I've just found out about the forgetful functor between the category of vector spaces and the category of abelian groups. It maps a vector space to it's additive abelian group.
My question is, is there an adjoint of this forgetful functor, and if so what is it?
If we considered the forgetful functor between the category of vector spaces and the category of sets that forgets all structure, I think the adjoint functor would take a set and create a generic free vector space generated by the elements of the set.
However, when forgetting only to the level of an abelian group, there still retains some additional structure, so I don't think it's that simple. For example, just by looking at the additive group, you know $v$ and $v+v$ are related by the scalar multiplication $v+v=2v$. Similarly, one could deduce that rational combinations like $6v$ and $9v$ are related through multiplication, since there would exist an element $3v$ that can be added to itself several times to create both. On the other hand, just by looking at the additive structure, I think there is no way to tell that $v$ and $\pi v$ were once related since $\pi$ is irrational.
Best Answer
If you're working over the field $k$, I think the functor $k \otimes_{\mathbb{Z}} -: \mathbf{Ab} \to \mathbf{Vect}_k$ should work.
Let $F$ be the forgetful functor $\mathbf{Vect}_k \to \mathbf{Ab}$, $A$ an abelian group and $V$ a vector space over $k$. Then, given $f: A \to F(V)$, define $f': k \otimes_{\mathbb{Z}} A \to V$ by setting $f'(s \otimes a) := s \cdot f(a)$ and extending additively. Given $g: k \otimes_{\mathbb{Z}} A \to V$, define $g^*:A \to F(V)$ by setting $g^*(a) := g(1 \otimes a)$. One can check that the assignments $f \mapsto f'$ and $g \mapsto g^*$ are mutually inverse and thus give bijections between $\mathrm{Hom}_{\mathbf{Ab}}(A,F(V))$ and $\mathrm{Hom}_{\mathbf{Vect}_k}(k \otimes_{\mathbb{Z}}A,V)$.