Universal property of the abelianisation

Statement

Let $G$ be a group, let $\pi \colon G \to G^{\mathrm{ab}}$ be the projection to its abelianisation, and let $A$ be an abelian group. For each group homomorphism $\varphi \colon G \to A$, there is exactly one group homomorphism $\bar{\varphi} \colon G^{\mathrm{ab}} \to A$ with $\bar{\varphi} \circ \pi = \varphi$.

Remarks

Each commutator goes to $e$ under $\varphi$, because $A$ is abelian, so $[G, G] \subseteq \ker \varphi$. Then the universal property of the quotient group gives $\bar{\varphi}$. Thus the homomorphisms from $G$ to $A$ correspond exactly to the homomorphisms from $G^{\mathrm{ab}}$ to $A$.

Explanations

L1 human

A homomorphism into an abelian group cannot see the commutators of $G$. So it factors through the abelianisation, where the commutators are gone. This makes the abelianisation the best abelian approximation of $G$.