Let $f: A \to B\ $ be an Abelian group homomorphism. Are there Abelian groups $G,\ H,\ K$ such that $K \subseteq H \subseteq G$ and the map $$\pi \circ i: H \to G/K$$ which is the composition of projection and inclusion is isomorphic to $f$? By isomorphic to $f\ $I mean there exist isomorphisms $\tau: H \to A\ $ and $\sigma: G/K \to B$ such that $f = \sigma \circ \pi \circ i \circ \tau^{-1}$.
I think it is a quite natural question, since the case where $f$ is epimorphism is a consequence of the fundamental homomorphism theorem. However, I can't prove the existence nor the uniqueness of the group extension. I'm not familiar at all with the general theory of group extension, but maybe the case where $A,\ B\ $are Abelian could be easier.
If such an extension really exists, it would be much more intuitive when dealing with chain complexes and exact sequences, imaging there is a "huge" group such that elements of all other groups are just its cosets.
The question has been answered in MO (link).