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I already have asked a question to the following article:

There appeared two further questions to it (the page and theorem numbers refer to the preprint version). Both questions refer to page 28 (proof of theorem 4.2). The general assumption is that $L$ is a non abelian limit group over a free group (of rank at least 2). But I have the same two questions also for the situation that L is not a limit group but a non-elementary hyperbolic group (page 30: proof of proposition 4.3).

  1. On page 28 we have the following sequence of proper epimorphisms:

$L_1 = L * F_t \to L_2 = L * F_{t-1} \to ... \to L_{t-1} = L * F_2 \to L_t = L*\mathbb{Z}\to L_{t+1} = L$

Fujiwara and Sela say that for every index $i$, $1 \leq i \leq t$ there is an approximating sequence of epimorphisms $v_n^i: L_i \to L_{i+1}$ that converges into the limit group $L_i$. Why does such a sequence exist? Maybe the answers of @HJRW to my other question could be helpfull:

Question to limit groups (over free groups)

  1. Further down on page 28 we have the following:

The sequence of epimorphisms $(v_n^1: L_1 \to L_2)_{n \in \mathbb{N}}$ conveges into $L_1$. Hence, by proposition 2.3 $\lim \limits_{n \to \infty} \mathrm{e}(L_2,v_n^1(S_1)) = \mathrm{e}(L_1,S_1)$. The maps $v_n^1$ are proper epimorphisms, and they converge into $L_1$. Hence, the pairs $(L_2,v_n^1(S_1))_{n \in \mathbb{N}}$ belong to infinitely many distinct isomorphism classes of pairs (of a limit group and its finite set of generators).

My question is why it is not possible that there are only finitely many distinct isomorphims classes? Somehow Fujiwara and Sela give already an answer. They say that this is the case because "The maps $v_n^1$ are proper epimorphisms, and they converge into $L_1$." But I don't see yet why the claim follows from this.

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    $\begingroup$ In your first question, I don't understand whether the claimed sequence $v^i_n$ is supposed to have anything to do with the initially given sequence of epimorphisms. If not, then you can just use my previous answer and take a free product with the identity map on $L$. $\endgroup$
    – HJRW
    Commented Sep 7, 2023 at 14:38
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    $\begingroup$ For the second question, if there were only a finite set then the convergent sequence would be eventually constant, and so would converge to $(L_2,v^1_n(S_1))$ for some $n$. Since the limit is also $(L_1,S_1)$, this means that $v^1_n$ is an isomorphism, which contradicts the claim that it's a proper epimorphism. $\endgroup$
    – HJRW
    Commented Sep 7, 2023 at 14:42
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    $\begingroup$ For 1: You're right, you need a more subtle construction to get the final two maps in the sequence. For instance, you need to understand why $L*\mathbb{Z}$ is fully residually $L$. This is a nice exercise you should definitely try to do for yourself. $\endgroup$
    – HJRW
    Commented Sep 8, 2023 at 12:35
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    $\begingroup$ For 2: I don't have the paper in front of me, so don't know what Propositions 2.3 or 3.2 say. But the argument looks right. $\endgroup$
    – HJRW
    Commented Sep 8, 2023 at 12:37
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    $\begingroup$ Yes, that’s right. $\endgroup$
    – HJRW
    Commented Oct 7, 2023 at 8:37

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