Free group on a set $X$ is generated by $X$, why?

Question

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 $F$?

Answer 1

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 $F=\langle i(X)\rangle$.

Answer 2

Here's another argument, of similar flavor to that of Mark but using properties of functions directly; it's really just the end that justifies itself a bit differently.

Let $G=\langle i(X)\rangle$; let $j\colon X\to G$ be the embedding (it is the same as $i$, but with different codomain), and let $k\colon G\to F$ be the subgroup embeddings.

The map $j$ induces a morphism $\Phi\colon F\to G$ with $\Phi\circ i = j$. Now consider the map $k\circ \Phi\colon F\to G\to F$. We have $$(k\circ \Phi)\circ i = k\circ j = i = \mathrm{id}_F\circ i.$$ By the uniqueness clause of the universal property, we deduce that $\mathrm{id}_F=k\circ\Phi$. It follows that $k$ is has a right inverse, so it is surjective (and $\Phi$ has a left inverse, so it is injective). But $k\colon G\to F$ is the subgroup inclusion map, so it was already injective. Thus $G=F$, as desired.