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Definition. A group $G$ splits freely if $G$ acts on a tree $T$ without global fixed point and such that every edge stailizer is trivial. If $G$ does not split freely, then $G$ is called freely indecomposable.

Example. $\mathbb Z=\pi_1(S^1)$. Equivalently, $\mathbb Z$ acts on $\mathbb R$ without global fixed points. So $\mathbb Z$ splits freely.

If $G \ncong \mathbb Z$ but $G$ splits freely, then $G=G_1 \ast G_2$ for $G_1, G_2 neq 1$.

Definition. The rank of $G$ is the minimal $r$ such that $F_r$ surjects $G$.

It is clear that $rank(G_1\ast G_2)\leq rank(G_1)+rank(G_2)$.

Grushko’s Lemma. Suppose $\varphi:F_r \longrightarrow G$ is surjective and $r$ is minimal. If $G=G_1 \ast G_2$, then $F_r=F_1 \ast F_2$ such that $\varphi(F_i)=G_i$ for $i=1,2$.

Pf. Let $X_i=K(G_i,1) (i=1,2)$ be simplicial and let $\mathfrak{X}$ be a graph of spaces with vertex spaces $X_1, X_2$ and edge space a point. So $G=\pi_1(X_{\mathfrak{X}}, x_0)$ where $x_0=(*, \frac{1}{2})$.

Let $\Gamma$ be a graph so that $\pi_1(\Gamma)\cong F_r$ and realize $\varphi$ as a simplicial map $f: \Gamma \longrightarrow X_{\mathfrak{X}}$. Let $y_0 \in f^{-1}(x_0)$. Because $r$ is minimal, $f^{-1}(x_0)$ is a forest, contained in $\Gamma$. The goal is to modify $f$ by a homotopy to reduce the number of connected components of $f^{-1}(x_0)$.

Let $U \subseteq f^{-1}(x_0)$ be the component that contains $y_0$. Let $V \subseteq f^{-1}(x_0)$ be some other component. Let $\alpha$ a path in $\Gamma$ from $y_0$ to $V$.

Look at $f \circ \alpha \in \pi_1(X_{\mathfrak{X}}, x_0)$. Because $\varphi$ is surjective, there exists $\gamma\in \pi_1(\Gamma, y_0)$ such that $f \circ \gamma = f \circ \alpha$. Therefore if $\beta= \gamma^{-1} \cdot \alpha$, then $f \circ \beta$ is null-homotopic in $X_{\mathfrak{X}}$ and $\beta$ gives a path from $y_0$ to $V$.

We can write $\beta$ as a concaternation as $\beta=\beta_1 \cdot \beta_2 \cdot \cdot \cdot \beta_n$ such that for each $i$, $f \circ \beta_i \subseteq X_{\mathfrak{X}}\smallsetminus {x_0}$. By the Normal Form Theorem, there exists $i$ such that $f\circ \beta_i$ is null-homotopic in $X$.

We can now modify $f$ by a homotopy so that $im (f\circ\beta_i)={x_0}$. Therefore $\beta_i \subseteq f^{-1}(x_0)$ and the number of components of $f^{-1}(x_0)$ has gone down. By induction, we can choose $f$ so that $f^{-1}(x_0)$ is a tree. Now $f$ factors through $\Gamma'=\Gamma/ f^{-1}(x_0)$. Then $F_r\cong \pi_1(\Gamma')$ and there is a unique vertex of $\Gamma'$ that maps to $x_0$. So every simple loop in $\Gamma'$ is either contained in $X_1$ or $X_2$ as required. $square$

An immediate consequence is that $rank(G_1\ast G_2)=rank (G_1) + rank (G_2)$.

Grushko’s Theorem. Let $G$ be finitely generated. Then $G\cong G_1 \ast \cdots\ast G_m \ast F_r$ where each $G_i$ is freely indecomposable and $F_r$ is free. Furthermore, the integers $m$ and $r$ are unique and the $G_i$ are unique up to conjugation and reordering.

Pf. Existence is an immediate corollary of the fact that rank is additive.

Suppose $G=H_1\ast \cdots \ast H_n \ast F_s$. Let $\mathcal{G}$ be the graph of groups. Let $T$ be the Bass-Serre tree of $\mathcal{G}$.

