# Intuition behind Mac Lane's “subdivision category”

I've always felt that in proving that co/ends are co/limits, Mac Lane's CWM makes use of a category apparently coming out of nowhere.

Let $$C$$ be a category; I define the subdivision graph of $$C$$ to be the digraph having a vertex $$c^\S$$ for each object $$c\in C$$, and a vertex $$f^\S$$ for each morphism $$f : c\to c'$$ in $$C$$, and edges all the arrows $$s(f)^\S \to f$$ and $$t(f)^\S \to f^\S$$, as $$f$$ runs over morphisms of $$C$$.

Formally adding identities, and taking the trivial function as composition law (so that $$u\circ v$$ is only defined when at least one arrow is an identity) turns the subdivision graph of $$C$$ into a category, the subdivision category of $$C$$.

Now, you can say many things about this definition, but certainly not that it is easy to see where it comes from: Mac Lane himself stresses how outside of page 220 (1st edition), where it is used to show that every functor $$F : C°\times C\to D$$ induces a functor $$F^\S : C^\S \to D$$, and the end of $$F$$ is the limit of $$F^\S$$, there will be no other mention of $$C^\S$$. So

where does this definition come from? What is the intuition behind it? How did Mac Lane (or whoever else mentioned it first) come up with it?

• I cannot possibly know what Mac Lane thought, but it would seem to me that this definition is specifically designed to make the cited theorem about colimits and coends true. The very special form of the indexing category F^§ allows us to compute the colimit over F^§ as the coequalizer of two maps between the corresponding coproducts, and from there it is easy to get to coends. – Dmitri Pavlov Sep 2 '19 at 15:52
• Yes, I think clearly this category is cooked up precisely so as to make the theorem true. But as Tim's answer points out, there are also other categories that suffice to make the same theorem true and that are less ad hoc. – Mike Shulman Sep 4 '19 at 3:24
• I don't see why you say other : afaicu what Tim is pointing out is that you can consider the subdivision category as the image of $C$ under a functor, but this functor is better appreciated at the level of $\infty$-categories, because at that level it's a subdivision functor $$X\mapsto \text{sd}(X) = \int^n X_n\times N([n]\cup [n]^{op} )$$ – Fosco Sep 5 '19 at 21:46

I believe that the nerve of the subdivision category is the edgewise subdivision of the nerve. The edgewise subdivision is the left Kan extension of the functor $$[n] \mapsto Nerve([n] + [n]^{op})$$, $$\Delta \to sSet$$.