I asked a question on MathOverflow (https://mathoverflow.net/q/454012/513011) where the following lemma appeared:
Folklore lemma: Let $S$ be a finite subset of the free group $F_n$ of finite rank $n$. If $n>2$ then there is an epimorphism $\psi:F_n\to F_{n-1}$ that is injective on $S$.
I want to prove this lemma now but I have some problems with that. I have the following hints for the proof:
As you can see already from the question that I asked on MathOverflow, the proof of this lemma might have to do something with the theory of limit groups/fully residually free groups. Here is a good reference for that if you don't know anything about that topic:
An Introduction to Limit Groups by Wilton (pdf).
In particular we can see from Example 2.2 that free groups are fully residually free, so there is always a homomorphism from $F_n\to F_{n-1}$ that is injective on $S$. But we need an epimorphism.
But maybe this theory is not needed at all and there exists an elementary proof for that.
Let $F_n = F\{x_1, \dots, x_n\}$. For $w \in F_{n-1}$, define $\psi_w:F_n \to F_{n-1}$ by $x_i \mapsto x_i$ for $i = 1, \dots, n-1$ and $x_n\mapsto w$. Obviously $\psi_w$ is an epimorphism for any $w$, since the image of $\psi_w$ contains all the generators $x_1, \dots, x_{n-1}$. Therefore it suffices to prove the following.
Claim 1: Let $n > 2$. For any finite subset $S \subset F_n$ we can choose $w \in F_{n-1}$ so that $\psi_w$ is injective on $S$.
The following claim is obviously equivalent (apply it to $S' = \{uv^{-1} : u, v \in S, u \ne v\}$).
Claim 2: Let $n > 2$. For any finite subset $S \subset F_n \setminus \{1\}$ we can choose $w \in F_{n-1}$ so that $\psi_w(v) \ne 1$ for each $v \in S$.
There must be many ways to prove this claim. I will outline two.
Method 1: (probabilistic group theory) This method appeals to my own proclivities, so I find it intuitive -- maybe you will too. It suffices to know a sequence of finite groups $G_N$ with the following two properties:
(1) A random homomorphism $F_2 \to G_N$ is surjective with probability tending to $1$ as $N \to \infty$.
(2) For any fixed nontrivial word $v \in F_n$, a random homomorphism $F_n \to G_N$ is nontrivial on $v$ with probability tending to $1$ as $N \to \infty$.
Suppose we have such a sequence of groups. Then it follows that, for $N$ large enough, almost all homomorphisms $\phi_N : F_n \to G_N$ have the property that $\phi_N(F_2)=G_N$ and $\phi_N(v) \ne 1$ for each $v \in S$. Choose any $\phi_N : F_n \to G_N$ with this property. Since $\phi_N(F_2)=G_N$, we can choose $w \in F_2$ such that $\phi_N(x_n) = \phi_N(w)$. Then $\phi_N = \phi_N \circ \psi_w$. Since $\phi_N$ is nontrivial on $S$ it follows that $\psi_w$ is nontrivial on $S$.
Examples of such sequences of finite groups include for example $\mathrm{Alt}(N)$ and $\mathrm{SL}_2(p)$. The properties (1) and (2) are reasonably well-known for these groups, though not trivial. Property (1) for $\mathrm{Alt}(N)$ is called Dixon's theorem (1969). Property (1) for $\mathrm{SL}_2(p)$ is related to something called Dickson's theorem (1901). This is a coincidence.
Method 2: (combinatorial group theory) This is admittedly easier overall, but a bit fiddly. Forget finite groups. We claim we can just set $w = x_1 x_2^N$ for any sufficiently large integer $N$ (many other choices are also reasonable). For any $v \in S$ we have some reduced expression $$v = x_{i_1}^{e_1} \cdots x_{i_k}^{e_k},$$ where $k$ (the number of "syllables") is positive, each factor $x_{i_j}^{e_j}$ is a nontrivial element of $\langle x_{i_j}\rangle$, and $i_j \ne i_{j+1}$ for $1 \le j < k$. Now if we apply $\psi_w$ then we replace all instances of $x_n$ with $w$. The result may not be reduced but the cancellation comes from neighboring factors like $x_n x_2^{-1}$ or $x_1^{-1} x_n$ or their inverses. Now one can argue that there only finitely many choices for $N$ (depending on $v$) for which $\psi_w(v)$ has fewer syllables than $v$. Since this holds for each $v \in S$, it follows that there are only finitely many choices for $N$ such that $\psi_w(v) = 1$ for some $v \in S$.