8
$\begingroup$

In Gurskis definition (page 32 of his thesis) of a strict functor $F$ of tricategories he requires that

$F$ maps the adjoint equivalences $a,l,r$ in the source tricategory to the same adjoint equivalences in the target tricategory.

This means for example $F (a_{fgh}) = a_{FfFgFh}$. (Note that these morphisms are indeed parallel, because of the strictness of $F$.)

However it is not required, that the modifications in the definition of a tricateory $(\pi,\lambda,\mu,\rho)$ are preserved as well. For example $ F (\pi _{fghk}) $ is not required to be equal to $ \pi_{FfFgFhFk} $. So my question is:

Why isn't it required that $(\pi,\lambda,\mu,\rho)$ are preserved? Is this because this is already implied by the other requirements? Or are there strict functors which doesn't preserve the modifications $(\pi,\lambda,\mu,\rho)?$

EDIT

Let me draw some parallel to the definition of strict functors between bicategories: ($*$ denotes composition along objects)

There it is only required that the constraint-2-cells $$ F(f)*F(g) \to F(f*g) $$ and $$ 1_{Fa} \to F(1_a) $$ are identities. That $a,l,r$ are preserved is not demanded by definition, but it follows easily from the functor axioms.

Therefore one might hope that in the case of a strict functor of tricategories it follows somewhat analoge that the "highest" constraint data $\pi,\lambda,\mu,\rho$ are preserved, although it is not demanded by definition.

$\endgroup$
2
$\begingroup$

The axioms in the definition of a non-strict trihomomorphism (Definition 3.3.1, page 31-32) already require that the modifications $\pi,\lambda,\mu,\rho$ are preserved "up to" the coherences $\pmb{\chi}, \pmb{\iota}, \omega,\gamma,\delta$. The definition of a strict functor (3.3.3, p33) then requires that $\pmb{\chi}$ and $\pmb{\iota}$ are identities while $\omega,\gamma,\delta$ are "as close to identities as possible" (they can't literally be identities because their domain is not equal to their codomain), which therefore implies that $\pi,\lambda,\mu,\rho$ are "preserved as much as possible".

$\endgroup$
3
  • $\begingroup$ Please tell me if I am mistaken: The local strictness of $F$ together with the requirement that $\chi$ is an identity ensures that $F$ preserves all kinds of composition "on the nose", i.e. $F(x \blacksquare y) =F(x) \blacksquare F(y) $ for whatever composition $\blacksquare$ stands for. This, together with the requirement that $a$ gets preserved on the nose ensures that $F(\pi_{fghk})$ and $\pi_{FfFgFhFk}$ are parallel morphisms (same source and target) in the target tricategory. Therefore "$\pi$ is preserved as much as possible" would mean "$\pi$ is preserved on the nose"? $\endgroup$ – Peter Guthmann Apr 20 '18 at 8:22
  • 1
    $\begingroup$ @PeterGuthmann sounds possible, but I'm not confident enough to say definitely yes. I suggest asking Nick directly. $\endgroup$ – Mike Shulman Apr 21 '18 at 2:34
  • $\begingroup$ I followed your advise and Nick Gurski confirmed that strict tricat-functors indeed preserve also the top level coherence-cells strictly. At the moment I don't understand the reason well enough to write a real answer, though. $\endgroup$ – Peter Guthmann Apr 23 '18 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.