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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?

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

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$\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 $$F = g(H) = g(\Span{\phi(S)}) = \Span{g(\phi(S))} = \Span{\phi(S)}.$$

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