Our goal is to understand the structure of subgroups of graphs of groups. This will enable us to prove things like: $F \ast_{z} F$ is LERF.

Exercise 22. A group is coherent if every fg subgroup is fp. Prove that $\pi_1$ of a finite graph of groups with coherent vertex groups and cyclic edge group is coherent.

Recall, that if $H \subseteq \pi_1 \mathcal{G}=G$ is fg then H has an induced graph of groups structure. $\mathcal{H} =H \backslash T_H \subseteq H \backslash T$ where T is the Bass-Serre tree of $\mathcal{G}$.

Topologically, $H$ defines a covering space $X^H \rightarrow X_g$, which inherits a graph of spaces structure with underlying graph $H\backslash T$.

$\Gamma' =H \backslash T_H \subseteq H \backslash T$ is a finite connected subgraph and the image of $\Gamma'$ in $X^H$ is a connected subcomplex $X'$ such that $\pi_1 X' = \pi_1 X^H =H$. $X'$ has the structure of a graph of space with underlying graph $\Gamma'$

There is an explicit algebraic description of $H\backslash T$, which follows immediately from Lemmas 16,17,18.

Lemma 23. (a) The vertices of $H\backslash T$ are in bijection with $\coprod_{v \in V(\Gamma)} H \backslash G/G_v=\coprod_{v \in V(\Gamma)}\{HgG_v\mid g \in G\}$.

The vertex $HgG_v$ is labelled by: $H \cap gG_vg^{-1}$, well-defined up to conjugation in $H$.

(b) The edges of $H\backslash T$ are in bijection with $\coprod_{e \in E(\Gamma)} H \backslash G/G_e$. The vertex $HgG_e$ is labelled with $H \cap g G_e g ^{-1}$.

(c) The edges of $H\backslash T$ adjoining the vertex corresponding to $HgG_v$  are in bijection with $\coprod_{e~\mathrm{adjoining}~v} (g^{-1} H g \cap G_v ) \backslash G_v/G_e$.

Now we will try to understand this topologically. In particular, we will understand the edge maps of a covering space of a graph of spaces.

Let $f:X \rightarrow Y$ be a continuous map and $Y' \rightarrow Y$ a covering map. If $f':X \rightarrow Y'$ makes the diagram commutes, then $f'$ is a lift of $f$.

Lemma 24. Fix basepoint $x \in X$, $y=f(x) \in Y$, and $y' \in Y'$ mapping to $y$. There is a lift $f':X \rightarrow Y'$ with $f'(x)=y'$ if and only if $f_{\ast} \pi_1(X,x) \subseteq \pi_1(Y',y')$.  Furthermore, if this lift exists it is unique.

It may be impossible to lift at $y'$. But it is possible if we pass to a covering space. Eg., if $\tilde{X} \rightarrow X$ is the universal cover.  Intuitively, an elevation is a minimal lift.

Definition: Let $X,Y,Y', x,y,y'$ be as above. A based connected covering space $(X',x') \rightarrow (X,x)$ together with a based map $f':(X',x')\to (Y',y')$ such that

commutes is an elevation (of $f$ at $y'$) if whenever the diagram

commutes and $X' \rightarrow \bar{X}$ is a covering map of degree larger than 1 then the composition $(\bar{X},\bar{x}) \rightarrow (X,x) \rightarrow (Y,y)$ does not lift to $Y'$ at $y'$.

The unbased covering map $X' \rightarrow X$ or equivalently the conjugacy class of the subgroup $\pi_1(X') \subseteq \pi_1(X)$ is called the degree of the elevation $f'$.