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Is every group isomorphic to an inductive colimit (that is, directed colimit, also called inductive limit, or directed limit) of free groups?

I guess, the answer is no. In that case: Is there a characterization of the groups that are isomorphic to inductive colimits of free groups?

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  • $\begingroup$ Directed limit often refers to inductive limit of injective homomorphisms. Anyway... what do you think of a group with 2 elements, in this case too? $\endgroup$
    – YCor
    Commented Dec 27, 2017 at 19:06
  • $\begingroup$ With injective connecting homomorphisms the question will be very different. In fact, it seems to me that an inductive limit of free groups with injective connecting homomorphisms is again a free group. $\endgroup$
    – Hannes Thiel
    Commented Dec 27, 2017 at 19:11
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    $\begingroup$ No, for instance $\mathbf{Q}$ or $\mathbf{Q}\ast\mathbf{Q}$ are not free but are inductive limits of free groups with injective homomorphisms. These are not free. Inductive limits of free groups with injective homomorphisms are called "locally free groups". Another non-free example is the $\pi_1$ of the Hawaiian earring space. $\endgroup$
    – YCor
    Commented Dec 27, 2017 at 19:14
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    $\begingroup$ Actually I think it's true (but nontrivial) that all inductive limits of free groups are locally free. This reduces to the finitely generated case, in which case it follows from the fact that any sequence of epimorphisms between finitely generated free groups stabilizes (because the rank decreases until it stabilizes, and then we have isomorphism by Hopfianness). $\endgroup$
    – YCor
    Commented Dec 27, 2017 at 19:19
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    $\begingroup$ Every group that is a directed colimit of free groups has homological dimension at most 1 and cohomological dimension at most 2. In particular, $\mathbb{Z}/2$ is not a directed colimit of free groups. $\endgroup$
    – Yonatan Harpaz
    Commented Dec 27, 2017 at 19:22

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Let $G$ be a group. Equivalences:

  1. $G$ is locally free (i.e., all its finitely generated subgroups are free)
  2. $G$ is a filtering inductive limit of free groups with injective connecting homomorphisms
  3. $G$ is a filtering inductive limit of free groups.

The equivalence between (1) and (2) is clear (given that subgroups of free groups are free - otherwise "locally free" would be a bit imprecise); (3) is tautologically implied by (2).

Conversely suppose (3), and let's prove (1). Passing to a finitely generated subgroup of $G$ reduces to $G$ finitely generated. Then passing to a cofinal subsystem allows to assume that the filtering system has an initial element $H$ with $H\to G$ surjective. Then some finitely generated subgroup $K$ of $H$ has onto image in $G$ and then replacing each group in the filtering system, we realize $G$ as a filtering inductive limit of finitely generated free groups with surjective connecting homomorphisms. Then the rank of these free groups decreases until it stabilizes, and since finitely generated free groups are Hopfian, this means that there's a cofinal subsystem made up of isomorphisms. So $G$ is free.

Note: for more other classes of groups, usually such an implication as (3)$\Rightarrow$(2) fails. For instance, every group is a filtering inductive limit of finitely presented groups (this is not specific to groups!), but usually this cannot be done with injective connecting homomorphisms (e.g., for an infinitely presented finitely generated group).

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