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How one can use the universal mapping property of free abelian groups to prove that $\sum_{b\in S}<b>$ is a free abelian group generated by $S$?

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up vote 3 down vote accepted

Let $A$ be an arbitrary Abelian group and $f:S\to A$ a function.

Then there is a unique way to extend $f$ to a homomorphism $\bar f:\sum_{b\in S}\langle b\rangle\,\to A$, namely, we have to set $$\bar f(n_1b_1+n_2b_2+\dots+n_kb_k):=n_1\,f(b_1)+n_2\,f(b_2)+\dots +n_k\,f(b_k)\,.$$

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