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Let $F_n := \langle x_1,\dotsc,x_n\rangle$ be the free group on $n$ generators. Let $w_1,\dotsc,w_n\in F_n$ and consider the endomorphism $\varphi:F_n\to F_n, x_i\mapsto w_ix_iw_i^-$.

  1. I conjecture that $\varphi$ is always injective. Is this true?
  2. In some cases, $\varphi$ is not surjective, e.g. for $w_1=x_1x_2$. Can we say in general in which cases $\varphi$ is surjective? Maybe if $w_i$ does not contain $x_i$?
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  • $\begingroup$ Am I missing something: isn't conjugation always an automorphism? $\endgroup$ – Aniruddh Agarwal May 1 '19 at 16:21
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    $\begingroup$ @AniruddhAgarwal: They are conjugating each generator by a different word. $\endgroup$ – Arturo Magidin May 1 '19 at 16:31
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    $\begingroup$ Yes it's injective, because the image in the abelianization of $F_n$ of $\varphi(F_n)$ is all of $\mathbf{Z}^n$, and hence $\varphi(F_n)$ has rank $\ge n$, and every surjective endomorphism of $F_n$ is injective. $\endgroup$ – YCor May 1 '19 at 16:49
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1. About injectivity: Yves already answered about injectivity in the comments. Below is an alternative argument which works more generally for free products:

Proposition 1: Let $G=A_1 \ast \cdots \ast A_n$ be a free products and let $w_1, \ldots, w_n \in G$ be $n$ elements. Then $$H:=\langle w_1A_1w_1, \ldots, w_nA_nw_n^{-1} \rangle = w_1A_1w_1^{-1} \ast \cdots \ast w_nA_nw_n^{-1}.$$

Sketch of proof. Think of $G$ as the fundamental group of the graph of groups having a central vertex labelled by $\{1\}$ linked to $n$ vertices labelled by $A_1, \ldots, A_n$ and whose edges are all labelled by $\{1\}$. Let $T$ denote the corresponding Bass-Serre tree.

Choose $z_1, \ldots, z_n \in G$ such that $H= \langle z_1A_1z_1^{-1}, \ldots, z_nA_nz_n^{-1} \rangle$ with $|z_1| + \cdots + |z_n|$ minimal. For every $1 \leq i \leq n$, let $v_i$ denote the vertex of $T$ stabilised by $z_i A_i z_i^{-1}$. Also, let $v$ denote the vertex of $T$ corresponding to the coset $\{1\}$.

I claim that, for every $i \neq j$, $v_i$ does not belong to $[v_j,v]$. Otherwise, take a $z \in \mathrm{stab}(v_i)$ sending the first edge of $[v_i,v_j]$ to the first edge of $[v_i,v]$. Then the length of $zz_j$ is smaller than the length of $z_j$ because $z \cdot v_j$ is closer to $v$ than $v_j$. This contradicts the choice of the $z_i$'s.

As a consequence, there exists a subtree $S \subset T$ whose leaves are the $v_i$'s. Now a ping-pong argument leads to the desired conclusion. $\square$

2. About surjectivity: Given an $n$-tuple $(a_1, \ldots, a_n)$ of words written over the generators (and their inverses) of $\mathbb{F}_n$, consider the following elementary transformations:

  1. if $a_i$ is not reduced, then reduce it;
  2. if $x_i^{\pm 1}$ is the last letter of $a_i$, remove it;
  3. if $a_jx_j^{\pm 1}$ is a prefix of $a_i$, remove from $a_i$ the last letter of this prefix.

Notice that the total length $|a_1|+ \cdots +|a_n|$ decreases when an elementary transformation is applied. Call $(a_1, \ldots, a_n)$ a reduction of $(w_1, \ldots, w_n)$ if $(a_1, \ldots, a_n)$ is obtained from $(w_1, \ldots, w_n)$ by applying elementary transformations until it is no longer possible.

Proposition 2: Let $(a_1, \ldots, a_n)$ a reduction of $(w_1, \ldots, w_n)$. Then $\varphi$ is surjective if and only if $a_1= \cdots a_n=1$.

Sketch of proof. Let $\Gamma_0$ be a bouquet of balloons, such that each wire is an oriented path of length $|w_i|$ labelled by $w_i$ and its corresponding balloon is a single oriented edge labelled by $x_i$. Let $\Gamma_1, \ldots, \Gamma_k$ be a sequence of graphs such that $\Gamma_i$ is obtained from $\Gamma_{i-1}$ by applying a Stallings' fold, and such that no such a fold applies to $\Gamma_k$. Notice that $\Gamma_k$ looks like a bouquet of balloons in which the wires are glued together like a rooted tree. Let $a_1, \ldots, a_n$ denote the words labelling the oriented paths from the root to the balloons.

By looking at how a Stallings' fold works, we find that $(a_1, \ldots, a_n)$ is a reduction of $(w_1, \ldots, w_n)$.

If $\Gamma$ is a bouquet of $n$ circles, because no Stallings' fold applies to $\Gamma_k$, the canonical map $\Gamma_k \to \Gamma$ is a local isometry*. As shown by Stallings, it is possible to add edges to $\Gamma_k$ in order to get a new graph $\Psi$ such that the local isometry $\Gamma_k \to \Gamma$ extends to a covering map $\Psi \to \Gamma$. Consequently, $\varphi$ is surjective if and only if $\Psi = \Gamma_k$, which amounts to saying that the tree-like part of $\Gamma_k$ is empty; in other words $a_1= \cdots = a_n=1$. $\square$

Below is an illustration of the method for $\mathbb{F}_2= \langle a,b \rangle$, $w_1 = ab$ and $w_2=a$:

enter image description here

*As a consequence, $\Gamma_k \to \Gamma$ is $\pi_1$-injective, whence a third argument showing the injectivity of $\varphi$.

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