Abelianization of free group is the free abelian group How does one prove that if $X$ is a set, then the abelianization of the free group $FX$ on $X$ is the free abelian group on $X$?

Here is an algebraico-topological proof, using :

Hurewicz's theorem. For a topological space $X$, the natural morphism
  $$ \pi_1(X)^{\rm ab} \to H_1(X) $$
  is an isomorphism.

The fundamental group of $\bigvee_{s \in S} \mathbb S^1$ is the free group on the set $S$ (using Van Kampen for example). The $1$-homology group of $\bigvee_{s \in S} \mathbb S^1$ is the free $\mathbb Z$-module on $S$ (using Mayer-Vietoris, or another long exact sequence-wise proof). So Hurewicz's theorem concludes. 
