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?

oneway 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 anidentityas long as the unit constraints are weak enough. $\endgroup$ – Mike Shulman Jun 8 '10 at 20:27andeat it) $\endgroup$ – David Roberts Mar 27 '12 at 22:59