# When is a functor of bicategories part of an equivalence?

Assuming the axiom of choice (I think), $F: \mathcal{C} \to \mathcal{D}$ is part of an equivalence of categories iff $F$ is fully faithful and essentially surjective on objects. This is sometimes handy if you want to show an equivalence, but the weak inverse functor is too cumbersome to define.

What is the generalisation of this to equivalences of bicategories?

• It is in fact equivalent to the axiom of choice. – Zhen Lin Jan 27 '16 at 14:42

Weak equivalences of (either strict or weak) $n$-categories are defined inductively:

$F : \mathcal{C} \to \mathcal{D}$ is a weak equivalence of $n$-categories if it has the following properties:

• For every object $Y$ in $\mathcal{D}$, there is an object $X$ in $\mathcal{C}$ and an equivalence $F (X) \simeq Y$ in $\mathcal{D}$.
• For every pair $(X_0, X_1)$ of objects in $\mathcal{C}$, $F : \mathcal{C} (X_0, X_1) \to \mathcal{D} (F (X_0), F (X_1))$ is a weak equivalence of $(n-1)$-categories.

Of course, the base case is $n = -2$: every functor between $(-2)$-categories is a weak equivalence, by definition.

Warning: While it is true that a strict functor between 1-categories that is a weak equivalence has a quasi-inverse that is a strict functor, this fails already for 2-categories; but if you want to work with bicategories, you should not be thinking in terms of strict functors anyway.

• Right, but it has a "pseudofunctor" as inverse? Is there a reference with this definition? – Turion Jan 28 '16 at 9:40
• Yes, there will be a pseudofunctor quasi-inverse. It's briefly mentioned in [Lack, A 2-categories companion]. – Zhen Lin Jan 28 '16 at 9:55