Adjoint functors on categories of monoids induced by a strong monoidal adjunction

Solution 1:

$Mon$ is a 2-functor from the 2-category of monoidal categories, monoidal functors, and monoidal natural transformations to the 2-category of categories. Given $\alpha:F\to G:C\to D$ in the latter 2-category, $Mon(\alpha)$ is just $\alpha$, which is a natural monoid homomorphism between the images under $F$ and $G$ of any monoid in $C$. The respect of $\alpha$ for the actions of $F,G$ on tensor products gives rise to its preservation of monoid multiplications, and similarly for units. 2-functors preserve adjunctions, so that $Mon(F)$ and $Mon(G)$ are always adjoints.

In fact, we didn't use strongness of $F,G$ here: it appears they could be lax. However, this extra generality is partially spurious, because if the unit and count of the adjunction are monoidal natural transformations, as is required for the 2-functoriality claim, then $F$, though not $G$, is automatically strong.

An example of such a monoidal adjunction with lax right adjoint is the free-forgetful adjunction between sets and abelian groups: the product of the underlying sets of two abelian groups is certainly not isomorphic to the underlying set of their tensor product, but there is certainly a natural map $A\times B\to A\otimes B$-namely, the universal bilinear map which characterizes $A\otimes B$. Since the free abelian group functor is strong monoidal, this is a monoidal adjunction, so that we get an adjunction between $Mon(Set)$ and $Mon(Ab)$, that is, between ordinary monoids and (unital, associative) rings. This is just the adjunction between the free ring on a monoid-whose base abelian group is the free abelian group on the monoid-and the forgetful functor sending a ring to its multiplicative monoid.