The definition of free group given in class was: Given a non-empty set $X$ and an inclusion $i:X\rightarrow F$. We say that $F$ is a free group on $X$ if, given any group $G$ and a set map $\varphi:X\rightarrow G$, there is a unique group-homomorphism $\Phi:F\rightarrow G$ such that $$ \Phi\circ i=\varphi.$$
Then, it was given an exercise to prove that $i(X)$ generates whole $F$, with use of uniqueness of $\Phi$.
But, when going into proof, I got stuck! If we assume that $i(X)$ generates a proper subgroup of $F$, then I was unable to think of different extension of $\Phi|_{i(X)}$.
Why I got stuck: Since even in family of finite groups, some examples can be easily found, where a homomorphism from a subgroup $H$ of $G$ into another group $G_0$ may have unique extension, or extension may not exist also. Then in free group, how can we assure of getting other extensions if $i(X)$ is proper subgroup of $F$?
Any hint for proceeding to prove that $i(X)$ generates $F$?
Best Answer
Define the map $j:X\to\langle i(X)\rangle$ by $j(x)=i(x)$. So $j$ is exactly $i$, we just changed the range. By assumption there is a unique homomorphism $\Phi:F\to\langle i(X)\rangle$ such that $\Phi\circ i=j$. Since $j$ and $i$ are the same function, we can also write that $\Phi$ is a homomorphism $F\to F$ which satisfies $\Phi\circ i=i$.
On the other hand, $i:X\to F$ itself can be "extended" to the identity map $id_F:F\to F$ which satisfies $id_F\circ i=i$. Since the extension is unique, it follows that we must have $\Phi=id_F$! And so the image of $id_F$ is contained in $\langle i(X)\rangle$. Since the image of $id_F$ is obviously $F$, this means that $F=\langle i(X)\rangle$.