Words which are not inverted by any endomorphism

Question

Let $w$ be a word in a free group $F_2$ of two generators $x_1, x_2$ such that there does not exist any endomorphism of free group which takes $w$ to $w^{-1}$. Let $w_1, w_2$ be two words in the same free group such that they generate rank $2$ free subgroup. Let $\phi$ be an endomorphism of $F_2$ generated by sending $x_i$ to $w_i$. I want to prove that for the word $u = \phi(w)$, there does not exist any endomorphism of $F_2$ inverting $u$.

I have tried but did not get anywhere. This is what I have tried: Suppose there exists an endomorphism $\psi$ of $F_2$such that $\psi(u) = u^{-1}$ but $\psi \circ \phi (w) = \phi(w^{-1}).$ $\phi(F_2)$ has rank 2 and $\psi$ inverts $\phi(w).$ By Sela and Kharlampovich-Myasnikov, non abelian free group have the same elementary theory. And it may happen that $\psi$ may not be invariant for $\phi(F_2).$ Certainly rank of $\psi(F_2) \cap \phi(F_2)$ is either 1 or 2. In both cases, I am stuck.

Different methods are also appreciated. Thanks in advance!

Answer 1

This partial answer just covers the case of $\psi$ non-surjective, i.e. not an automorphism.

Just to set some notation: Define $A=\langle w_1, w_2\rangle=\operatorname{im}(\phi)\leq F_2$. We will show that if $\psi(u)=u^{-1}$ then there is an endomorphism $\hat{\phi}:A\to A$ such that $\hat\phi(u)=u^{-1}$. The result (for non-surjective endomorphisms) follows by contradiction.

Now, a (proper) retract of a group $G$ is a (proper) subgroup $H\leq G$ such that there exists an endomorphism $\mu:G\to G$, called the retraction map, such that $\operatorname{im}(\mu)=H$ and $\mu(h)=h$ for all $h\in H$. (Equivalently, $G$ decomposes as $N\rtimes H$.) For example, free factors are retracts.

A result of Turner (from Test words for automorphisms of free groups, Bull Lond. Math. Soc. 1996) says that if a non-surjective endomorphism $\varphi: F\to F$ of the free group $F$ fixes a word $v\in F$, then $v$ is contained in a proper retract $H$ of $F$. (If $\varphi$ is injective, $H$ is infact a free factor of $F$.)

Now, suppose that $\psi:F_2\to F_2$ is non-surjective such that $\psi(u)=u^{-1}$. Then $\psi^2(u)=u$, so $u$ is contained in a proper retract $H$ of $F_2$. Set $K=H\cap A$, and note $u\in K$ and also that $\mu|_A:A\to A$ is a retraction map for $A$, with image $K$. As $H$ is a proper retract of the free group of rank $2$, it is cyclic. Hence, $K$ is cyclic and so the map $\iota: K\to K$ defined by $\iota(k)=k^{-1}$ is an automorphism of $K$. Therefore, the map $\hat\psi:A\to A$ defined by $a\mapsto \iota\circ \mu|_A(a)$ is an endomorphism of $A$ with the required property that $\hat\psi(u)=u^{-1}$.