In the category of modules, the tensor product functor is the left adjoint of the covariant Hom functor. Similarly in the category of sets, the Cartesian product functor is the left adjoint of the covariant Hom functor. I’m wondering whether this can be generalized.
Let $C$ be a category where the covariant Hom functor has a left adjoint. My question is, does this left adjoint make $C$ into a monoidal category? That is, does it induce a tensor product on $C$, in the sense that it is the tensor product functor of some tensor product on $C$?
If not, does anyone know of a counterexample?
Sets and modules over a commutative ring share the property that they are closed monoidal. This means they're not only (symmetric) monoidal, but the monoidal product has a right adjoint, which equips them both with a notion of "internal hom," or equivalently an enrichment over themselves.
If you start with just a category $C$, the Hom functor takes values in $\text{Set}$, so a left adjoint to $\text{Hom}(c, -) : C \to \text{Set}$ is, if it exists, a functor $\text{Set} \to C$. This adjoint, if it exists, turns out to be
$$\text{Set} \ni X \mapsto \bigsqcup_X c \in C;$$
in other words it takes a set $X$ and outputs the coproduct of $X$ copies of $c$; you can think of this as a tensoring $X \otimes c$, but note that $X$ is a set, not another object in $C$. This tensoring equips $C$ with the structure, not of a monoidal category, but of a module over the (cartesian) monoidal category $\text{Set}$.
Edit: There is a notion of closed category, which is a category equipped with just an internal hom functor. If the covariant internal hom has an enriched left adjoint, that reconstructs a monoidal structure; see the nLab.