Let $C$ and $D$ be monoidal categories and let $F:C\rightarrow D:G$ be a strong monoidal adjunction (adjunction in which $F$ and $G$ are strong monoidal functors). There are induced functors $Mon(F):Mon(C)\rightarrow Mon(D)$ and $Mon(G):Mon(D)\rightarrow Mon(C)$ on categories of monoids (actually this holds more generally when $F$ and $G$ are lax monoidal functors). When are $Mon(F)$ and $Mon(G)$ adjoint?


1 Answer 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.

  • $\begingroup$ Helpful answer. Do you know if the functor $Mon$ preserves (sufficiently filtered) 2-colimits? $\endgroup$
    – user84563
    Aug 24, 2016 at 0:01
  • $\begingroup$ Late response, but since I just came back to this:it preserves filtered colimits, since monoids and their homomorphisms are finitary concepts. $\endgroup$ Sep 11, 2017 at 16:49

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