I am currently working on an exercise in homology, whose consequence seems to me, that for every morphism of free abelian groups $f: A \rightarrow B$, there exists a subgroup $C$ of $A$, such that $A=ker(f) \oplus C$. I am not completely sure if this is true, but I have managed to prove it for finitely generated Abelian groups and am not able to find any counter-examples for larger groups.
Could you please tell me whether this result is true? If not could you provide a counter-example? Having one would help me quite a bit.
Yes it's true. To see this, notice that $f(A)$, as a subgroup of $B$, is a free abelian group. Now the restriction of $f$ to its image is a surjective homomorphism, so using the fact that $f(A)$ is free, you can prove that $\bar{f}:A\to f(A)$ has a section $s$. Then it is well-known that $\ker(f)\oplus sf(A)=A$.