Let $F = F(A)$ be a free group, and let $f : A \rightarrow G$ be a set-function from the set $A$ to a commutative group $G$, and where $[F,F]$ is the subgroup of F generated by all elements of the form $aba^{-1}b^{-1}$. I believe that is what is commonly referred to as the commutator subgroup.
Prove that $f$ induces a unique homomorphism $F/[F, F] \rightarrow G$ and conclude that $F/[F, F] \cong Fab(A)$.
Hints:
== In any group $\;G\;$, the quotient $\;G/[G:G]\;$ is abelian , and the commutator subgroup $\;[G:G]\;$ is characterized for being the minimal such subgroup, meaning that if $\;N\lhd F\;$ is s.t. $\,G/N\;$ is abelian then $\,[G:G]\le N\;$ .
== The universal property of free groups tells us that any set function $\,A\to G\;$, with $\;G\;$ a group, can be uniquely extended to a group homomorphism $\;F(A)\to G\;$ .
Work out the above, which btw can be found in any decent group theory book.