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 $X$?

Another approach: 
Try the following cute 
Lemma: An element in $\;F(X)\;$, written as a word $\;w(x)=x_i^{a_i}\cdot\ldots\cdot x_j^{a_j}\;,\;\;x_k\in X\;,\;\;a_k\in\Bbb Z$ belongs to the commutator group $\;F(X)'\;$ iff the exponent sum of each element in $\;X\;$ is zero, meaning: if we denote by 
$$\;E_x^w:=\{n\in\Bbb Z\;;\;n\;\;\text{appears as exponent of the letter}\;\;x_i\;\;\text{in the word}\;\;w\}\;$$
then
$$\sum_{n\in E^w_x}n=0\;\;,\;\;\forall\,x\;\;\text{in the word}\;\;w$$
For example, the element $\;x^2yx^{-1}zx^{-1}y^2z^{-1}y^{-3}\;$ belongs to $\;F(x,y,z)'\;$ , but $\;xzy^{-2}x^{-3}\;$ doesn't.
Letting $\;F(X)\,,\,A(X)\;$ be the free (free abelian) group on the set $\;X\;$ , resp. , define
$$f:X\to A(X)\;,\;\;f(x):=x$$
By the universal property of $\;F(X)\;$ there exists a unique homomorphism $\;\phi: F(X)\to A(X)\;$ extending $\;f\;$ , and if $\;w(x)\;$ is a word in the "letters" (elements) of $\;X\;$ ,  we get that (writing all multiplicatively)
$$1=\phi(w(x)) = w(f(x)) \iff \;\text{ the exponent sum of each letter in the word $\;w\;$ is zero}$$
