# 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

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\renewcommand{\phi}{\varphi}I am starting from the homomorphism f : F \to H you have constructed, where H = \Span{\phi(S)}, which satisfies f(\phi(s)) = \phi(s) for s \in S. Note that this implies that H is also free on S. In fact, given any group G and any map \rho : S \to G, one obtains immediately a map t : H \to G such that t(\phi(s)) = \rho(s) for s \in S. This is unique, because t \circ f : F \to G is the unique homomorphism such that t(f(\phi(s))) = \phi(s) for s \in S. Clearly there is also a homomorphism g : H \to F such that g(\phi(s)) = \phi(s) for s \in S. This is just inclusion. The composite g \circ f : F \to F maps \phi(s) to \phi(s) for s \in S, and thus by uniqueness is the identity. Since H is also free on S, also f \circ g : H \to H is the identity. Thus g is an isomorphism of groups. In particular$$F = g(H) = g(\Span{\phi(S)}) = \Span{g(\phi(S))} = \Span{\phi(S)}.$\$