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The coherence theorem for bicategories, as usually stated, reads

Any bicategory $B$ is biequivalent to a (strict) 2-category.

It is possible to give an explicit construction of the strictification as the full image of its Yoneda embedding $y:B\rightarrow [B,\text{Cat}]$, see for instance this reference.

This seems like a natural construction, so I would expect an equivalence of tricategories

$$ \text{Bicat} \leftrightarrows 2\text{-Cat}$$

where the rightgoing functor is the full image of the yoneda embedding, and the leftgoing functor is the inclusion. However, I cannot find such a statement in the literature. If it is true, a reference would be appreciated.

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    $\begingroup$ If you want a quick overview of the situation, you could try the slides here, especially pages 8 and 14: maths.ed.ac.uk/~tl/toronto $\endgroup$
    – Tom Leinster
    Commented Mar 2, 2015 at 14:06

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Probably you had some trouble finding this because the search term $2$-$\text{Cat}$ is not accurate enough; you want not the cartesian monoidal product on $2$-$\text{Cat}$, but rather what is called the Gray monoidal product; the tricategory you want then is denoted $\text{Gray}$, the tricategory of strict 2-categories, strict 2-functors, pseudonatural transformations, and modifications between them, but most usefully considered as equipped with the Gray tensor product.

Anyway, see this paper by Steve Lack: Bicat is not triequivalent to Gray. This actually corrects a slight misstatement in the monograph by Gordon-Power-Street. See the end of Lack's paper for the punchline. I'm not sure what the state of the art is, but perhaps you could email Lack (or perhaps Street).

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    $\begingroup$ Just adding to Todd's answer, we the op writes "2-cat" it is ambiguous. The objects are certainly strict 2-cats, but what are the bicats of morphisms? If they are all weak functors, transformations, and modifications, then the answer to the op's question is "yes". However if what is meant is that we use strict functors and strict transformations (and modifications) then the answer is "no, these are not equivalent", as Lack shows. The key problem is that there can be weak functors between strict 2-categories which are not equivalent to strict functors between the same 2-categories. (cont) $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 2, 2015 at 14:36
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    $\begingroup$ However something like the above statement is true if you restrict the class of 2-categories even further. There is a class of "cofibrant" 2-categories which do have the property that every weak functor out of them is equivalent to a strict functor, and similarly for transformations. Most proofs of the coherence theorem actually show that every bicategory is equivalent to a cofibrant 2-category (for example the proof via Yonneda embedding does this, though not obviously). So is "2-cat" means cofibrant 2-categories, strict functors, strict transformations, and modifications then (cont) $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 2, 2015 at 14:41
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    $\begingroup$ the answer is again yes. See remark 3 at the end of Lack's paper which Todd mentions. $\endgroup$
    – Chris Schommer-Pries
    Commented Mar 2, 2015 at 14:43

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