The higher-categorical Dold-Kan correspondence II

The past few posts have been focused on a discussion of Lurie’s version of the Dold-Kan correspondence in stable $\infty$ -categories. I’ve made these posts more detailed than usual: while I’ve been trying to treat such category theory as a black box on this blog, it should be interesting (at least for me) to see how the machines work beneath the surface, in some specific examples. In previous posts, I stated the result, and described an important lemma on the structure of simplicial objects in a stable $\infty$ -category, which depended on the combinatorics of cubes.

The goal of this post is to (finally) prove the result, an equivalence of ${\infty}$ -categories

$\displaystyle \mathrm{Fun}(\Delta^{op}, \mathcal{C}) \simeq \mathrm{Fun}( \mathbb{Z}_{\geq 0}, \mathcal{C}),$

valid for any stable ${\infty}$ -category ${\mathcal{C}}$ . As before, the intuition behind this version of the Dold-Kan correspondence is that a simplicial object determines a filtered object by taking successive geometric realizations of the ${n}$ -truncations. The fact that one can go in reverse, and reconstruct the simplicial object from the geometric realizations of the truncations, is specific to the stable case.

1. Formalizing the equivalence

Let’s make this precise. To do so introduces a little extra notation, but the ideas are exactly those as above.

There is a chain of equivalences which is strictly functorial in the stable ${\infty}$ -category ${\mathcal{C}}$ (functorial for functors which are exact, i.e. preserve finite limits and colimits). Namely, let’s take the category ${\Delta^{op, +} \times \mathbb{Z}_{\geq 0}}$ . As always, categories such as ${\mathbb{Z}_{\geq 0}}$ are considered as simplicial sets via their nerves (this is admittedly a slight abuse of notation). There is a subcategory ${\mathcal{I} \subset \Delta^{op, +} \times \mathbb{Z}_{\geq 0}}$ which consists of all pairs ${([m], n)}$ with ${m \leq n}$ . Let ${\mathcal{I}^0 \subset \mathcal{I}}$ be the subcategory not including elements with ${m = -1}$ . We can draw ${\mathcal{I}}$ and ${\mathcal{I}^0}$ as subcategories of ${\Delta^{op, +} \times \mathbb{Z}_{\geq 0}}$ (which we may represent in a 2-dimensional plane) by those elements lying below a “diagonal.”

Our strategy to get from a simplicial object to a filtered object will be, with more precision, as follows:

A simplicial object ${f: \Delta^{op} \rightarrow \mathcal{C}}$ determines a functor ${\Delta^{op} \times \mathbb{Z}_{\geq 0} \rightarrow \mathcal{C}}$ , and thus a functor ${\mathcal{I}^0 \rightarrow \mathcal{C}}$ . This functor just forgets about the second factor, so it is not the most interesting functor, and in particular one can recover the original functor ${f}$ from it. Actually, ${\Delta^{op}}$ does not quite sit inside ${\mathcal{I}^0}$ , so this requires a bit of an argument.
We left Kan extend to ${\mathcal{I}}$ , i.e. to the points ${([-1], n)}$ for ${n \geq 0}$ . At each ${([-1], n)}$ , this is, roughly speaking, a colimit over all ${f([m])}$ with ${m \leq n}$ : that is, the geometric realization of the ${n}$ -truncation of ${f}$ .
We restrict to ${\mathbb{Z}_{\geq 0}}$ . This just remembers the filtered object.

We have four relevant categories and morphisms between them:

$\displaystyle \mathrm{Fun}(\Delta^{op}, \mathcal{C}) \stackrel{f_1}{\rightarrow} \mathrm{Fun}'(\mathcal{I}^0, \mathcal{C}) \stackrel{f_2}{ \leftarrow} \mathrm{Fun}'(\mathcal{I}, \mathcal{C}) \stackrel{f_3}{\rightarrow} \mathrm{Fun}(\mathbb{Z}_{\geq 0}, \mathcal{C})$

where all the morphisms are given by forgetful functors of some sort. The primes denote conditions which will be specified shortly. Namely, ${f_1}$ sends a simplicial object to a functor ${\mathcal{I}^0 \rightarrow \mathcal{C}}$ —a functor which pays no attention to the second factor. The functor ${f_2}$ is retriction (an inverse to the left Kan extension mentioned above, except ${f_2}$ is functorial in a 1-categorical sense). Here ${f_3}$ is ordinary restriction.

Anyway, the advantage of writing everything out this way is that it is very rigorous: we are working with natural transformations of simplicial sets, in the ordinary 1-categorical sense. The ${\infty}$ -categorical claim is that all the ${f_i}$ are equivalences, when ${\mathcal{C}}$ is stable.

Here are the conditions on the “primes:”

${\mathrm{Fun}'(\mathcal{I} ^0, \mathcal{C})}$ refers to functors ${\mathcal{I}^0 \rightarrow \mathcal{C}}$ which only see the first variable: that is, they take all maps ${([m], n) \rightarrow ( [m] , n')}$ which are identity on the first variable to equivalences.
${\mathrm{Fun}'(\mathcal{I}, \mathcal{C})}$ refers to functors whose restrictions to ${\mathcal{I}^0}$ satisfy the previous condition and which are left Kan extensions of their restrictions to ${\mathcal{I}^0}$ (i.e., the associated filtered object comes from taking partial geometric realizations).

