In the comments section of [this question](https://mathoverflow.net/questions/301882/on-functors-preserving-monoid-objects#comment879728_301882) 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



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> 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?
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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.



Functors that preserve monoids
A simple class of counterexamples can be found among [thin](https://ncatlab.org/nlab/show/thin+category) [small](https://ncatlab.org/nlab/show/small+category) [strict monoidal categories](https://ncatlab.org/nlab/show/strict+monoidal+category), 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](http://www.tac.mta.ca/tac/reprints/articles/1/tr1abs.html), 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](https://math.stackexchange.com/questions/353160) non-abelian partially ordered groups.


