Abelianization of free group is the free abelian group

Solution 1:

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.

Solution 2:

Since both the free group and abelianization are left adjoint (and hence preserve coproduts) we shall have up to canonical isomorphism $(FX)^{\mathrm{ab}}=(F\coprod_{x\in X}\left\lbrace x\right\rbrace)^\mathrm{ab}=(\ast_{x\in X}F_1)^{\mathrm{ab}}=\bigoplus_{x\in X}\mathbf{Z}$, since $F_1=\mathbf{Z}$.