please help me to solve the following problem.


Let $F$ be a non-abelian free group and $w(x)=1$ be an equation in one variable $x$ ($w(x)$ may contain elements of $F$ as constants). Clearly, one can consider $w(x)$ as an element of free product $F\ast \langle x\rangle$. 


**Suppose** $w(a)=1$ for all $a\in F$. 


**Is it true** that $w(x)$ equals $1$ in $F\ast \langle x\rangle$?


Do you know a simple proof? Probably, you can remember the papers which can be useful for this propblem?



Equations in free groups satisfying all elements
Yes. Denote by $a,b$ the first two free generators of $F$.


The case $w\in F\cup\langle x\rangle$ is clear. So we can suppose, after conjugation, that $w=u\_1x^{n\_1}\dots u\_kx^{n\_k}$ with $k\ge 1$, $u\_i\in F\smallsetminus\{1\}$, and $n\_i\in\mathbf{Z}\smallsetminus\{0\}$. 


Choose $n$ large enough such that the reduced form of $a^nu\_ia^{-n}$ starts with $a$ and finishes with $a^{-1}$ for all $u\_i$ that is not a power of $a$. In all cases, the reduced form $v\_i$ of $a^nu\_ia^{-n}$ starts and finishes with $a^{\pm 1}$. Then $q=\prod v\_ib^{n\_i}$ is a reduced form and hence $\neq 1$. We have
$$a^{-n}qa^n=a^{-n}\left(\prod a^nu\_ia^{-n}b^{n\_i}\right)a^n=\prod u\_ia^{-n}b^{n\_i}a^n=w(a^{-n}ba^n)\neq 1.$$


Added: This actually shows that the canonical map $F\ast\langle x\rangle\to F^F$, $w\mapsto (t\mapsto w(t))$ is injective. Indeed this is a group homomorphism ($F^F$ being a group for the target operation: infinite power of the group $F$ by the index set $F$). Of course this fails when $F$ has rank $r\le 1$: for $r=0$, $x$ is the kernel, while for $r=1$ $axa^{-1}x^{-1}$ is in the kernel. The above shows injectivity of the projection to $F^X$, where $X=\{a^nba^{-n}:n\in\mathbf{Z}\}.$


