3
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 30 at 17:29
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 30 at 17:49
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 30 at 17:55
  • 2
    $\begingroup$ "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 $F'$?). [...] $\endgroup$ Jul 1 at 10:08
  • 2
    $\begingroup$ 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 $F'$? $\endgroup$ Jul 1 at 10:09
6
$\begingroup$

A simple class of counterexamples can be found among thin small strict monoidal categories, i.e. preordered monoids. These are just sets equipped with a preorder and a monoid structure such that the multiplication is increasing in each variable. For simplicity, let us only look at partial orders.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.