2 oops

The "free group" proof rests on proving that that the fundamental group of a graph is free. For the analogue we'd need to essentially prove that the fundamental group of a "torus" (something that looks like a quotient of a vector space by a discrete subgroup) is free abelian. A sketch:

Given a real vector space V, we can put the direct limit topology on it (so that subsets are closed if and only if their intersection with any finite dimensional subspace is closed). This is a contractible topological group.

If A is a free abelian group, then A is a discrete subgroup of the associated real vector space (ℝ ⊗ A) and the quotient space has fundamental group A. Any covering space is a quotient of (ℝ ⊗ A) by a discrete subgroup B of A.

So the question boils down to showing: Any discrete subgroup of a vector space (with the direct limit topology) is free abelian.

Let's say that a partial basis is a set S of elements of B such that

• S is linearly independent, and
• S generates B ∩ Span(S).

Then partial bases are a partial order under containment, and Zorn's lemma implies that there is a maximal element S. I claim that S is a basis of B as a free abelian group.

S is linearly independent by construction, so it generates a free abelian group, and hence it suffices to show that it generates all of B. If b in B is not in S, then it is not in Span(S). Let S' be (S ∪ {b}). Then Span(S')/Span(S) is a 1-dimensional vector space and the image of B ∩ Span(S') must be discrete, because otherwise Span(S') would contain an element (rb + v) for v in Span(S), r not rational Span(S) that we could use to generate a non-discrete subset of B. (If v is a combination of w1...wn in S, then B it suffices to check that any subgroup of the finite-dimensional space Span(w1...wn,b) contains (b+v), and the images of b and rb+v are generate an indiscrete subspace iff r requiring more than n generators is not a rational number.indiscrete.)

Thus any lift of a generator of B ∩ Span(S') would extend to a larger generating set, contradicting maximality.

(My apologies for the comment last night, which this morning looks snarkier than I intended. I'm a fan of using this topological reasoning for free groups myself, because it compartmentalizes the proof into much more understandable pieces. In particular, I don't think I'd really understand a purely algebraic proof that an index n subgroup of a free group on m generators is free on nm - n + 1 generators.)

1

The "free group" proof rests on proving that that the fundamental group of a graph is free. For the analogue we'd need to essentially prove that the fundamental group of a "torus" (something that looks like a quotient of a vector space by a discrete subgroup) is free abelian. A sketch:

Given a real vector space V, we can put the direct limit topology on it (so that subsets are closed if and only if their intersection with any finite dimensional subspace is closed). This is a contractible topological group.

If A is a free abelian group, then A is a discrete subgroup of the associated real vector space (ℝ ⊗ A) and the quotient space has fundamental group A. Any covering space is a quotient of (ℝ ⊗ A) by a discrete subgroup B of A.

So the question boils down to showing: Any discrete subgroup of a vector space (with the direct limit topology) is free abelian.

Let's say that a partial basis is a set S of elements of B such that

• S is linearly independent, and
• S generates B ∩ Span(S).

Then partial bases are a partial order under containment, and Zorn's lemma implies that there is a maximal element S. I claim that S is a basis of B as a free abelian group.

S is linearly independent by construction, so it generates a free abelian group, and hence it suffices to show that it generates all of B. If b in B is not in S, then it is not in Span(S). Let S' be (S ∪ {b}). Then Span(S')/Span(S) is a 1-dimensional vector space and the image of B ∩ Span(S') must be discrete, because otherwise Span(S') would contain an element (rb + v) for v in Span(S), r not rational that we could use to generate a non-discrete subset of B. (If v is a combination of w1...wn in S, then B ∩ Span(w1...wn,b) contains (b+v), and the images of b and rb+v are generate an indiscrete subspace iff r is not a rational number.)

Thus any lift of a generator of B ∩ Span(S') would extend to a larger generating set, contradicting maximality.

(My apologies for the comment last night, which this morning looks snarkier than I intended. I'm a fan of using this topological reasoning for free groups myself, because it compartmentalizes the proof into much more understandable pieces. In particular, I don't think I'd really understand a purely algebraic proof that an index n subgroup of a free group on m generators is free on nm - n + 1 generators.)