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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 $Aut(F(X_{n-1}))$?

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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: 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.

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 $x_2$).

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