The abelianization of a group $G$ is an abelian group $A$ and a homomorphism $\varphi: G \to A$ such if $B$ is any abelian group, and $\phi: G \to B$ is any homomorphism, there is a unique homomorphism $\psi: A \to B$ (which might depend on $\phi$) such that $\psi\varphi = \phi$.
Now, I am reading some lecture notes, and the following is asserted.
If $G = \langle e_1, e_2, \ldots, e_n \mid w_1, w_2, \ldots, w_m\rangle$ is a finitely presented group, then$$A = \langle e_1, e_2, \ldots, e_n \mid w_1, \ldots, w_m, [e_1, e_2], \ldots, [e_i, e_j], \ldots, [e_{n - 1}, e_n]\rangle$$is a presentation of the abelianization of $G$, where the homomorphism $\varphi: G\to A$ sends the equivalence class of $w$ in $G$ to the equivalence class of $w$ in $A$ for each word $w \in G$.
To me, this is not a priori clear at all. Could anybody tell me why this is true?
Actually, in the category of Abelian groups, the same presentation $\langle e_1,\dots\,|\, w_1,\dots\rangle$ works for the abelianization of $G$.
But stepping back to general groups, we have to encode the Abelianness, i.e. that each pair of elements commute. It is enough to pose that each pair of generating elements commute.
And that's exactly the extra data in in the formula of the presentation of abelianization among groups.