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3 added 59 characters in body

My guess is that $G$ is forced to be projective, hence free, in this situation. To show this, we need to verify that $\mathrm{Hom}(G,-)$ is an exact functor. As we have an identification of functors $\mathrm{Hom}(G,-) \simeq \lim_i \mathrm{Hom}(F_i,-)$, applying $\mathrm{Hom}(G,-)$ to an exact sequence

$0 \to A \to B \to C \to 0$

of abelian groups, we get an induced sequence

$0 \to \lim_i \mathrm{Hom}(F_i,A) \to \lim_i \mathrm{Hom}(F_i,B) \to \lim_i \mathrm{Hom}(F_i,C) \to R^1 \lim_i \mathrm{Hom}(F_i,A) \to ...$

So it suffices to show that $R^1 \lim_i \mathrm{Hom}(F_i,A)$ vanishes for any abelian group $A$. Is this true?

If

Here's a not-so-well-thought-out idea: if I chased elements correctly, this an affirmative answer to the question above follows from the bijectivity (in fact, surjectivity) of the natural map $\mathrm{Hom}(F_j,A) \to \lim_{i < j} \mathrm{Hom}(F_i,A)$, for j sufficiently big. We may also further assume After making the harmless assumption that the system $(F_i)$ consists of all finitely generated saturated subgroups of $G$. The G$, the preceding bijectivity question then translates to: given a free abelian group F of finite rank, when is the natural map$\mathrm{colim}_i H_i \to F$an isomorphism, where the indexing set I is the poset of all proper saturated subgroups$H_i \subset F$. I think the answer to this question is yes when the rank of$F$is at least$3$(which is enough for the application at hand), but I'm not sure. 2 Changed content; added 35 characters in body I think such a group Edit: The answer below has been modified to reflect the comments. My guess is necessarily a that$G$is forced to be projective, hence free,$\mathbf{Z}$-module. in this situation. To show projectivitythis, we need to verify that$\mathrm{Hom}(G,-)$is an exact functor. Note that As we have an identification of functors$\mathrm{Hom}(G,-) \simeq \lim_i \mathrm{Hom}(F_i,-)$. Applying mathrm{Hom}(F_i,-)$, applying $\mathrm{Hom}(G,-)$ to an exact sequence

$0 \to A \to B \to C \to 0$

of abelian groups, we get an induced sequence

$0 \to \lim_i \mathrm{Hom}(F_i,A) \to \lim_i \mathrm{Hom}(F_i,B) \to \lim_i \mathrm{Hom}(F_i,C) \to R^1 \lim_i \mathrm{Hom}(F_i,A) \to ...$

So it suffices to show that $R^1 \lim_i \mathrm{Hom}(F_i,A)$ vanishes for any abelian group $A$. This Is this true?

If I chased elements correctly, this follows from the bijectivity (in factthat , surjectivity) of the transition maps natural map $\mathrm{Hom}(F_j,A) \to \mathrm{Hom}(F_i,A)$ are surjective (lim_{i < j} \mathrm{Hom}(F_i,A)$, for j sufficiently big. We may also further assume that the system$i \leq j$)(F_i)$ consists of all finitely generated saturated subgroups of $G$. The preceding bijectivity question then translates to: given a free abelian group F of finite rank, which follows from (and when is equivalent the natural map $\mathrm{colim}_i H_i \to ) F$ an isomorphism, where the fact that indexing set I is the poset of all proper saturated subgroups $F_i H_i \subset F$. I think the answer to F_j$this question is a direct summandyes when the rank of$F$is at least$3$(which is enough for the application at hand), but I'm not sure. 1 I think such a group is necessarily a projective, hence free,$\mathbf{Z}$-module. To show projectivity, we need to verify that$\mathrm{Hom}(G,-)$is an exact functor. Note that we have an identification of functors$\mathrm{Hom}(G,-) \simeq \lim_i \mathrm{Hom}(F_i,-)$. Applying$\mathrm{Hom}(G,-)$to an exact sequence$0 \to A \to B \to C \to 0$of abelian groups, we get an induced sequence$0 \to \lim_i \mathrm{Hom}(F_i,A) \to \lim_i \mathrm{Hom}(F_i,B) \to \lim_i \mathrm{Hom}(F_i,C) \to R^1 \lim_i \mathrm{Hom}(F_i,A) \to ...$So it suffices to show that$R^1 \lim_i \mathrm{Hom}(F_i,A)$vanishes for any abelian group$A$. This follows from the fact that the transition maps$\mathrm{Hom}(F_j,A) \to \mathrm{Hom}(F_i,A)$are surjective (for$i \leq j$), which follows from (and is equivalent to) the fact that$F_i \to F_j\$ is a direct summand.