Sequence of epimorphisms of residually finite groups stabilizes Let $G_1 \to G_2 \to \cdots$ be a sequence of epimorphisms of finitely generated residually finite groups. Does it eventually stabilize? That is, are all but finitely many epimorphisms actually isomorphisms?
Note that finitely generated residually finite groups are Hopfian, so this excludes the simple counterexample of each $G_i$ being a fixed group and each epimorphisms being a fixed one onto itself.
The analogous result holds when the groups are residually free: this is Proposition 6.8 in Charpentier Guirardel "Limit groups as limits of free groups". The proof only uses the fact that residually free groups are residually $SL_2(\mathbb{C})$, and it seems that it can be adapted to the case where each $G_i$ is residually $GL_n(\mathbb{C})$ for a fixed $n$. It seems unlikely that this holds for general residaully finite groups: the Jordan-Schur Theorem implies that for a general finite group the minimal degree $n$ such that it embeds into $GL_n(\mathbb{C})$ can be arbitrarily large.
Is there another way to adapt the proof? Is there a counterexample?
 A: In the same vein as dodd's answer, a counterexample can also be deduced from the second Houghton group $H_2$, which is defined as the group of bijections $L^{(0)} \to L^{(0)}$ that preserves adjacency and non-adjacency for all but finitely pairs of vertices in the bi-infinite line $L$. A presentation of $H_2$ is
$$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle$$
where $t$ corresponds to a unit translation and $\sigma_i$ to the permutation $(i,i+1)$. Now, truncate the presentation and define $G_n$ via
$$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ n \geq |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle.$$
By using the relations $t\sigma_it^{-1}=\sigma_{i+1}$ in order to remove the generators $\sigma_0,\sigma_{-1},\ldots$ and $\sigma_{n+2},\sigma_{n+3},\ldots$, we find the following presentation of $G_n$:
$$\left\langle \sigma_1, \ldots, \sigma_{n+1}, t \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \\ t\sigma_it^{-1}= \sigma_{i+1}, \ 1 \leq i \leq n} \right. \right\rangle.$$
Observe from this presentation that $G_n$ decomposes as an HNN extension of
$$\left\langle \sigma_1,\ldots, \sigma_{n+1} \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \right. \right\rangle,$$
which turns out to be isomorphic to the symmetric group $\mathfrak{S}_{n+2}$, where the stable letter conjugates $\langle \sigma_1,\ldots, \sigma_n \rangle$ to $\langle \sigma_2, \ldots, \sigma_{n+1} \rangle$. Thus, as the HNN extension of a finite group, $G_n$ must be virtually free.
The conclusion is that the canonical quotient maps $G_1 \twoheadrightarrow G_2 \twoheadrightarrow \cdots$ defines a sequence of epimorphisms between virtually free groups that does not stabilise.
Remark: By reproducing the above argument almost word for word with the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$ instead of the Houghton group $H_2$ provides the same conclusion. The reason is that these groups have a similar structure: they are of the form $C \rtimes \mathbb{Z}$ for some locally finite Coxeter group $C$ where $\mathbb{Z}$ acts on $C$ via an isometry of the graph defining $C$. (Loosely speaking, all the other groups of this form can be recovered from $\mathbb{Z}_2 \wr \mathbb{Z}$ and $H_2$, so there is no other interesting examples in this direction.)
A: The answer is "no". The lamplighter group (which is infinitely presented) is a limit of a sequence of virtually free groups and surjective homomorphisms (see, for example, this question and answers there). All virtually free groups are residually finite.
