Is the $Hom (-,k)$ functor exact?

Question

$\DeclareMathOperator{\Hom}{Hom}$ Let $k$ be a field of characteristic $0$. I want to know if $\Hom(-,k)$ is an exact functor from the category of abelian groups to itself. If it's true can you give a sketch of a proof ?

I already know that it's left-exact in general. I don't know to much about homological algebra and category theory. I'm new to these fields.

Answer 1

$\DeclareMathOperator{\Hom}{Hom}$ We need the following lemma :

Lemma : Let $k$ be a field of characteristic $0$. If $G$ is an abelian group, any group homomorphism from a subgroup of $G$ to $k$ admit an extension to $G$.

In other words $k$ is an injective $\mathbb{Z}$-module. Thus $\Hom(-,k)$ is exact.

Answer 2

In complement to your self-answer:

The injective objects in $\Bbb Z$-Mod are the divisible groups. The additive group $(k,+)$ associated to the characteristic zero field $k$ is a divisible group, since for any $g\in k$ and $n\in\Bbb Z$ there is an $h\in k$ such that $hn=g$. This is clear since $\Bbb Q\subseteq k$. You might find the structure theorem for divisible groups interesting, noting that every field of characteristic zero $\Bbb Q\subseteq k$ has underlying $\Bbb Q$-module structure (necessarily free since $\Bbb Q$ is a field) in which case $k$ appears in the classification there as a sum of copies of $\Bbb Q$.