Let $G\_1$ and $G\_2$ be countable abelian groups, and let $\iota\colon G\_1\to G\_2$ be an injective group homomorphism (so that we may regard $G\_1$ as a subgroup of $G\_2$). Suppose that for every finitely generated subgroup $K$ of $G\_2$, there is a map $\pi\_K\colon K\to G\_1$ such that $\iota \circ \pi\_K\circ \iota= \mbox{id}\_{\iota(G\_1)\cap K}$ (or if you want to omit $\iota$ from the notation, this would be $\pi\_K(g)=g$ whenever $g\in G\_1\cap K \subseteq G\_2$. 


In other words, for every finitely generated subgroup of $G\_2$, one can find a splitting. Does it follow that there is a "global" splitting $\pi\colon G\_2\to G\_1$ for $\iota$? 


I don't want to assume that the "partial" splittings $\pi\_K$ have any sort of coherence (by this I mean that I don't want to assume that whenever $K\subseteq K'$, the splitting $\pi\_{K'}$ can be chosen so that $\pi\_{K'}$ restricted to $K$ is just $\pi\_K$). I also don't want to assume that $G\_1$ is finitely generated.


Of course the conclusion would be that $G\_1$ is a direct summand in $G\_2$.



Does an injective group homomorphism between countable abelian groups that splits over every finitely generated subgroup, necessarily split?
Here is another concrete example constructed as in Yves Cornulier's comment. Let $G\_2$ be the group defined by the (abelian group) presentation


$G\_2=\langle\ x,y\_i\ (i \ge 0) \mid y\_i^2 = y\_{i-1}x\ (i \ge 1)\ \rangle$


and $G\_1 = \langle x \rangle$. 


Any finitely generated subgroup is contained in $\langle y\_i,x\rangle$ for some $i$, which is free abelian of rank $2$, and any subgroup of that splits over its intersection with $\langle x \rangle$.


But a complement of $G\_1$ in $G\_2$ would be generated by elements $y\_i' = y\_ix^{k\_i}$ for some integers $k\_i$ with $y\_i'^2 = y'\_{i-1}$ for all $i > 1$. But $y\_i'^2 = y\_{i-1}x^{2k\_i+1}$, so $k\_{i-1} = 2k\_i+1$ for all $i>1$, which is clearly not possible.


