The Question : Let $F$ be a group and let $j : X \rightarrow F $ be a mapping.
Assume that for every mapping $f$ of $X$ into a group $G$, there is a homomorphism $\phi$ of F into G unique such that $f = ϕ ◦ j$ . Show that $F_{X} \simeq F $.
My Attempt : I used the universal property : Let $η : X \rightarrow F_{X}$ be the canonical injection. For every mapping $f$ of $X$ into a group $G$, there is a homomorphism $ϕ'$ of $F_{X}$ into $G$ unique such that $f = ϕ' ◦ η$, namely $ϕ' (a_{1}, a_{2}, . . ., a_{n}) = f (a_{1}) f (a_{2}) · · · f (a_{n})$.
I searched a homomorphism , I found this one $\zeta : F_{X} \rightarrow F$ such that $\zeta(())=0_{F} $ (The image of the empty word) and $ \zeta(a_{1}, a_{2}, . . ., a_{n}) = j(a_{1}) j (a_{2}) · · · j (a_{n}) $ . I couldn't find the kernel of this homomorphism. if $Ker \zeta=\{0\}$ then $F_{X} \simeq F $. Otherwise finding another homomorphism will be a key to solve this problem.
Thank you for reading my question !
If you know about the universal property of $F_X$, then you know that you can define a unique homomorphism $\phi\colon F_X\to F$ such that, for every $x\in X$, $$ \phi(x)=j(x) $$ On the other hand, the property of $F$ allows to build a unique homomorphism $\psi\colon F\to F_X$ such that $\eta=\psi\circ j$.
Now, for $x\in X$, $$ \psi\circ\phi(x)=\psi(\phi(x))=\psi(j(x))=\psi\circ j(x)=\eta(x) $$ Hence, by uniqueness, $\psi\circ\phi$ is the identity on $F_X$.
Similarly, $$ \phi\circ\psi(j(x))=\phi(\eta(x))=\phi(x)=j(x) $$ By uniqueness, $\phi\circ\psi$ must be the identity on $F$.