On functors preserving monoid objects

Question

If $C$ is a monoidal category, we can define the category $Mon(C)$ of monoids in $C$; call $U_C : Mon(C) \to C$ the forgetful functor. I'm interested in functors between categories of monoids:

Suppose $(C, I, \otimes)$, $(D, J, \boxtimes)$ are (strict) monoidal categories, and call $Mon(C)$, $Mon(D)$ the corresponding monoid categories. It is well known that any lax monoidal functor $F : C \to D$ preserves monoids; in other words we can define another functor $\tilde{F} : Mon(C) \to Mon(D)$ such that $U_D \circ \tilde{F} = F \circ U_C$.

My question: Suppose we have a functor $\tilde{G} : Mon(C) \to Mon(D)$: is it possible to produce a functor $G : C \to D$ such that the above equality holds? If not, are there some results in the literature providing conditions under which it exists (or might exist)?

My failed attempts: At first I thought that such an "extension" would always exist (and be a lax monoidal functor), but when I tried to compute it in some special cases, my strategy failed.

All the cases where I failed use the existence of a coproduct in some other category $B$; let me elaborate on one of such cases:

Example of failure: Suppose $B$ has finite coproducts $+$, as well as a terminal object $\ast$. Consider the following:

$$\Phi : [B, B] \to [B, B]$$ $$F(-) \mapsto F(- + \ast)$$

In other words, $\Phi(F)(a) = F(a + \ast)$; it acts in the obvious way on morphisms (both in $B$ and in $[B, B]$). It can be shown that if $M : [B, B]$ is a monoid (that is, a monad on $B$) then so is $\Phi(M)$:

The unit (of the monoid structure on $\Phi(M)$) is given by the composition: $$ (\epsilon_{\Phi(M)})_a =: a \xrightarrow{\iota_1} a + \ast \xrightarrow{\epsilon_{M(a + \ast)}} M(a + \ast) $$

The multiplication is quite more involved to define: $$(\mu_{\Phi(M)})_a =: M(M(a + \ast) + \ast) \xrightarrow{M(\psi)} M(M(a + \ast)) \xrightarrow{\mu_{M(a + \ast)}} M(a + \ast)$$ It is defined as long as we can define $\psi : M(a + \ast) + \ast \to M(a + \ast)$.

But such a $\psi$ can be defined: consider the universal property of $+$. By that universal property, given any couple of morphisms $f : a \to c$ and $g : b \to c$, we can produce a morphism $<f, g> : a + b \to c$.

Now, consider the case where $b = \ast$ and $c = a + \ast$: by the monoid structure on $M$ we have a map $$\phi : \ast \xrightarrow{\iota_2} a + \ast \xrightarrow{\epsilon_{a + \ast}} M(a + \ast)$$ apply the construction given before to $1_{M(a + \ast)} : M(a + \ast) \to M(a + \ast)$ and $\phi : \ast \to M(a + \ast)$ to produce a morphism $$\psi := <1_{M(a + \ast)}, \phi> : M(a + \ast) + \ast \to M(a + \ast)$$ This concludes the construction of the monoid structure on $\Phi(M)$ (since the naturality of $\psi$ is quite straightforward to prove).

As some of the readers will have noticed, the morphism $M(\psi)$ does exactly what the "laxness of $\Phi$" would have done, did it exist: $M(\psi) : \Phi(M) \circ \Phi(M) \to \Phi(M \circ M)$. However, we had to assume $M$ to be a monoid in the first place: were it not, $\psi$ couldn't be constructed.

This does not mean that $\Phi$ is not a lax monoidal functor: it could well have different laxness structural morphisms, that "reduce" to the above construction when the argument $M$ is in fact a monoid in $[B, B]$ (monad on $B$).

The other two cases where this happened are:

• $\Phi(F)(a) = F(a^*)$, where $a^*$ is the "Kleene Star" (or, in category-theory speech, the free monoid on $a$); this requires the further assumption that finite product exist, countable coproduct exist (in addition to finite ones), and finite products distribute over countable coproducts, in $B$

• $\Phi(F)(a) = F(b + a)$ for some object $b : B$; this does not even require the existence of a terminal object.

The above construction carries over to these other two cases, with slight modifications.

Motivation: I am particularly interested in the case where the two categories (and their monoidal structures) are the same, being an endofunctor category with the obvious monoidal structure (as in the "failed attempt")

$$(C, I, \otimes) = (D, J, \boxtimes) = ([B, B], 1_B, \circ)$$

where $Mon([B, B]) = Monads(B)$ the category of monoids in $[B, B]$ is the category of monads on $B$: as many will have guessed, this question arised from haskell programming and monad transformers.

Rephrasing the question: Given a functor between categories of monoids (internal to some monoidal categories), when can it be extended as a functor between the underlying monoidal categories? What properties should I expect such an extension to have (if such an extension does indeed exist)?

Lastly, is the "failed attempt" really a counter-example? If so, is there a counter-example that does not rely on functor categories? In the last few days, I searched for a counter-example in the category of non-commutative rings (as many will know, non-commutative rings are monoid objects in the category of abelian groups) without success.

Answer 1

Here is a simple counterexample: take $C=D=\mathrm{Set}$, and let $\tilde{G}$ be the abelianization of a monoid. Then $\tilde{G}$ cannot arise from any $G:\mathrm{Set}\to\mathrm{Set}$, because the cardinality of the underlying set of $\tilde{G}(M)$ depends on the monoid structure of $M$, not just on its cardinality. For instance, take an abelian monoid and a nonabelian monoid of the same cardinality, say the symmetric group $S_3$ and the cyclic group $C_6$: then $U(S_3) \cong U(C_6)$, but $U \tilde{G}(S_3) \not\cong U\tilde{G}(C_6)$ since the latter still has 6 elements but the former has fewer; thus $U \tilde{G} \not\cong G U$ for any functor $G$.

Answer 2

@Mike Shulman's answer is already quite interesting, and answers one of my main perplexities. Here I want to add a little note on why my "proposed counterexample" really is a counterexample (again, this carries over to the other two cases with little modifications):

Set $\Phi(F) (a) := F(\ast + a)$ as in the question.

We want to show that there are $C$ a category, $F, G : C \to C$ functors such that there is no natural transformation $\Phi(F) \circ \Phi(G) \to \Phi(F \circ G)$.

Take $C = Set$, so that $\ast$ is any singleton, $F = 1_C$, and $G=\Delta_{\emptyset}$ to be the functor constant on $\emptyset$ the empty set; obviously $$\Phi(F) \circ \Phi(G) = \Delta_\ast$$ $$\Phi(F \circ G) = \Delta_\emptyset$$ and we all know that there is no function from any set to the empty set.