# Functors that preserve monoids

In the comments section of this question there was a question that I don't know if it has been asked on the site. It is well-known and easily proved that lax monoidal functors preserve monoids. So the question, which is about the converse, in my words is the following

Given a functor between monoidal categories that preserves monoids and morphisms of monoids on these categories, does it follow that the functor is lax monoidal?

I am interested in this question because in my research I've encountered a functor between the underlying collections of operads that do preserve the operad structure as well as the morphisms of operads. Since operads are monoids on the category of collections (or $$\mathbb{S}$$-modules for the symmetric case) with the composite of collections as monoidal product, I was expecting this functor to be lax monoidal, but I haven't been able to prove it.

If the answer to my above question is positive it must mean that I have made some mistake and I would post a question related to that particular case. If not, I am fine with my functor preserving just operads and its morphisms. In fact, I believe that this implication is not going to be true in general, because the associativy axiom of monoidal categories relates 3 (in general different) objects while the axioms of monoids only involve a single object. But maybe there are some nice sufficient condition for this implication to hold.

A similar question that I might post separated is related to the fact that symmetric lax monoidal functors not only preserve monoids on the category but also operads on the category (here I am using the monoidal product of the underlying category, not the composite, which is eventually defined in terms of the monoidal product of the underlying category). So a natural question again is whether a functor that preserves operads and their morphisms is automatically symmetric lax monoidal.

• It seems unlikely to me, surely you could cook up a monoidal category that had no non-trivial monoids, and then the preserves monoids condition tells you almost nothing and just about any functor will do. Jun 30 at 17:29
• The problem is that monoids only talk about tensor powers of an object, but a lax monoidal structure incorporates all tensor products. Surely there will be counterexamples. Maybe you can do something when coproducts exist which are compatible with tensors. Jun 30 at 17:49
• Also there are some technical inaccuracies here. It does not make sense to say that a functor preserves monoids, since being a monoid is not a property. Similarly, it does not make much sense to say that the functor is lax monoidal. You probably want to have a functor between monoid objects which lifts the given functor, maybe with some properties which need to be specified. Jun 30 at 17:55
• "monoid" is an extra structure, not property, so that "maps monoids to monoids" is not defined; same for "lax monoidal". Here is one version of your question how I understand it: Let $(\mathcal{C},\otimes),(\mathcal{D},\otimes)$ be two monoidal categories, let $F : \mathcal{C} \to \mathcal{D}$ and $F ' : \mathrm{Mon}(\mathcal{C},\otimes) \to \mathrm{Mon}(\mathcal{D},\otimes)$ be two functors such that $U_{\mathcal{D}} F'=FU_{\mathcal{C}}$, where $U_{\mathcal{C}}$ denotes the forgetful functor $\mathrm{Mon}(\mathcal{C},\otimes) \to \mathcal{C}$. (Do you want to assume more about $F'$?). [...] Jul 1 at 10:08
• Question. Is there a lax monoidal functor $(F,\eta,\mu) : (\mathcal{C},\otimes) \to (\mathcal{D},\otimes)$ (with underlying functor $F$) such that $\mathrm{Mon}(F,\eta,\mu) : \mathrm{Mon}(\mathcal{C},\otimes) \to \mathrm{Mon}(\mathcal{D},\otimes)$ is equal to $F'$? Jul 1 at 10:09

Let $$P$$ be a partially ordered monoid. A monoid in $$P$$ is an element $$a$$ of the underlying set satisfying $$1 \leq a$$ and $$a^2 \leq a$$ (and hence $$a^2=a$$). (So here, a monoid structure is merely a property.) If $$1$$ happens to be the largest element of $$P$$, it follows that $$1$$ is the only monoid in $$P$$. It follows that if $$f : P \to Q$$ is an increasing map with $$f(1)=1$$, it preserves monoids. The question is if we can conclude that $$f$$ is lax monoidal, which means that $$f(a) \cdot f(b) \leq f(a \cdot b)$$ holds for all $$a,b \in P$$. We cannot.
For example, let $$P = Q = ([0,\infty],\geq,+,0)$$ (this example appears in the definition of Lawvere metric spaces, which are just $$P$$-enriched categories). The monoidal unit is $$0$$ and it is the largest element with respect to $$\geq$$. Consider the map $$f(a):=a^2$$. The lax monoidal condition would mean $$a^2 + b^2 \geq (a+b)^2$$, which is not true.
Edit. Alternatively, if $$P$$ is any partially ordered group, $$1$$ is the only monoid in $$P$$ (since $$a^2=a \implies a=1$$). The map $$f : P \to P$$, $$f(a):=a^2$$ is increasing and satisfies $$f(1)=1$$, but it is lax monoidal iff $$a^2 b^2 \leq (ab)^2$$ or all $$a,b$$ iff $$ab \leq ba$$ for all $$a,b$$ iff $$ab=ba$$ for all $$a,b$$ iff $$P$$ is commutative. But there are non-abelian partially ordered groups.