Suppose $X$ is a set and $F$ is a free object on $X$ (with $i:X\rightarrow F$) in the category of groups. Prove that $i(X)$ is a set of generator for the group $F$.
I have the following hint: If $G$ is the subgroup of $F$ generated by $i(X)$, then there is a homomorphims $\phi:F\rightarrow G$ such that $\phi\circ i=i$. Show that $F\overset \phi \rightarrow G\overset\subseteq\rightarrow F$ is the identity map.
I can prove that for every $x\in F$ there exist a $\bar x\in G$ such that $\phi(x)=\phi(\bar x)=\bar x$ but I need to get $x=\bar x$ or something like that.
Thanks!
By the universal property of being "free", there is at most one homomorphism $f\colon F \to F$, such that $i \circ f = i$ (any set map $X \to H$ into an arbitrary group [here $H = F$] extends to an unique homomorphism). You have two maps with this property, namely ${\rm id}_F$ and ${\rm i}_{G \subseteq F} \circ \phi$. So this two maps must be equal, hence ${\rm i}_{G\subseteq F} \circ \phi = {\rm id}_F$.