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 $\text{Aut}(F\_n)$.



Does the injection $\text{Aut}(F_n) \hookrightarrow \text{Aut}(F_{n+1})$ split?
[Bridson and Vogtmann](http://arxiv.org/abs/math/0209191) 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.'


