Let $C_1$ and $C_2$ be monoidal categories (not necessarily symmetric or strict) and let $\Psi : C_1 \rightarrow C_2$ be a strong monoidal functor. Is it possibly to construct a strict monoidal functor $\Psi'$ which is naturally isomorphic to $\Psi$? If not, is this possible if we assume that the $C_i$ are strict?
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$\begingroup$ Not an answer, but in the general context of 2-monads, a (strict) algebra $A$ with the property that every pseudo morphism with domain $A$ (and strict codomain) is isomorphic to a strict one is called semi-flexible. See Blackwell-Kelly-Power, 2-dimensional monad theory, Theorem 4.7. In general, not every algebra is semi-flexible, and that is probably also the case here. $\endgroup$– Mike ShulmanCommented Jun 29, 2014 at 6:29
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3 Answers
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Take $C_2$ to be a non-strict monoidal category, $C_1$ to be its strictification and $\Psi$ to be the equivalence. Since $C_2$ is not strict there's no strict monoidal equivalence between $C_1$ and $C_2$, so in particular $\Psi$ is not naturally isomorphic to a strict functor.
answered Jun 27, 2014 at 18:15
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$\begingroup$ Great, I should have noticed that. I'm going to hold off on accepting it to see if someone answers the second part of my question (namely, what if we assume that both of the $C_i$ are strict). $\endgroup$– JillCommented Jun 27, 2014 at 18:49
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1$\begingroup$ Why does $C_2$ not being strict imply that there's no strict monoidal equivalence between $C_1$ and $C_2$? The strictness of the functors between $C_1$ and $C_2$ is independent of the strictness of $C_1$ and $C_2$. $\endgroup$– varkorCommented Sep 25 at 9:15
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I think the answer to the second part (both $C_i$ strict) is negative.
Fix a finite group $G$ with a non-trivial 2-cocycle $\gamma \in H^2(G, \mathbb{C}^{\times})$. Let $C$ be the strict $\mathbb{C}$-linear monoidal category with simple objects $\{g: g \in G\}$ and tensor product $g \otimes h = gh$. Then the 2-cocycle $\gamma$ can be used to define a monoidal functor $(1_C, \gamma): C \to C$. This functor is not naturally isomorphic to a strict functor (the only possibility is it is isomorphic to the (strict) identity functor, but this cannot be the case since $\gamma$ is non-trivial).
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$\begingroup$ Could you please say a little bit more how the functor $(1_C,\gamma):C\to C$ is defined? $\endgroup$– jijijojoCommented May 7, 2020 at 21:35
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3$\begingroup$ The functor $(1_C, \gamma)$ is the identity on objects and morphisms, and the monoidal structure maps $g \otimes h \to g \otimes h$ are given by $\gamma(g,h)$ times the identity of the object $gh$ (the cocycle condition ensures that the axioms for a monoidal functor are satisfied, and two cocycles which differ by a coboundary give monoidally naturally isomorphic functors) $\endgroup$ Commented May 8, 2020 at 22:34
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Here is another example that shows the answer to the second part (where both $C_i$ are strict) is negative.
Let $C_1$ be the discrete strict monoidal category whose monoid of objects is the Booleans. Let $C_2$ be the following strict monoidal category: its underlying monoid of objects is the naturals, and all objects > 0 have a unique morphism between them. There are no other nonidentity morphisms.
There is a unique strict monoidal functor from $C_1$ to $C_2$, but there are other (strong) monoidal functors (equivalences, all isomorphic to each other) that are not isomorphic to it.
