If $A$ is an abelian group, then $\mbox{Hom}(A,\mathbb{Z}) \ne 0$ if and only if $A$ has an infinite cyclic direct summand.
The hint is to use
If $F$ is a free abelian group and $g:B \to F$ is a surjective homomorphism from some abelian group $B$ then B = \mbox{ker} g \oplus F', where F' \simeq F.
I guess I am misinterpreting the question. Can't I just take $A = \mathbb{Z}/2\mathbb{Z}$, and define $\phi(z) = \begin{cases} 0 &\mbox{ if } z = 0 \\ 1 &\mbox{ if } z = 1 \end{cases}$