**Lemma 29: ***Suppose is a finite set of infinite degree elvations and is compact. Then for all sufficiently large , there exists an intermediate covering such that*

*(a)* * embeds in *

*(b) every * *descends to an elevation of degree *

*(c) the are pairwise distinct *

**Proof: **We claim that the images of * *never share an infinite ray (a ray is an isometric embedding of *). *Neither do two ends of the same elevation* **. * Let’s claim by passing to the universal cover of* **, *a tree *.*

For each , lift * *to a map . If and share an infinite ray then there exists such that and overlay in an infinite ray. The point is that correspond to cosets and . But this implies that

This implies that . So . A similar argument implies that the two ends of do not overlap in an infinite ray. This proves the claim.

Let be the core of . Enlarging if necessary, we can assume that

(i) ;

(ii) is a connected subgraph;

(iii) for each , for some , ;

(iv) for each , .

For each identifying with so that is identified with and is identified with . Let

For all sufficiently large ,

Now, the restriction of factors through , where . This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required.

**Theorem 19:*** is LERF.*

Recall the set-up from the previous lecture. We built a graph of spaces for .

**Proof: **Let be finitely generated. Let be the corresponding covering space of and let be compact. Because is finitely generated, there exists a subgraph of spaces such that . We can take large enough so that . We can enlarge so that it contains every finite-degree edge space of . Also enlarge so that

for any . For each let and let incident edge map of infinite degree .

Applying lemma 29 to , for some large , set . (Here we use the fact that vertex groups of are finitely generated)

Define as follows:

For each , the edge space is the that the lemma produced from the corresponding .

Now, by construction, can be completed to a graph of spaces so that the map

factors through and embeds. Let be identical to except with +’s and -’s exchanged. Clearly satisfies Stallings condition, as required.

## Leave a comment

Comments feed for this article