For $n \in\mathbb{N}$ define:


* $X\_n=\{x\_1,\ldots,x\_n\}$,
* $F(X\_n)$ the free group on $X\_n$,
* $\varphi:F(X\_n)\to F(X\_{n-1})$ an epimorphism defined by $x\_i\stackrel{\varphi}{\mapsto} x\_i$ for $1\le i\le n-1$ and $x\_n\stackrel{\varphi}{\mapsto} 1$.


Is it true that for every IA automorphism $\alpha\in Aut(F(X\_n))$ the map $x\_i \mapsto \varphi(\alpha(x\_i))$ for $1\le i\le n-1$ defines an element in $Aut(F(X\_{n-1}))$?



Property of IA automorphisms of free groups
No, consider the map $F\_3 \to F\_3$ given by
$$
x\_1 \mapsto x\_1, x\_2 \mapsto x\_2 [x\_1, x\_3[x\_2, x\_1]], 
x\_3 \mapsto x\_3[x\_2, x\_1].
$$
It is an [IA automorphism](http://groupprops.subwiki.org/wiki/IA-automorphism): it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x\_1, x\_3[x\_2, x\_1], x\_2 [x\_1, x\_3[x\_2, x\_1]] \}$ generates and $F\_3$ is [Hopfian](http://en.wikipedia.org/wiki/Hopfian_group).


Further, if you kill $x\_3$ (that is, $n = 3$ in the OP's question), then the resulting map under consideration in the question is:
$$
x\_1 \mapsto x\_1, x\_2 \mapsto x\_2 [x\_1, [x\_2, x\_1]],
$$
which is clearly not an isomorphism (the word $x\_2 [x\_1, [x\_2, x\_1]] = x\_2x\_1^{-1} (x\_2^{-1} x\_1^{-1} x\_2 x\_1) x\_1 (x\_1^{-1} x\_2^{-1} x\_1 x\_2)$ in reduced form begins and ends with $x\_2$).


