For what kind of a $\bf{Z}$-Module there can always be found a $\bf{Z}$-Module homomorphism s.t. the diagram commutes

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What kind of a $\bf{Z}$-Module should $A$ be so that the following condition holds?

If $B$ and $C$ are $\bf{Z}$-Modules such that there is a surjective $\bf{Z}$-Module homomorphism $f : C → B$ and there is a $\bf{Z}$-Module homomorphism $g : A → B$ there can be found a $\bf{Z}$-Module homomorphism $h : A → C$ such that $f(h(a)) = g(a)$ for all $a ∈ A$.

What I have done: I understand that $\bf{Z}$-Modules are just abelian groups and in order such an $h$ to be found it should be that $h(A) ⊂ f^{-1}(g(A))$. So I can focus my attention as assuming (without loss of generality) $C = f^{-1}(g(A))$ and $B = g(A)$.

Then the condition in the problem becomes (or implies? I am not sure): $A$ is an abelian group such that if $B$ and $C$ are abelian groups and there are surjective homomorphisms $g : A → B$ and $f : C → B$ then $C/Ker(f)$ and $A/Ker(g)$ are isomorphic.

I could not move any further.