Quantcast
Channel: Can equivalences be strictified to isomorphisms? - MathOverflow
Viewing all articles
Browse latest Browse all 3

Can equivalences be strictified to isomorphisms?

$
0
0

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?


Viewing all articles
Browse latest Browse all 3

Latest Images

Trending Articles





Latest Images