direct limit of free complemented subgroups Consider the following property of an abelian group $G$:
S: $G$ is torsionfree and a directed limit of finitely generated (hence free) subgroups $\{F_i\}_i$, such that for all $i \leq j$, $F_i$ is a direct summand of $F_j$.
Clearly, every free abelian group satisfies S. Does also the converse hold? If not, is S satisfied by $A^\mathbb{N} / A^{(\mathbb{N})}$, where $A=\mathbb{Q}_+=\mathbb{Z}^{(\mathbb{P})}$?
 A: Edit: The answer below has been modified to reflect the comments.
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?
Here's a not-so-well-thought-out idea: if I chased elements correctly, an affirmative answer to the question above follows from the bijectivity of the natural map $\mathrm{Hom}(F_j,A) \to \lim_{i < j} \mathrm{Hom}(F_i,A)$, for j sufficiently big. After making the harmless assumption that the system $(F_i)$ consists of all finitely generated saturated subgroups of $G$, the preceding bijectivity question 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.
