Showing a free abelian group is generated by its basis

Question

I'm following Massey's "Basic Course in Algebraic Topology" and I'm stuck on his section explaining free abelian groups. He seems to be using a definition of free abelian groups that differs from most others' definitions, and then he leaves it to the reader to show that they're the same, but I can't figure out the proof. More specifically:

He defines a free abelian group on an arbitrary set $S$ as an abelian group $F$ together with a function $\phi:S\to F$ such that for any abelian group $A$ and any function $\psi:S\to A$, there exists a unique homomorphism $f:F\to A$ such that $f\circ\phi=\psi$. He shows that free abelian groups over a given set are unique up to isomorphism, and then leaves the following as an exercise:

"Prove directly from the definition that $\phi(S)$ generates $F$. [Hint: Assume not; consider the subgroup $F'$ generated by $\phi(S)$.]"

I can't figure out this exercise, but it seems really important to understanding this particular definition of free abelian groups. My guess is to let $\langle\phi(S)\rangle$ be $A$ and let $\psi$ be $\phi$ in the above definition, but I can't produce a contradiction.

Thank you for your help.

Edit: I just realized this question is very related to this other question, but I'm not convinced by the responses. The responses show that $\langle\phi(S)\rangle$ is isomorphic to $F$, but that doesn't seem (to me...) to show that they are equal, since $2\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ without being equal.

Answer 1

This is not what the hint tells you to do, but I think that it is easier to show $<\phi(S)>$ satisfies the universal property. Then, you'll have that $<\phi(S)>$ is isomorphic to $F$ and that the isomorphism is given by the inclusion.

Answer 2

Your goal is to show the inclusion map $i:F'\to F$ is an isomorphism. As you suggest, take $A=F'$ and $\psi=\phi$. to get a homomorphism $f:F\to F'$ which restricts to the identity on $\phi(S)$. Since $\phi(S)$ generates $F'$, this implies $f$ is the identity on all of $F'\subseteq F$, i.e. that the composition $fi:F'\to F'$ is the identity. To conclude that $i$ is an isomorphism, it suffices to show that $if:F\to F$ is the identity. For this, you want to use the uniqueness part of the universal property of $F$, in the case where $A=F$ and $\psi=\phi$.

(This argument is actually equivalent to Nitrogen's answer--this is just an explicit implementation of the proof that any two groups satisfying the universal property are canonically isomorphic.)