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In category theory there are lots of examples of isomorphisms that cannot be strictified to become identities. For instance, every monoidal category is equivalent to a strict monoidal category, where the associativity and unit isomorphisms are identities, but not every braided monoidal category is equivalent to a strictly braided one where the braiding is an identity.

In higher category theory, constraints are in general (adjoint) equivalences rather than isomorphisms. For instance, the associativity and unit 2-cell constraints for a tricategory (weak 3-category) are equivalences, but the 3-cell constraints (such as the interchanger) are still isomorphisms, since there is "no room" for anything weaker (there being no 4-cells in a tricategory). Every tricategory is equivalent, not to a strict 3-category where all constraints are identities, but to a Gray-category where the associators and unitors are identities but the interchanger is not.

I am looking for an example of a higher-categorical structure containing constraint equivalences which cannot be strictified even to become isomorphisms (not necessarily identities). Since the only nontrivial constraints in a Gray-category are top-dimensional and hence isomorphisms, the first place to look for this would be in some sort of 4-category. But we can also make it more manageable by being somewhat degenerate. A triply degenerate 4-category (exactly one 0-, 1-, and 2-cell) would be (by the delooping hypothesis) a symmetric monoidal category, with no room for any equivalences that aren't isomorphisms, so the next level of complexity seems the first place to look.

A doubly degenerate 4-category should be the same as a braided monoidal bicategory, and by the coherence theorem for tricategories, everyone of those is equivalent to a braided Gray-monoid (a Gray-monoid being a Gray-category with one object). Howver, the braiding in a braided Gray-monoid is, a priori, still only an equivalence, so one way to make this question precise would be:

Is every braided Gray-monoid equivalent to one whose braiding is an isomorphism, rather than merely an equivalence?

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  • $\begingroup$ A priori, the final question you ask seems a shade stronger than what you started off with. There are other situations where it's know that you can make aspect A or aspect B of something strict, but not both at once; and it would seem conceivable that, say, every braided Gray-monoid could have its braiding strictified to an equivalence, but that this process might destroy the Gray-ness, giving just a braided monoidal bicategory, and then Gray-ifying it back might turn the braiding back into just an equivalence? Or does Gray-ification preserve the property of braiding being iso? $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jun 8, 2010 at 15:42
  • $\begingroup$ Well, I said "one way to make this question precise would be..." (-: I think the question is most interesting if we ask for some other semistrictness in addition, though, since we know that (in low dimensions at least) interchange can actually be turned into an identity as long as the unit constraints are weak enough. $\endgroup$
    – Mike Shulman
    Commented Jun 8, 2010 at 20:27
  • $\begingroup$ Also, the notion of "1-cell isomorphism" doesn't make a whole lot of sense in a bicategory that's not a strict 2-category. Even identities in a bicategory are not usually isomorphisms! But one could ask about a braided monoidal strict-2-category which is weaker than a Gray-monoid, I guess. $\endgroup$
    – Mike Shulman
    Commented Jun 8, 2010 at 20:29
  • $\begingroup$ I think it is true that every Gray-group (a Gray-groupoid with one object) is equivalent to one arising from a 2-crossed module. One could see what the braiding becomes under this equivalence, but this isn't exactly an impossibility result, more like a conditional impossibility result (can't have your cake and eat it) $\endgroup$
    – David Roberts ♦
    Commented Mar 27, 2012 at 22:59

2 Answers 2

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I'm surprised that I didn't notice this immediately, and that no one else pointed it out either.

Every category is equivalent to a skeletal one. Therefore, every bicategory is (bi)equivalent to one whose hom-categories are skeletal. But in a bicategory of this sort, every 1-cell equivalence is an isomorphism. Therefore, every (braided, symmetric, ...) monoidal bicategory is monoidally equivalent to one in which all the 1-cell equivalence constraints are isomorphisms.

This is definitely an instance of the "whack-a-mole" aspect of coherence that Peter mentioned, though, since we can't expect to make a bicategory both strict (i.e. a 2-category) and locally skeletal. So it doesn't answer the "one way to make the question precise" that I asked, but it says something about the imprecise version.

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  • $\begingroup$ The equivalence between a category and a skeletal one, in general, uses Choice. It may be important in some cases to know whether the equivalence/isomorphism involved has computational content or not. $\endgroup$
    – Jacques Carette
    Commented Jul 6, 2015 at 15:33
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It appears that the answer to the corresponding question for symmetric Gray-monoids is yes; this was shown by Schommer-Pries and cleanly reformulated by Gurski-Johnson-Osorno. Nick says that he thinks the braided case should work similarly.

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