Consider free groups $F(A)$ and $F(B)$ on finite generating sets $A,B$. Write $A$ and $B$ as the disjoint unions $A=A_1\sqcup A_2$ and $B=B_1\sqcup B_2$. We consider the free groups $F(A_i)$ and $F(B_i)$ naturally as subgroups of $F(A)$ and $F(B)$ respectively. Suppose $f:F(A)\to F(B)$ is a group homomorphism such that

  1. $f(F(A_1))\subseteq F(B_1)$ and $f|_{F(A_1)}:F(A_1)\to F(B_1)$ is injective,
  2. if $p_2:F(B)\to F(B_2)$ is canonical the projection, then $p_2\circ f|_{F(A_2)}:F(A_2)\to F(B_2)$ is injective.

Must $f$ be injective?

The answer is certainly affirmative in some simple cases. However, in general, $f$ may map a generator $a\in A_2$ to a word in $F(B)$ having many letters from $B_1$ and so it is not immediately clear to me how to prove this using explicit word cancellations. It is possible I am missing something simple here or am just unaware of some free group theory that can address this question.


1 Answer 1


This seems to be true and, like many facts about free groups, can be proved using Stallings' famous folding technique. Here's a sketch of the argument.

Think of $F(B)$ as the fundamental group of a bouquet of circles $Y=Y_1\vee Y_2$, where $F(B_i)=\pi_1(Y_i)$. The map $f$ can now be realised as a morphism of graphs $\phi:X\to Y$, where $X$ also decomposes as a wedge $X_1\vee X_2$, with $F(A_i)=\pi_1(X_i)$. Since $f|_{F(A_i)}$ is injective for both $i$, the restriction of $\phi$ to each $X_i$ can be taken to be an immersion. The image $f(F(A))$ is the fundamental group of the graph $\overline{X}$ obtained by folding the map $\phi$. Since $\phi|_{X_1}$ is an immersion, $\phi$ embeds $X_1$ into $\overline{X}$, and we may therefore identify $X_1$ with its image in $\overline{X}$.

We now need to interpret hypothesis 2. Let $Z$ be the graph obtained by collapsing each edge in $\overline{X}$ that maps to $Y_1$, and let $\psi:Z\to Y_2$ be the map induced by $\phi$. The image of $p_2\circ f|_{F(A_2)}$ is identified with the fundamental group of $\overline{Z}$, the graph obtained by folding the map $\psi$.

The second hypothesis now tells us that $b_1(\overline{Z})=b_1(X_2)$, where $b_1$ denotes first Betti number, i.e. the rank of the fundamental group.

We can now conclude by estimating first Betti numbers. Since $\overline{Z}$ is obtained by folding $\psi$, we have

$b_1(X_2)=b_1(\overline{Z})\leq b_1(Z)$ .

Since $Z$ was obtained from $\overline{X}$ by collapsing a subgraph that contained $X_1$, we have

$b_1(Z) + b_1(X_1)\leq b_1(\overline{X})$,

and since $\overline{X}$ was obtained by folding $\phi$, we have

$b_1(\overline{X})\leq b_1(X)$ .

Putting these together gives

$b_1(X)=b_1(X_1)+b_1(X_2)\leq b_1(\overline{X})\leq b_1(X)$

so $b_1(\overline{X})= b_1(X)$. This tells us that the rank of the image of $f$ is equal to the rank of $F(A)$. Since free groups are Hopfian, it follows that $f$ is injective, as required.

  • $\begingroup$ Thank you. I am trying to unpack some of this and I cannot seem to find what "folding" of a map means (though I understand now what folding of a graph is). $\endgroup$
    – J.K.T.
    Nov 18, 2021 at 16:24
  • $\begingroup$ @J.K.T. The background for this is Stallings’ 1983 paper “Topology of finite graphs”, which underlies the modern point of view on free groups. In §3.3 of that paper, Stallings observes that any map of finite graphs $f:\Gamma\to\Delta$ factors uniquely as $\Gamma\to\overline{\Gamma}\to\Delta$, where $\Gamma\to\overline{\Gamma}$ is a finite composition of folds and $\bar{f}:\overline{\Gamma}\to\Delta$ is an immersion. So when I write, say, that $\overline{X}$ is obtained by folding the map $\phi$, I mean that $\overline{X}$ is the middle term of this canonical factorisation of $\phi$. $\endgroup$
    – HJRW
    Nov 18, 2021 at 16:32

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