs of spaces equipped with maps $\Xi' \to \Xi$, $\phi_{v'} : X_{v'} \to X_v$ and $\phi_{e'} : X_{e'} \to X_e$ as before.  Recall that in Stallings’ proof of Hall’s Theorem, we completed an immersion to a covering map by gluing in edges.  We will aim to do the same thing with graphs of spaces.

Definition: Let $\mathcal{X}$ be a graph of spaces, and let $\eta : X_{\mathcal{X}} \to \Xi$ be the map to the underlying graph.  If $\Delta \subseteq \Xi$ is a subgraph, then $\eta^{-1}(\Delta) \subseteq X_{\mathcal{X}}$ has a graph-of-spaces structure $\mathcal{Y}$ with underlying graph $\Delta$.  Call $\mathcal{Y}$ a subgraph of spaces of $\mathcal{X}$.

We’re seeking a condition on $\mathcal{X}'$ such that $\mathcal{X}'$ is realized as a subgraph of spaces of some $\widehat{\mathcal{X}}$ with a covering map $\widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}}$ such that the following diagram commutes:

Definition: For each edge map $\partial_e^{\pm} : X_e \to X_v$ of $\mathcal{X}$, and each $v' \mapsto v$ a vertex of $\mathcal{X}'$, let

$\mathcal{E}^{\pm}(e) = \bigcup_{v' \mapsto v} \mathcal{E}^{\pm}(e, v')$.

For each possible degree $\mathcal{D}$, let $\mathcal{E}_{\mathcal{D}}^{\pm}(e) \subseteq \mathcal{E}^{\pm}(e)$ be the set of elevations of degree $\mathcal{D}$.  We will say $\mathcal{X}'$ satisfies Stallings’ condition if and only if the following two things hold:

(a) Every edge map of $\mathcal{X}'$ is an elevation of the appropriate edge map of $\mathcal{X}$.
(b) For each $e \in E(\Xi)$ and $\mathcal{D}$, there is a bijection $\mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e)$.

So in Figure 1, the graph of spaces $\mathcal{X}'$ is something you might be able to turn into a covering.  In the picture, $\mathcal{E}_{\mathcal{D}}^+(e)$ is represented by the blue circles, and $\mathcal{E}_{\mathcal{D}}^-(e)$ is represented by the green circles.  Observe that the blue circles are in bijection with the green circles.

Corollary: $\mathcal{X}'$ satisfies Stallings’ condition if and only if $\mathcal{X}'$ can be realized as a subgraph of spaces of some $\widehat{\mathcal{X}}$ such that

(a) $V(\Xi ') = V(\widehat{\Xi})$, and
(b) there is a covering map $\widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}}$ such that the following diagram commutes:

Proof of Corollary. First we’ll show that if $\Phi$ can be extended to a covering map as described above, then $\mathcal{X}'$ satisfies Stallings’ condition.  By Theorem 17, every edge map of $\widehat{\mathcal{X}}$ is an elevation.  So there are inclusions

Furthermore, these maps are surjective, and clearly degree-preserving.

Now assume that $\mathcal{X}'$ satisfies Stallings’ condition.  Then we build $\widehat{\mathcal{X}}$ as follows.  Let $V(\widehat{\Xi}) = V(\Xi ')$.  As above, we have degree-preserving inclusions (this time, not surjections)

Extend these inclusions to bijections $\mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e)$.  Now we set $E(\widehat{\Xi}) = \bigcup_{e \in E(\Xi)} \mathcal{E}^+(e)$.  Each of these $\widehat{e}$ is an elevation

This defines an edge space $X_{\widehat{e}}$ and an edge map $\partial_{\widehat{e}}^+$.  Consider the corresponding elevation in $\bigcup_{e \in E(\Xi)} \mathcal{E}^-(e)$:

Because $\partial_{\widehat{e}}^+$ and $\partial_{\widehat{e}}^-$ are of the same degree, we have a covering transformation $X_{\widehat{e}} ' \to X_{\widehat{e}}$.  So we can identify them, and use $\partial_e^-$ as the other edge map.  By construction, $\widehat{\mathcal{X}}$ satisfies the conditions of Theorem 17, so there is a suitable covering map $\widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}}$.

Exercise 25: (This will be easier later, but we have the tools necessary to do this now.)  Prove that if $G_1$ and $G_2$ are LERF groups, then so is $G_1 * G_2$.