A terminological question regarding bicategories

Question

My question is very simple. I have come accross some literature regarding bicategories (Benabou, etc) and I am a little confused on the terminology:

Could you please tell me if bicategories and 2-categories are exact synonyms? If not, I would appreciate some clarification regarding the differences.

I guess the kind of answer will also be suitable for 3-categories.....

Thanks in advance.

Answer 1

A $2$-category is just a category enriched over the category $\mathsf{Cat}$ of small categories and functors between them. The typical example is the $2$-category $\underline{\mathsf{Cat}}$ whose objects are small categories and the category $\underline{\mathsf{Cat}}(A,B)$ between two such objects is the category of functors $A \to B$ and natural transformations between them. Incidentally, the underlying ordinary category of $\underline{\mathsf{Cat}}$ is $\mathsf{Cat}$.

Because of the way enriched categories are defined, a $2$-category $\mathcal K$ comes with composition functors $$ c_{A,B,C} : \mathcal K(A,B) \times \mathcal K(B,C) \to \mathcal K(A,C) $$ such that the following commutes in $\mathsf{Cat}$:

Look long enough at it and you'll see it says that associativity of composition in $\mathcal K$ holds on the nose. In particular, given $1$-cells $f:A\to B$, $g:B\to C$ and $h:C\to D$, we have an equality : $$h\circ(g\circ f) = (h \circ g)\circ f$$ where $\circ$ stands for the correct composition.

Now, recall that equality between functors is not very handy and is better replaced by natural isomorphisms: that is why we prefer to work with equivalences of categories instead of isomorphisms of categories. So why don't we do the same conceptual leap here and ask the big diagram to commute only up to natural isomorphism ? This is what bicategories do ! Of course, if you do that for the associativity diagram, you want to do that also for the identity laws. And what to do with all those natural isomorphisms ? How do they interact with each other ? By fiddling around enough, you will eventually get back to Mac Lane's triangles and pentagone that appears in the definition of a bicategory. Archetypal examples of bicategories depend on what are your mathematical preferences but I like the following one:

For every sets $A,B$ you have a category $\mathrm{Span}(A,B)$ whose objects are the pairs $A\overset s\leftarrow X \overset t\to B$ of functions from a common set, and a morphism $(X,s,t) \to (Y,s',t')$ is just a function $f:X\to Y$ such that $s = s'f$ and $t=t'f$. Of course there is a distinguished element $(A,\mathrm{id}_A,\mathrm{id}_A)$ in each $\mathrm{Span}(A,A)$. Finally there is a way to shape the data of $A\overset s\leftarrow X \overset t\to B$ and $B\overset {s'}\leftarrow X' \overset {t'}\to C$ into a span $A\overset {s''}\leftarrow X'' \overset {t''}\to C$ by just taking $X''$ to be the pullback of $X$ and $X'$ above $B$.

So great ! We should be able to make a $2$-category out of it with objects the sets and the category between $A$ and $B$ being $\mathrm{Span}(A,B)$. But no. The problem is that we have chosen pullbacks in order to define the composition of two spans and that pullbacks are only defined up to (unique) isomorphism. So given $A\overset {s_0}\leftarrow X_0 \overset {t_0}\to B$, $B\overset {s_1}\leftarrow X_1 \overset {t_1}\to C$ and $C\overset{s_2}\leftarrow X_2 \overset {t_2}\to D$, I can only prove $$ (X_2,s_2,t_2) \circ ((X_1,s_1,t_1) \circ (X_0,s_0,t_0)) \simeq ((X_2,s_2,t_2) \circ (X_1,s_1,t_1)) \circ (X_0,s_0,t_0) $$ and I can not replace $\simeq$ by $=$... If I am very thorough though, I can prove that all those natural isomorphisms respect the pentagone law. Similarly for the identity laws. So even if I don't have a $2$-category, I still have a bicategory $\mathrm{Span}$ to work with!

But behold : every bicategory is $2$-equivalent to a $2$-category. What it means is more or less that you have to be aware that in a bicategory you don't have associativity and identities on the nose, but it does not matter too much to manipulate a bicategory as if it were a $2$-category for most of the things that you can tell in the language of bicategories.

(If you are familiar with monoidal category, the same happens there : you have to recall the isomorphisms for associativity and unital laws but when dealing with monoidal properties, you can more or less work as if you were in a strict monoidal category. Actually, strictification of monoidal categories is a special case of the strictification of bicategories.)

Answer 2

Short answer: Bicategories are weak 2-categories, in the sense that the laws for composition of $1$-cells hold up to coherent isomorphism in bicategories but hold strictly in 2-categories.

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Long answer: Bicategories and 2-categories both have $0$-cells, $1$-cells and $2$-cells. Where they differ is that composition of 1-cells is weak in a bicategory and strict in a 2-category, in the following sense: let $$A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} D$$ be three $1$-cells. Then

• In a 2-category, we have $(h \circ g) \circ f = h \circ (g \circ f)$ and $\mathrm{id}_B \circ f = f = f \circ \mathrm{id}_A$;

• In a bicategory, we have invertible $2$-cells (called coherence isomorphisms): $$\alpha : (h \circ g) \circ f \Rightarrow h \circ (g \circ f)$$ $$\lambda : \mathrm{id}_B \circ f \Rightarrow f$$ $$\rho : f \circ \mathrm{id}_A \Rightarrow f$$ called the associator, left unitor and right unitor, respectively, which are required to satisfy certain axioms called coherence axioms.

Equivalently, a 2-category is a bicategory whose coherence isomorphisms are identities.

For tri- and 3-categories it's even trickier; there's a whole book about it (Coherence in three-dimensional category theory by Nick Gurski), where it takes several pages to even write down the definition of a tricategory.