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My question refers to the following article (to page 26: proof of Theorem 4.1):

In the article we find the statement that for a non-abelian limit group $L$ we always find a sequence of epimorphisms to $F_2$ that converges to $L$. But by definition of a (non-abelian) limit group we find -a priori- only a sequence of homomorphisms. Why can the homomorphisms (in this situation) always be chosen to be surjective?

From the article An Introduction to Limit Groups by Wilton (pdf) I could find out that limit groups are fully residually free and because $L$ is not abelian it has to contain a free group of rank at least 2. Maybe this helps somehow to prove the statement.

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You need to prove the following folklore lemma, which is well known to researchers in the field but perhaps not written down anywhere. The proof is a nice exercise.

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 $S$.

Note that the hypothesis that $n>2$ cannot be extended to $n\geq 2$, since any homomorphism $F_2\to F_1\cong\mathbb{Z}$ kills the commutator subgroup.

Using the lemma, it's not difficult to complete the claim. Let $S$ be a finite subset of the non-abelian limit group $L$; we need to find an epimorphism $L\to F_2$ that injects $S$.

By adding elements to $S$, $S$ can be taken to include both the identity and some non-trivial commutator (since $L$ is non-abelian). Let $f:L\to F_2$ be a homomorphism that injects $S$. The image of $f$ is some finitely generated, non-abelian free group $F_n$. By the lemma and induction on $n$, there is an epimorphism $\phi:F_n\to F_2$ that injects $f(S)$. Now $\phi\circ f$ is the required epimorphism.

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  • $\begingroup$ I think you want to consider $f:L \to F_n$ and not $f:L \to F_2$, right? But I don't see yet why this answeres my question. You constructed an epimorphism $L \to F_2$ that injects $S$. But at the end we want to have a stable sequence $(\varphi_n : L \to F_2)_{n \in \mathbb{N}}$ of epimorphisms (not only one epimorphism) that converges into $L$. $\endgroup$
    – TheMathematician
    Commented Sep 5, 2023 at 12:49
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    $\begingroup$ I'll address the two distinct parts separately. 1. In your question, you said you saw why you can consider homomorphisms to $F_2$. One may replace a homomorphism $f:L\to F_2$ by an epimorphism $f:L\to F_n$, just by replacing $F_2$ by the image of $f$. I let $f$ denote both of these maps. $\endgroup$
    – HJRW
    Commented Sep 5, 2023 at 12:56
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    $\begingroup$ 2. For your larger worry, look at Example 3.2 in the old notes of mine you refer to in your question. $\endgroup$
    – HJRW
    Commented Sep 5, 2023 at 12:57
  • $\begingroup$ Thank's for the answer, I think I understand now. Maybe one further question: Is the statement also true if we consider a limit group $L$ over a non elementary hyperbolic group $\Gamma$? In more detail: To have a limit group over a non elementary hyperbolic group means that we have a group $G$ and a stable sequence $(\varphi_n: G \to \Gamma)_{n \in \mathbb{N}}$ that converges into $L$. In this situation, does there exist a sequence of epimorphisms $(\psi_n: L \to \Gamma)_{n \in \mathbb{N}}$ that converges into $L$? $\endgroup$
    – TheMathematician
    Commented Sep 6, 2023 at 8:03
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    $\begingroup$ @TheMathematician: No. For instance, $F_2$ is a limit group over $F_3$, but there are no epimorphisms $F_2\to F_3$. And there are many further examples: any fg subgroup of $\Gamma$ that doesn’t surject $\Gamma$ provides an example. $\endgroup$
    – HJRW
    Commented Sep 6, 2023 at 8:23

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