TITLE: Categorical proof subgroups of free groups are free?
QUESTION [20 upvotes]: This is a crossport of this question from MSE.

Is there a categorical proof that subgroups of free groups are free?
How about the result that subgroups of free abelian groups are free abelian?
What is it about $\mathsf{Grp}$ and $\mathsf{Ab}$ that makes subobjects of free objects free? Can one characterize such categories? In particular, are there any other ones apart from $\mathsf{Grp}$ and $\mathsf{Ab}$?

I have received some comments on MSE, one claiming such a proof exists, one doubting it, and another claiming it cannot be proved in $ZF$.

REPLY [12 votes]: Let me elaborate on my comment. I think freeness is a red herring. The content of the standard topological proof is that a covering space of a $1$-dimensional CW complex is again a $1$-dimensional CW complex. Of course this naturally generalizes to the case where $1$ is replaced by any positive integer $k$, which suggests the following definition.
Say that a group $G$ has homotopy dimension at most $k$ if $BG$ can be presented by a $k$-dimensional CW complex (e.g. take $BG$ to be a compact hyperbolic $k$-manifold). Then the same argument about covering spaces shows that every subgroup of a group with homotopy dimension at most $k$ again has homotopy dimension at most $k$. Homotopy dimension at most $1$ is equivalent to freeness in the presence of choice, but using it instead of freeness in the argument removes the need to choose a spanning tree and that should address concerns about uses of choice. 
As the long list of examples in Todd's answer also suggests, we should be looking at some property other than freeness in general. I think some kind of cohomological dimension is a better candidate; as stated in Matthias Wendt's comment below, the nontrivial free groups are precisely the groups of cohomological dimension $1$. And $1$-dimensionality also shows up in Ben Webster's answer.