# Showing a free abelian group is generated by its basis

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.

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.

• 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. – Bernard Oct 27 '15 at 2:10

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.
• 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. – homotop Oct 27 '15 at 2:45
• 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). – Nitrogen Oct 27 '15 at 2:49
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 $\psi=\phi$.
• 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. – homotop Oct 27 '15 at 4:07