I have a question that is related to the topic of limit groups:
Let $G$ and $H$ be finitely generated groups and let $(\varphi_n: G \to H)_{n \in \mathbb{N}}$ be a sequence of group epimorphisms. Does there exist a stable subsequence of $(\varphi_n)_{n \in \mathbb{N}}$?
This is the definition of "stable sequence":
Let G be a group and $(\varphi_n) _{n \in \mathbb{N}}⊂ Hom(G, H)$. The sequence $(\varphi_n) _{n \in \mathbb{N}}$ is stable if for any $g ∈ G$ either $\varphi_n(g) = 1$ for almost all $n$ or $\varphi_n(g)\neq 1$ for almost all $n$.
I conjecture that the answer to the question is positive, because the statement of the question is used on page 27 of the following article (in the article we have a sequence of epimorphisms $g_n: F_l \to L$ to which the statement of my question is applied to):
https://arxiv.org/pdf/2002.10278.pdf
Edit: In the above article we have the situation that G is a free group of rank at least two and H is a non-abelian limit group. So it would be sufficient if somebody can answer my question for this situation. Of course you can still say something to the general situation.
Edit: "Almost all" means that the statement ist true for all natural numbers with possibly a finite number of exceptions.
The answer is yes by a standard diagonal argument.$\def\phi{\varphi}$ We need only assume $G$ is countable (which is weaker than assuming it is finitely generated). Let $G = \{g_1, g_2, g_3, \dots\}$ be a countable group and let $(\varphi_n : G \to H)_{n \in \mathbf N}$ be a sequence of homomorphisms. There is a subsequence $(\varphi^{1}_n)$ of $(\varphi_n)$ such that $\varphi^1_n(g_1)$ is identically $1$ or identically $\ne 1$. There is a further subsequence $(\varphi^2_n)$ of $(\varphi^1_n)$ such that $\varphi^2_n(g_2)$ is identically $1$ or identically $\ne 1$. Continue, definining an infinite sequence of subsequences $(\phi^1_n), (\phi^2_n), (\phi^3_n), \dots$ such that, for all $i$, $(\phi^i_n)_n$ is a subequence of $(\phi^{i-1}_n)_n$ such that $\phi^i_n(g_i)$ is identically $1$ or identically $\ne 1$. Finally consider diagonal subsequence $\psi_n = \phi^n_n$ and observe that $\psi_n(g_i)$ is identically $1$ or identically $\ne 1$ for $n \ge i$.
This argument is perfectly analogous to the proof that $\{0,1\}^{\mathbf N}$ is compact. Actually the claim reduces to that fact. Define $f_n : \mathbf N \to \{0,1\}$ by $f_n(i) = 0 \iff \varphi_n(g_i) = 1$. Since $\{0,1\}^\mathbf N$ is compact (and metrizable), $(f_n)$ has a convergent subsequence, and if you unpack the definitions that is precisely what you want.