# How does one Segal-subdivide a 2-category?

Let $\mathcal{C}$ be a small category. Then, its Segal subdivision $\text{sd }\mathcal{C}$ is a new category whose objects are morphisms of $\mathcal{C}$, and a morphism from $f:x \to y$ to $g: w \to z$ is a pair of morphisms $a: x \to w$ and $b: y \to z$ in $\mathcal{C}$ so that $f = b\circ g \circ a$. It is easy to see (by a contractible-fiber argument) that there is a homotopy equivalence of classifying spaces $B\text{Sd }\mathcal{C} \simeq B\mathcal{C}$. My first encounter with this construction was in the preprint on Morse theory and classifying spaces (link here).

Here's the question:

What is the analogue of Segal subdivision for small (strict) $2$-categories?

In particular, from a given small $2$-category $\mathcal{D}$, I would like to construct a new $2$-category $\text{sd}_2\mathcal{D}$ whose one-skeleton coincides with the ordinary Segal subdivision of the one-skeleton of $\mathcal{D}$. Some mysterious and powerful $2$-morphisms should exist, and their addition should magically make the classifying spaces of $\text{sd}_2\mathcal{D}$ and $\mathcal{D}$ homotopy-equivalent.

What, if anything, is the precise collection of $2$-morphisms which achieve the homotopy equivalence, and where can I find this written down?

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I'd say this construction was first considered by MacLane in his book, who called it twisted arrow category. Then by Baues and Wirsching in their work on cohomology of categories, under the name of factorization category, which I like very much because it reflects exactly what it is. –  Fernando Muro Jun 29 '13 at 14:39
In fact, the construction of sd as you sketch was also in early work on extensions of categories:Charles Wells, Extension theories for categories (preliminary report), (available from cwru.edu/artsci/math/wells/pub/pdf/catext.pdf), 1979. –  Tim Porter Jun 29 '13 at 15:33
My guess is that you have to see what properties of this factorisation category do you need and look at those. There is related work, I think, by Cegarra et al on the nerve of a 2-category. –  Tim Porter Jun 29 '13 at 15:35
Tim: I assume you are referring to the 10 different (but homotopy equivalent!) constructions of 2-categorical nerves found here: arxiv.org/abs/0903.5058‎ ? I'm not quite sure how it relates to the question. –  Vidit Nanda Jun 29 '13 at 18:49
By the way, $BSdC$ and $BC$ have homeomorphic realizations. Each $n$-cell is subdivided into $2^n$ little $n$-cells. –  Tom Goodwillie Dec 11 '13 at 14:55

Let me make a proposal. I won't check that the property you demand about nerves is satisfied. I suspect it holds, but the checking may require some work, so if I did it I would be tempted to do something with it. Since you may really want to do something with it, I leave it to you ;-)

Given a $2$-category $\mathcal C$, define the 2-category of factorizations $F(\mathcal C)$ as follows. Objects in $F(\mathcal C)$ are morphisms in $\mathcal C$, $f\colon X\rightarrow Y$. A morphism in $F(\mathcal C)$ from $f\colon X\rightarrow Y$ to $g\colon U\rightarrow V$ is a diagram in $\mathcal C$ as follows

$$\begin{array}{ccccc} &&h\\ &X&\leftarrow&U\\ f&\downarrow&\Rightarrow&\downarrow&g\\ &Y&\rightarrow&V\\ &&k \end{array}$$

A $2$-cell in $F(\mathcal C)$ from this morphism to

$$\begin{array}{ccccc} &&h'\\ &X&\leftarrow&U\\ f&\downarrow&\Rightarrow&\downarrow&g\\ &Y&\rightarrow&V\\ &&k' \end{array}$$

consists of $2$-cells in $\mathcal C$, $h\Rightarrow h'$ and $k\Rightarrow k'$ such that the pasting of

$$\begin{array}{ccccc} &&h\\ &X&\leftarrow&U\\ f&\downarrow&\Rightarrow&\downarrow&g\\ &Y&\stackrel{k}\rightarrow&V\\ &||&\Downarrow&||\\ &Y&\stackrel{k'}\rightarrow&V \end{array}$$

coincides with the pasting of

$$\begin{array}{ccccc} &X&\stackrel{h}\leftarrow&U\\ &||&\Downarrow&||\\ &X&\stackrel{h'}\leftarrow&U\\ f&\downarrow&\Rightarrow&\downarrow&g\\ &Y&\rightarrow&V\\ &&k' \end{array}$$

Horizontal and vertical composition in $F(\mathcal C)$ are defined in the obvious way.

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Thanks again, Fernando. I will work out whether this construction provides a classifying space that is homotopy equivalent to the original 2-category or not. –  Vidit Nanda Jun 30 '13 at 3:44
You're welcome. –  Fernando Muro Jun 30 '13 at 6:49

Another, somewhat similar, suggestion, in a cubical setting, which produces a double category:

In the picture I wrote the one morphisms as subdivisions, but if you want to see it as a double category, then these should be the vertical morphisms, and the top and bottom faces are the horizontal morphisms.

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I’m finding it a bit tricky to fill in the omitted labels on your commutative cube. Could you possibly elaborate on how and why your suggestion differs from Fernando Muro’s? –  Peter LeFanu Lumsdaine Dec 11 '13 at 16:21
The missing labels are the data of the 2 morphism, I.e. Arbitrary edges and faces which make the whole thing commute. The idea was to subdivide the 2 morphism on the face into a composition of 2 morphisms, mimicking segals construction for 1 morphisms. –  Adam Gal Dec 12 '13 at 0:25
As far as the difference from Fernando's suggestion - his suggestion is of the globular kind, I.e. a 2 morphism is between two 1 morphisms with the same source and target, and my suggestion is of the cubical kind, where this restriction is relaxed somewhat. –  Adam Gal Dec 12 '13 at 0:28
Ah, thanks; now you say that you’re describing a cubical category, your diagrams make sense. May I suggest editing your answer to state that upfront? It’s a bit confusing as it stands currently. (Precisely, I guess that what you’re constructing here is something like a double bicategory in the sense of Verity — is that right?) –  Peter LeFanu Lumsdaine Dec 12 '13 at 0:48
I'm not sure what verity's definition entails, but I think that what I described can be seen as just a double category. –  Adam Gal Dec 12 '13 at 0:52