Existence of injective homomorphism

The group $S^1$ of complex numbers of modulo $1$ is divisible, so it is an injective object in the category of abelian group, which is tantamount to saying that the functor $\operatorname{Hom}({-},S^1)$ is exact.

A group $D$ (written additively) is divisible if, for every $x\in D$ and every integer $n\ne0$, there is $y\in G$ such that $x=ny$.

A group $E$ is injective if the functor $\operatorname{Hom}(-,E)$ is right exact (or exact, since it is left exact for every group).

Being divisible is a necessary condition for being injective. Indeed, let $E$ be injective and consider the injection $n\mathbb{Z}\to\mathbb{Z}$. Since $E$ is injective, the induced map $\operatorname{Hom}(\mathbb{Z},E)\to\operatorname{Hom}(n\mathbb{Z},E)$ is surjective.

Take $x\in E$ and define $f\colon n\mathbb{Z}\to E$ by $f(nk)=kx$. Then, by assumption, this map is the restriction to $n\mathbb{Z}$ of a homomorphism $g\colon\mathbb{Z}\to E$. Then $x=f(n)=g(n)=ng(1)$, so taking $y=g(1)$ we have provided the element $y$ such that $x=ny$.

The converse is known as Baer's criterion: a right module $E$ over the ring $R$ is injective if and only if, for every right ideal $I$ of $R$ and every homomorphism $f\colon I\to E$, there exists $y\in E$ such that $f(r)=ry$, for every $r\in I$.

I think you can do this directly, since you are working in the category of finite abelian groups, because in this case, $F=\bigoplus^n_{i=1} Z_{q_i}$, where the $q_i$ are powers of primes. So, you can take $z:F\to S^1$ to be the homomorphism that sends each generator $\gamma_i$ of $Z_{q_i}$ to a particular $q_i$-th root of unity.