Let $F_n$ be the free group on $n$ letters.

The question is as in the title: letting $i:\text{Aut}(F_n) \hookrightarrow \text{Aut}(F_{n+1})$ be the natural injection, does there exist a homomorphism $\phi: \text{Aut}(F_{n+1}) \rightarrow \text{Aut}(F_n)$ such that $\phi \circ i = \text{id}$? My guess is "no", but I have no idea how to prove it.

As "evidence", here are two similar situations in other analogous contexts:

  1. Let $S_n$ be the symmetric group on $n$ letters. Then the injection $S_n \hookrightarrow S_{n+1}$ does not split (at least for $n$ sufficiently large). This follows easily from the simplicity of the alternating group.

  2. The injection $\text{SL}(n,\mathbb{Z}) \hookrightarrow \text{SL}(n+1,\mathbb{Z})$ does not split. One way to see this is to use the fact that for $n \geq 3$, all normal subgroups of $\text{SL}(n,\mathbb{Z})$ are either finite or finite-index (this is a consequence of the Margulis normal subgroup theorem, but it can be proved in more elementary ways as well; I do not know who to attribute it to).

The above two proofs work because we understand normal subgroups of $S_n$ and $\text{SL}(n,\mathbb{Z})$. Such an understanding seems entirely out of reach for $\text{Aut}(F_n)$.


Bridson and Vogtmann proved a much stronger result. From the abstract: 'If $m$ is less than $n$ then [the image of] a homomorphism $\mathrm{Aut}(F_n)\to\mathrm{Aut}(F_m)$ can have cardinality at most 2.'

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    $\begingroup$ Theorem A is clearer: "If $n\geq 3$ and $n>m$, then every homomorphism $\mathrm{Aut}(F_n)\to\mathrm{Aut}(F_m)$ is either trivial, or has image of order $2$." $\endgroup$ – Arturo Magidin May 19 '16 at 20:40
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    $\begingroup$ @ArturoMagidin, other than the slightly unfortunate reference to the cardinality of a homomorphism, I don't see the difference. The result is, after all, also true when $n=2$. $\endgroup$ – HJRW May 19 '16 at 20:52
  • $\begingroup$ @HJRW: Yes, I meant just to clarify what "homomorphism [has] cardinality at most 2" meant. I was confused by it on reading your quote (which is verbatim from the abstract, but still unfortunate). Sorry for not being clear. $\endgroup$ – Arturo Magidin May 20 '16 at 2:22

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