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I am reading a proof of the following statement:

Let $F_n$ be the free group of rank $n$, generated by a basis $\{x_1, \ldots, x_n\}$ and $\Phi$ an automorphism of $F_n$. Let $F_{n-1}$ be the subgroup generated by $\{x_1, \ldots, x_{n-1}\}$ and let $F_{n-1}$ be $\Phi$-invariant, then $\Phi(x_n)$ contains $x_n$ or $x_n^{-1}$ exactly once.

The statement is proven in Lemma 3.2.1 of "The Tits alternative for $\operatorname{Out}(F_n)$ I: Dynamics of exponentially-growing automorphisms"* by Bestvina, Feighn and Handel, but I do not completely understand the prove. I was wondering if anyone would know an other reference or could explain the prove in the reference I have.

*(link) M. Bestvina, M. Feighn and M. Handel, The Tits alternative for $\operatorname{Out}(F_n)$ I: Dynamics of exponentially-growing automorphisms, Annals of Mathematics, 151 (2000), 517-623

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  • $\begingroup$ $\Phi$ induces an automorphism on the abelianization which can be represented by an integer matrix. Since $F_{n-1}$ is invariant, the matrix has a row of all zeroes until the last entry, which must then be $\pm 1$. $\endgroup$ – Max Apr 30 '18 at 3:02
  • $\begingroup$ @Max: I know that we can represent the automorphism by an integer matrix, but if I recall this correct, the entries of the matrix are the sum of exponents each generator in the image of a generator. Hence the entry being $\pm 1$ does only tell us that the number of positive exponents and negative exponents differ by $1$... $\endgroup$ – Student Apr 30 '18 at 7:28
  • $\begingroup$ What about the proof do you not understand? Also, in the future, you should actually cite where it is in that paper. $\endgroup$ – user29123 May 8 '18 at 21:25
  • $\begingroup$ @PaulPlummer the immersion part: i only knew what an immersion is for manifolds... I think I understand the part when $f$ is an immersion, however the part where it is not is unclear to me (last part of the proof). $\endgroup$ – Student May 9 '18 at 7:10
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    $\begingroup$ Also make sure that you put relavent tags on your questions so that the people who can answer them can find them. If it is a question about this paper geometric group theory probably belongs there, also group theory probably wouldn't hurt. You probably wouldn't have needed a bounty to try to get more attention (I didn't find the question because of the bounty, so not sure if that would have helped either)... $\endgroup$ – user29123 May 9 '18 at 11:13
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Generic stuff on Stallings' foldings and immersions

A map of graphs $f:A\rightarrow B$ is an immersion if it is locally injective (injective at the vertices). An example of an immersion is if you take a square and pinch two opposite sides together to get a line with a loop at each end. Then this is not an injection, but it is an immersion.

The proof of this lemma is based on an really neat idea, due to John Stallings. The idea basically says that instead of looking at covers to understand subgroups of $\pi_1(A)$, $A$ a graph, we can look at immersions (that is, subgraphs). As $\pi_1(A)$ is free, this gives an elegant, and extremely important, way of looking at subgroups of free groups. The key phrase is ""Stallings' foldings". These are the "folds" mentioned in the paper (see also section 2.4 on p528/p12), and were introduced in the (extremely readable) paper "J. Stallings, Topology of finite graphs, Invent. Math. 71 (1983), 551-565" (reference [Sta] in your paper).

Stallings' proves that every map of graphs $f:A\rightarrow B$ factors as $A\rightarrow C\rightarrow B$ where $C\rightarrow B$ is an immersion and $A\rightarrow C$ is a "folding" map. He also proves that this folding map is unique, although the individual folding moves may not be.

Stuff specific to the paper

In the proof in this paper we are considering a map $f:A\rightarrow A$. This factors as folds $p:A\rightarrow C$ and an immersion $f':C\rightarrow A$. Therefore, if the sentence "The only fold that can take place is between the initial and terminal ends of $e_n$" is true then the result follows from standard results about Stallings' foldings. So, (a) do you understand this sentence, and (b) do you understand that this is the key sentence and everything else is standard (for some value of "standard"...)?

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  • $\begingroup$ I think I understand. I'll have to read this more carerully, but hzve classes the whole day. Is it okay with you if I'll answr tomorrow? This way, i can look up the paper by stallings and reread the proof this question talks about (I have to check the details again and see if they make sense to me given your answer) $\endgroup$ – Student May 9 '18 at 8:51
  • $\begingroup$ @Student No problem. I am travelling tomorrow though so may not be as quick as you would like at responding. $\endgroup$ – user1729 May 9 '18 at 8:58
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    $\begingroup$ I had some spare time due to a class ending early. Thank you for pointing me towards this paper, it settles my problem. Many thanks! $\endgroup$ – Student May 9 '18 at 10:11
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    $\begingroup$ I believe that the answer to your last question is that if $f' : G' \to G$ is a homotopy equivalence of marked graphs, and if $f'$ is an immersion, then $f'$ is a homeomorphisms. To see why, if $f'$ were not surjective then it could not be a homotopy equivalence... $\endgroup$ – Lee Mosher May 11 '18 at 23:02
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    $\begingroup$ ... And if $f'$ were not injective then there would exist a connected graph $G''$, a proper inclusion $G' \hookrightarrow G''$ such that $G''$ deformation retracts to $G'$, and an extension $f'' : G'' \to G$ of $f' : G' \to G$, such that $f''$ is a nontrivial covering map. However, $f'$ being a homotopy equivalence implies that $f''$ is a homotopy equivalence, which contradicts $f''$ being a nontrivial covering map. $\endgroup$ – Lee Mosher May 11 '18 at 23:02

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