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

Let be graphs of spaces equipped with maps , and 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 be a graph of spaces, and let be the map to the underlying graph. If is a subgraph, then has a graph-of-spaces structure with underlying graph . Call a *subgraph of spaces of *.

We’re seeking a condition on such that is realized as a subgraph of spaces of some with a covering map such that the following diagram commutes:

**Definition: **For each edge map of , and each a vertex of , let

.

For each possible degree , let be the set of elevations of degree . We will say satisfies *Stallings’ condition* if and only if the following two things hold:

**(a)** Every edge map of is an elevation of the appropriate edge map of .

**(b)** For each and , there is a bijection .

So in Figure 1, the graph of spaces is something you might be able to turn into a covering. In the picture, is represented by the blue circles, and is represented by the green circles. Observe that the blue circles are in bijection with the green circles.

**Corollary:** satisfies Stallings’ condition if and only if can be realized as a subgraph of spaces of some such that

**(a)** , and

**(b)** there is a covering map such that the following diagram commutes:

**Proof of Corollary.** First we’ll show that if can be extended to a covering map as described above, then satisfies Stallings’ condition. By Theorem 17, every edge map of is an elevation. So there are inclusions

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

Now assume that satisfies Stallings’ condition. Then we build as follows. Let . As above, we have degree-preserving inclusions (this time, not surjections)

Extend these inclusions to bijections . Now we set . Each of these is an elevation

This defines an edge space and an edge map . Consider the corresponding elevation in :

Because and are of the same degree, we have a covering transformation . So we can identify them, and use as the other edge map. By construction, satisfies the conditions of Theorem 17, so there is a suitable covering map .

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

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6 April 2009 at 8.01 am

Henry WiltonFabulous picture!