Universal Mapping Property of Free Abelian Groups Let S be a set and $F=F_S$ the free group on S. Let $F'$ be the commutator subgroup of $F$. Set $A=A_S = F/F'$, and call it the free Abelian group on $S$. Prove the universal mapping property of the free Abelian group: for any function $f:S \rightarrow G$, where G is an Abelian group, there exists a unique group homomorphism $\varphi:A \rightarrow G$ so that the diagram
$S \xrightarrow{f} G$,
$S \xrightarrow{a \mapsto [a]} A$,
$A \xrightarrow{\varphi} G$
commutes.
We did a proof of this for free groups in class but I am not sure how to apply it to this. I think part of my issue is that I am having a hard time figuring out what the elements of A look like. Will $A=\{[s](a^{-1}b^{-1}ab) \vert a,b,s \in S\}$? If so, then could I just send $\varphi(w)$ to $f(w)$, with $w \in F$?

Well, it is now pretty standard to denote a general commutative (and associative) operation as $+$.
So, with this notation (keeping in mind that it's commutative), the set you're after is $A=\{"x+y"\mid x,y\in S\}$ where thus $x+y=y+x$ now.  
The natural morphism $\theta:F\to A$ is a surjection and it sends a word to the unordered 'sum' of its letters.
Since any permutation (of the letters of the words) can be composed of inversions (i.e. swaps), the kernel of $\theta$ will equal to the commutator subgroup $F'$.
