# Dropping rank of IA automorphisms

Is there a natural way to map a given IA automorphism $\alpha\in Aut(F(X_n))$ to $Aut(F(X_{n-1}))$?

Think about braids. A pure braid on $n$ strands can be naturally mapped to a braid on $n-1$ strands by pulling out the $n$th strand. Can a similar transformation be done on IA automorphisms?

I would say no, as such a map would not be induced by a single map $F_n \to F_{n-1}$. Indeed, any map $F_n \to F_{n-1}$ is, by a Theorem 3.3, p. 132, in Magnus-Karrass-Solitar's Combinatorial Group Theory, essentially just the map $\varphi : F_n \to F_{n-1}$ got by killing a coordinate in $F_n$ (that is, by selecting a different free generating set for $F_n$, we may suppose the map is precisely $\varphi$ on these new generators). Thus, one can use the example in my previous answer to give an IA automorphism whose image is not an isomorphism.
Thus, we cannot hope that anything very close to the Braid group case works here. However, it may be possible to show that for any IA automorphism, $f$, there is some map $F_n \to F_{n-1}$, depending on $f$, that in turn sends $f$ to an automorphism, but I'm not sure if I'd call this natural.