Consider the action of $G_i$ on $T$. Because $G_i$ is freely indecomposable, $G_i$ stabilize a vertex of $T$. Therefore $G_i$ is conjugate into some $H_i$.

Now consider the action of $F_r$ on $T$. $F_r\smallsetminus T$ is a graph of groups with underlying graph $\Delta$, say, and $\pi_1(\Delta)$ is a free factor in $F_r$. But there is a covering map $F_r\smallsetminus T \longrightarrow \mathcal{G}$ that induces a surjection $\pi_1(\Delta) \longrightarrow F_s$. Therefore, $r\geq s$. The other inequality can be obtained by switching $F_r$ and $F_s$. $\square$

As before, are $\chi'$ and $\chi$ are graphs of spaces equipped with maps $\Phi \colon X_{\chi'} \to X_{\chi}$, $\varXi' \to \varXi$, $\phi_{v'} \colon X_{v'} \to X_v$, and $\phi_{e'} \colon X_{e'} \to X_e$ such that

commutes.

Lemma 28: Suppose that every edge map of $\chi'$ is an elevation.  Then the map $\Phi$ is $\pi_1$-injective.

Proof: The idea is to add extra vertex spaces to $\chi'$ so that $\chi'$ satisfies Stalling’s condition.  As before, we have inclusions:

If $\chi'$ does not satisfy Stalling’s condition then one of these maps is not surjective.  Without loss of generality, suppose that

does not arise as an edge map of $\chi'$.  Suppose $\partial_{e}^{-} \colon X_e \to X_u$.  Then $(\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'}))$ is a subgroup of $\pi_1(X_u)$.  Let $X_{u'}$ be the corresponding covering space of $X_u$.  Since $\pi_1(X_{u'})$ contains $(\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'}))$, there is a lift $\partial_{e'} \colon X_{e'} \to X_{u'}$ of $\partial_{e'} \circ \phi_{e'}$.  Now replace $\chi'$ by $\chi' \cup X_{u'}$ and repeat.  After infinitely many repetitions, the result $\hat{\chi}$ satisfies the hypothesis of Stalling’s condition, and $\chi'$ is contained in a subgraph of spaces.

Theorem 18: If $G_{+}$ and $G_{-}$ are residually finite groups then $G_{+} * G_{-}$ is also residually finite.

Proof: Let $X_{\pm} = K(G_{\pm}, 1)$, let $e = *$ be a point, and fix maps $\partial^{\pm} \colon e \to X_{\pm}$.  This defines a graph of spaces $\chi$.  By the Seifert-van Kampen Theorem, $\pi_1(X_{\chi}) \cong G_{+} * G_{-}$.  Let $\tilde{\chi}$ be the graph of spaces structure on the universal cover of $X_{\chi}$, and let $\Delta \subseteq X_{\tilde{\chi}}$ be a compact subset.  We may assume that $\Delta$ is connected; we may also assume that if $\tilde{e} \in \tilde{\varXi}$ (Bass-Serre tree) and $\Delta \cap \tilde{e} \ne \emptyset$, then $\tilde{e} \subseteq \Delta$.  Let $\eta \colon X_{\tilde{\chi}} \to \tilde{\varXi}$ be the map to the underlying map of the Bass-Serre tree.  Then $\eta(\Delta)$ is a finite connected subgraph, $\varXi'$.  For each $v' \subseteq v(\varXi')$, let $\Delta_{v'} = \Delta \cap X_{v'}$, a compact subspace of $X_{v'}$.  Because $G_{\pm}$ are residually finite, we have a diagram

where $X_{\hat{v}} \to X_{\pm}$ is a finite-sheeted covering map and $\Delta_{v'}$ embeds in $X_{\hat{v}}$.  Let $\hat{\chi}$ be defined as follows.  Set $\hat{\varXi} = \varXi'$; for a vertex $\hat{v} \in \hat{\varXi}$, the vertex space is the $X_{\hat{v}}$ corresponding to $v'$.  For each $\hat{e} \in E(\hat{\varXi})$ corresponding to $e' \in E(\varXi')$, define $\partial_{\hat{e}}^{\pm}$ so that the diagram

commutes.  Now sum.

Exercise 26: If $G_1$ and $G_2$ are residually finite and $H$ is finite, prove that $G_1 *_H G_2$ is residually finite.

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 $G_1 * G_2$.