Lemma 29: Suppose $\{f_i':C_i\longrightarrow\Gamma^{H}\}$ is a finite set of infinite degree elvations and $\Delta \subseteq \Gamma^{H}$ is compact. Then for all sufficiently large $d>0$, there exists an intermediate covering $\Gamma_d$ such that

(a) $\Delta$ embeds in $\Gamma_d$

(b) every $f'_i$ descends to an elevation $\hat{f_i}:\hat{C_i}\longrightarrow \Gamma_d$ of degree $d$

(c) the $\hat{f_i}$ are pairwise distinct

Proof: We claim that the images of $f_i'$ never share an infinite ray (a ray is an isometric embedding of $[0,\infty)$). Neither do two ends of the same elevation $f_i'$. Let’s claim by passing to the universal cover of $\Gamma$, a tree $T$. For each $i$, lift $f_i'$ to a map $\tilde{f_i}:\mathbb{R}\longrightarrow T$. If $f'_i$ and $f'_j$ share an infinite ray then there exists $h\in H$ such that $\tilde{f_i}$ and $h\tilde{f_j}$ overlay in an infinite ray. The point is that $\tilde{f_i}, \tilde{f_j}$ correspond to cosets $g_if_{\ast}(\pi_1(C))$ and $g_jf_{\ast}(\pi_1(C))$. But this implies that $g_if_{\ast}(\pi_1(C))=hg_jf_{\ast}(\pi_1(C))$

This implies that $Hg_if_{\ast}(\pi_1(C))=Hg_jf_{\ast}(\pi_1(C))$. So $f'_i=f'_j$. A similar argument implies that the two ends of $f'_i$ do not overlap in an infinite ray. This proves the claim.

Let $\Gamma'$ be the core of $\Gamma^{H}$. Enlarging $\Delta$ if necessary, we can assume that

(i) $\Gamma'\subseteq\Delta$;

(ii) $\Delta$ is a connected subgraph;

(iii) for each $i$, for some $x_i\in C'_i$, $f'_i(x_i)\in\Delta$;

(iv) for each $i$, $|im(f'_i)\cap \delta\Delta|=2$.

For each $i$ identifying $C'_i$ with $\mathbb{R}$ so that $C$ is identified with $\mathbb{R}/\mathbb{Z}$ and $x_i$ is identified with $0$. Let $\Delta_d=\Delta\cup(\bigcup_i f'_i([-d/2,d/2]))$

For all sufficiently large $d$, $f'_i(\pm d/2)\notin\Delta$

Now, the restriction of $\Delta_d \longrightarrow \Delta$ factors through $\Delta_d/\sim\longrightarrow\Gamma$, where $f'_i(d/2)\sim f'_i(-d/2)$. This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required. $\Box$

Theorem 19: $D$ is LERF.

Recall the set-up from the previous lecture. We built a graph of spaces $\mathscr{X}$ for $D$.

Proof: Let $H\subset D$ be finitely generated. Let $X_H$ be the corresponding covering space of $X_{\mathscr{X}}$ and let $\Delta\subseteq X_H$ be compact. Because $H$ is finitely generated, there exists a subgraph of spaces $X'$ such that $\pi_1(X') =H$. We can take $X'$ large enough so that $\Delta \subseteq X'$. We can enlarge $\Delta$ so that it contains every finite-degree edge space of $X'$. Also enlarge $\Delta$ so that $\partial^{\pm}_{e'}(\Delta\times (\mathrm{interval}))\subseteq\Delta$

for any $e'\in E(\Xi')$. For each $v'\in V(\Xi')$ let $\Delta_{v'}=\Delta\cap X_{v'}$ and let ${f'_i}=\{$ incident edge map of infinite degree $\}$.

Applying lemma 29 to $\Gamma^H=X_{v'}$, for some large $d$, set $X_{\hat{v}}=\Gamma_d$. (Here we use the fact that vertex groups of $\mathscr{X}'$ are finitely generated)

Define $\mathscr{X}^+$ as follows: $\bullet \Xi^+=\Xi^-$ $\bullet$ For each $v^+\in V(\Xi^+)$, the edge space is the $X_{v^+}$ that the lemma produced from the corresponding $v'$.

Now, by construction, $\bigcup_{v^+}X_{v^+}$ can be completed to a graph of spaces $\mathscr{X}^+$ so that the map $X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}}$

factors through $X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}}$ and $\Delta$ embeds. Let $\mathscr{X}^-$ be identical to $\mathscr{X}^+$ except with +’s and -‘s exchanged. Clearly $\mathscr{X}^+\cup\mathscr{X}^-$ satisfies Stallings condition, as required. $\Box$