### Bicat is Not Triequivalent to Gray

#### Posted by David Corfield

Stephen Lack has a paper with that title out today. Abstract:Posted at December 13, 2006 8:43 AM UTCBicat is the tricategory of bicategories, homomorphisms, pseudonatural transformations, and modifications. Gray is the subtricategory of 2-categories, 2-functors, pseudonatural transformations, and modifications. We show that these two tricategories are not triequivalent.

## Re: Bicat is Not Triequivalent to Gray

Since Toby did me the favour of summarizing my paper here, I will do Steve the favour (?) of summarizing his paper here.

As it says in his abstract (in old-fashioned terminology), he compares the tricategories

(everything weak) and

(a mixture of strengths).

Steve proves that

BicatandGrayare not triequivalent. (I’ll drop the ‘tri’.) And Steve, being a precise person, really means that. In other words, he doesn’t just prove that the canonical inclusion $\mathbf{Gray} \rightarrow \mathbf{Bicat}$ is not an equivalence; he proves that there is no equivalence in the world.Actually, the first step

isto prove that this inclusion isn’t an equivalence. The proof boils down to exhibiting a weak functor between strict 2-categories that is not equivalent to any strict functor between them. This is not new (though may be due to Steve originally).The second step is to show that if there were

anyequivalence betweenBicatandGray, then the inclusion would be an equivalence. But we’ve just seen that it isn’t: QED. This is the new part.Steve then makes some comments about coherence for bicategories. As I understand it, the thought lurking in the background is the following: if you cite the coherence theorem as ‘every weak 2-category is equivalent to a strict one’ then you’re ignoring functors, transformations etc between 2-categories. To be more holistic, we should be looking for a statement of the coherence theorem that takes account of them. Steve’s result has a bearing on this.

I’ll probably say something about this in my Fields talk next month, though presumably Steve will present his result in his own talk.