gr.group theory - direct limit of free complemented subgroups

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.