# Universal property of free abelian group implies $\phi(S)$ generates F [duplicate]

I saw some definition of free abelian group of an arbitrary set $S$:

An free abelian group on $S$ is a abelian group $F$ and a map $\phi:S\to F$, s.t. for any ableian group $G$ and any map $\rho:S\to G$, there exists an unique homomorphism $f:F\to G$ s.t. $f\circ \phi=\rho$.

A problem ask me to prove that $\phi(S)$ generates $F$. The hint is consider $\phi(S)$ as a subgroup of $F$. This implies that there exists an unique homomorphism $f:F\to \phi(S)$ s.t. $f|_{\phi(S)}$ is identity function. But I don't know what to do next. Could you give me some hints?

## marked as duplicate by Eric Wofsey, David K, Vladhagen, Claude Leibovici, Martin SleziakJan 15 '17 at 9:52
