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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.

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  • $\begingroup$ This is a categorical definition of freeness. Think of the well-known property/characterisation of a basis in a vector space : as soon as you the images of the vectors of a basis, you have the linear map from the vector space into any other vector space by linearity. Similarly for $A$-algebra homomorphisms from the algebra of polynomials into any $A$-algebra: they're determined by the images of the indeterminates. $\endgroup$ – Bernard Oct 27 '15 at 2:10
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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.

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  • $\begingroup$ I thought about this path. I'm able to show that $\langle\phi(S)\rangle$ satisfies the definition of a free abelian group on $S$, and hence must be isomorphic to $F$. Better yet, I know (i.e., Massey tells me) that there is a unique isomorphism. But I'm not sure how to show that the inclusion is also an isomorphism from $\langle\phi(S)\rangle$ to $F$, since the inclusion might not be surjective. (Proving that it is surjective proves the claim.) Thanks for your help, I'll think about this more. $\endgroup$ – homotop Oct 27 '15 at 2:45
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    $\begingroup$ Look at the proof that a free group is unique up to isomorphism. You will see that the proof shows that the unique map $<\phi(S)> \to F$ such that it commutes with the maps from $S$ is an isomorphism. Now you just have to show that this map is actually the inclusion (which is easy). $\endgroup$ – Nitrogen Oct 27 '15 at 2:49
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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.)

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  • $\begingroup$ Ah! I think I get it now. But now I feel like we really just need to show that $i$ is surjective, which follows from showing that $i\circ f=\mathrm{id}_F$. This itself follows from noting (1) that $i\circ f\circ\phi=i\circ\phi=\phi$, (2) that $\mathrm{id}_F\circ\phi=\phi$, and (3) that there is (as you note) a unique homomorphism $F\to F$ such that the diagram commutes. So $i\circ\phi=\mathrm{id}_F$, hence $i$ is surjective. Please correct me if I'm wrong. $\endgroup$ – homotop Oct 27 '15 at 4:07
  • $\begingroup$ Yep, that's perfect. $\endgroup$ – Eric Wofsey Oct 27 '15 at 4:13

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