Our aim is to prove Howson’s theorem for free groups followings Stallings’ article, Topology of finite graphs; this note is more or less a sequel of How Stallings proved finitely-generated free groups are subgroup separable. The construction introduced here, the fiber product, will be also useful for the generalization of ideas of Stallings to special cube complexes (due to Wise).

Another interesting proof, based on regular languages, can be found in Meier’s book untitled Graphs, groups and trees.

Theorem: (Howson) Let $F$ be a free group of finite rank and $H,K$ be two finitely-generated subgroups. Then $H \cap K$ is also finitely-generated.

Let $\alpha : X \to Z$ and $\beta : Y \to Z$ be two immersions (ie. locally injective cellular maps) between graphs. We define the fiber product $X \otimes_Z Y$ as the graph whose vertices are $\{ (x,y) \in X \times Y \mid \alpha(x)= \beta(y) \}$

where two vertices $(x_1,y_1)$ and $(x_2,y_2)$ are linked by an edge if and only if $\alpha(x_1)=\beta(y_1)$ and $\alpha(x_2)=\beta(y_2)$ are linked by an edge in $Z$.

Notice that there are two obvious projections $p : X \otimes_Z Y \to X$, $q : X \otimes_Z Y \to Y$, and a natural map $\gamma : X \otimes_Z Y \to Z$ so that we have the following commutative diagram: Claim 1: $\gamma : X \otimes_Z Y \to Z$ is an immersion.

Proof. Let $(x_1,y_1)$ be a vertex of $X \otimes_Z Y$ and $e,f$ two edges starting from $(x_1,y_1)$; let $(x_2,y_2)$ and $(x_3,y_3)$ denote the ending points of $e$ and $f$. If $\gamma(e)=\gamma(f)$, then $\gamma(x_3,y_3)=\gamma(x_2,y_2)$ hence $(x_2,y_2)=(x_3,y_3)$. Therefore, if $e \neq f$ then $e$ and $f$ are two loops based at $(x_1,y_1)$, sent via $\gamma$ on two loops based at $\gamma(x_1,y_1)$. So $\gamma$ is locally injective. $\square$

Claim 2: $\gamma_* \pi_1( X \otimes_Z Y) = \alpha_* \pi_1(X) \cap \beta_* \pi_1(Y)$.

First, we need the following easy lemma:

Lemma: Let $\Gamma$ be a graph and $c_1,c_2$ be two reduced (ie. without round-trip) loops. If $c_1=c_2$ in $\pi_1(\Gamma)$ then $c_1=c_2$.

Sketch of proof. The result is clear if $\Gamma$ is a bouquet of circles, using the usual normal form for free groups. In fact, it is the only case to consider, since the quotient of $\Gamma$ by a maximal subtree is a bouquet of circles. $\square$

Proof of claim 2. Because $\alpha \circ p = \gamma = \beta \circ q$, the inclusion $\gamma_* \pi_1( X \otimes_Z Y) \subset \alpha_* \pi_1(X) \cap \beta_* \pi_1(Y)$

is clear. Conversely, let $c_0 \in \alpha_* \pi_1(X) \cap \beta_* \pi_1(Y)$, that is there exist $c_X \in \pi_1(X)$ and $c_Y \in \pi_1(Y)$ such that $\alpha(c_X)=c_0=\beta(c_Y)$ in $\pi_1(Z)$. Of course, $c_X$ and $c_Y$ may be chosen reduced so that $\alpha(c_X)$ and $\beta(c_Y)$ so are. We deduce that $\alpha(c_X)=\beta(c_Y)$ from our previous lemma. Therefore, by construction of $X \otimes_Z Y$, there exist a loop $c \in \pi_1(X \otimes_Z Y)$ such that $\gamma(c)=c_0$, hence the inclusion $\alpha_* \pi_1(X) \cap \beta_* \pi_1(Y) \subset \gamma_* \pi_1 (X \otimes_Z Y)$. $\square$

Proof of Howson’s theorem. As in the previous note, we see $F$ as the fundamental group of a finite bouquet of circles $B$, and we choose immersions of finite graphs $\alpha : X \to B$ and $\beta : Y \to B$ such that $\alpha_* \pi_1(X)=H$ and $\beta_* \pi_1(Y)= K$. Then the fiber product $\gamma : X \otimes_B Y \to B$ gives an immersion such that $\gamma_* \pi_1(X \otimes_B Y) = H \cap K$; because $X \otimes_B Y$ is a finite graph (since $X$ and $Y$ are itself finite), we deduce that $H \cap K$ is finitely-generated. $\square$