By strict equivalence, I mean a monoidal equivalence whose underlying monoidal functors are strict, and here I am looking for two monoidal categories which are not strictly equivalent.

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Here is a very universal/abstract example:

(The following is some very classical machinery, but I'm not very familiar with the literature on this so I'm not sure which reference should be quoted)

Let $\textbf{Mon}$ be the $2$-category of monoidal categories and (pseudo) monoidal functor between and $\textbf{SMon}$ the category of strict monoidal category and strict monoidal functor between them.

The forgetfull functor $\textbf{SMon} \rightarrow \textbf{Mon}$ has a left adjoint $F: \textbf{Mon} \rightarrow \textbf{SMon}$ defined as follow: given a monoidal category $C$, the set of objects of $F(C)$ is the free monoids on the objects of $C$. Given two objects $a,b \in F(C)$, formally written as $a_1 \otimes \dots \otimes a_n$ and $b_1 \otimes \dots \otimes b_m$ (where $\otimes$ denotes the product in the free monoid), one defines $Hom_{F(C)}(a,b)$ to be $Hom_C(\overline{a},\overline{b})$ where $\overline{a}$ and $\overline{b}$ denotes object obtain by computing the tensor product $a_1 \otimes \dots \otimes a_n$ and $b_1 \otimes \dots \otimes b_m$ in $C$, as these object are well defined up to unique isomorphisms, there is no ambiguity on what is the Hom set between them.

In particular, $F$ is a comonad on $\textbf{SMon}$ whose co-Kleisli category is the category of strict $2$-category and pseudo-monoidal functor between them.

Given a strict monoidal category $C$, the counit $F(C) \rightarrow C$ is a strict monoidal functor, and is always an equivalence of category. So it is an equivalence of monoidal category. But requiring it to have a strict inverse give a very strong property to $C$:

Lemma: For any strict monoidal category $C$, The following are equivalent:

  • The equivalence $F(C) \rightarrow C$ have a strict monoidal inverse.

  • Any pseudo-monoidal functor $C \rightarrow D$ is equivalent (as pseudo monoidal functor) to a strict monoid monoidal functor $C \rightarrow D$.

  • Any strict monoidal functor $E \rightarrow C$ which is an equivalence has a strict monoidal inverse.

(Note: Is there a standard name for this kind of objects ? maybe 'flexible' has been used for this ? )

Indeed, A pseudo-monoidal functor $C \rightarrow D$ is the same as a strict monoidal functor $F(C) \rightarrow D$, hence if $\theta:C \rightarrow F(C)$ is a strict monoidal inverse to $F(C) \rightarrow C$ precomposing $F(C) \rightarrow D$ with $\theta$ gives a strict monoidal functor $C \rightarrow D$ which as a pseudo-monoidal functor is equivalent to one we started from.

The third point follow from the second because any strict monoidal functor which is an equivalence has a pseudo-monoidal inverse, so if every pseudo-monoidal functor out of $C$ is equivalent to a strict one, one obtains a strict inverse. The first point immediately follow from the third.

So finding an example as asked by the question is exactly the same as finding a $C$ that do not satisifes these conditions: $F(C) \rightarrow C$ will be an example, and conversely given any example as in the question, then at least one of the two monoidal category involved do not satisfies these equivalent conditions.

Most strict monoidal categories works, but to give a very simple example:

Take $M$ to be a monoid, seen as discrete strict monoidal category, then $F(M) \rightarrow M$ has a strict inverse if and only if the map of monoid $LM \rightarrow M$ (where $LM$ denotes the free monoid on $M$) admits a retraction. Which I think only happen when $M$ is free, and anyway do not happen for most monoids.

  • $\begingroup$ Regarding your question, such monoidal categories are (by definition) the semi-flexible ones, which are (by a theorem) those monoidally equivalent to a truly flexible one. A flexible monoidal category is one cofibrant for the canonical model structure constructed by Lack, equivalently, one for which $F(C)\to C$ admits a strict monoidal section, which is automatically an equivalence inverse as $F(C)\to C$ is fully faithful. $\endgroup$ – Kevin Carlson Sep 13 at 20:26
  • $\begingroup$ Should've included a reference: arxiv.org/pdf/1112.1448.pdf, and above I should have said that the semi-flexibles are those strictly monoidally equivalent to the true flexibles. $\endgroup$ – Kevin Carlson Sep 13 at 20:47

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