The following result will complete the proof of the Dold-Kan correspondence.

Proposition 12 Each of the ${f_i}$ is an equivalence of ${\infty}$ -categories (for ${\mathcal{C}}$ stable).

Let’s start by noting that ${f_2}$ is an equivalence for formal reasons, because we only restricted to functors which were left Kan extensions of the restriction to ${\mathcal{I}^0}$ (and these left Kan extensions can always be made since we have finite colimits). What we really need to do is to handle ${f_3, f_1}$ .

2. The third equivalence

Let’s next show that ${f_3}$ is an equivalence. This is the harder part, and it requires an induction: it will show that the filtered object in ${\mathrm{Fun}(\mathbb{Z}_{\geq 0}, \mathcal{C})}$ uniquely determines a whole element of ${\mathrm{Fun}'(\mathcal{I}, \mathcal{C})}$ .

Let ${\mathcal{I}_{\leq k} \subset \mathcal{I}}$ be the truncated part of ${\mathcal{I}}$ consisting of elements ${( [m], n)}$ with ${m \leq n \leq k}$ or with ${m = -1}$ . We can think of ${\mathcal{I}}$ as the union,

$\displaystyle \mathcal{I} = \bigcup \mathcal{I}_{\leq k}.$

Let’s consider the ${\infty}$ -category ${\mathrm{Fun}'(\mathcal{I}_{\leq k}, \mathcal{C})}$ consisting of functors which satisfy a truncated version of the ${\prime}$ condition above: they see only the first variable and are left Kan extensions at each ${[-1], n}$ .

Proposition 13 ${\mathrm{Fun}'(\mathcal{I}_{\leq k+1}, \mathcal{C}) \rightarrow \mathrm{Fun}'(\mathcal{I}_{\leq k}, \mathcal{C})}$ is a trivial Kan fibration.

In particular, taking the inverse limit, we find that

$\displaystyle \mathrm{Fun}'(\mathcal{I}, \mathcal{C}) \rightarrow \mathrm{Fun}'(\mathbb{Z}_{\geq 0}, \mathcal{C}) = \mathrm{Fun}'(\mathcal{I}_{\leq -1}, \mathcal{C})$

is an equivalence.

Proof: We can think of ${\mathcal{I}_{\leq k+1}}$ as being obtained from ${\mathcal{I}_{\leq k}}$ by adding the object ${([k+1], k+1)}$ . Actually, we are also adding a bunch of other objects like ${([k], k+1)}$ . So let ${\mathcal{I}_{\leq k+1}^{\square}}$ be obtained by just adding ${([k+1], k+1)}$ . If one starts with a functor

$\displaystyle \mathcal{I}_{\leq k} \rightarrow \mathcal{C},$

to extend it to ${\mathcal{I}_{\leq k+1} \rightarrow \mathcal{C}}$ to satisfy the “prime” condition, one has to first extend to ${\mathcal{I}^{\square}_{\leq k+1}}$ (that is, at the object ${([k+1], k+1)}$ ) in such a way that it is a left Kan extension at ${( [-1], k+1)}$ , and then extend by left Kan extension to the points ${([i], k+1)}$ for ${i \leq k}$ (this just means by sending these points to the same image as in ${([i], k)}$ ).

Now we showed in a previous post that choosing a functor ${\mathcal{I}^{\square}_{\leq k+1} \rightarrow \mathcal{C}}$ such that it is a left Kan extension at ${([-1], k+1)}$ is the same thing as right Kan extending ${\mathcal{I}_{\leq k} \rightarrow \mathcal{C}}$ along ${\mathcal{I}^{\square}_{\leq k+1}}$ . In other words, this is a recaptiulation of the fact that once one has a ${k}$ -truncated simplicial object (that’s the functor ${\mathcal{I}_{\leq k} \rightarrow \mathcal{C}}$ when one forgets about ${([-1], m)}$ ) and a desired colimit for a ${k+1}$ -skeleton (that’s where ${( [-1], k+1)}$ goes) then it uniquely extends (via a right Kan extension process) to a ${k+1}$ -truncated simplicial object.

It follows from all this that there is (up to unique isomorphism) one way of extending a functor ${\mathcal{I}_{\leq k} \rightarrow \mathcal{C}}$ to ${\mathcal{I}_{\leq k+1} \rightarrow \mathcal{C}}$ so as to satisfy the primed conditions: it has to be a right Kan extension followed by a left Kan extension. Stated more precisely, this shows in fact that one has a trivial Kan fibration as stated.

3. The first equivalence

Next, we have to show that ${f_1: \mathrm{Fun}(\Delta^{op}, \mathcal{C}) \rightarrow \mathrm{Fun}'(\mathcal{I}^0, \mathcal{C})}$ is an equivalence. In other words, to give a simplicial object

$\displaystyle \Delta^{op} \rightarrow \mathcal{C}$

is equivalent to giving a morphism ${\mathcal{I}^0 \rightarrow \mathcal{C}}$ which only sees the first factor, informally. This is easier since no use of the stability of ${\mathcal{C}}$ is made—this is something valid for every ${\infty}$ -category ${\mathcal{C}}$ .

Anyway, one problem is that the map ${\mathcal{I}^0 \rightarrow \Delta^{op}}$ doesn’t admit a section. However, if we truncate, and restrict to pairs ${([m], n)}$ with ${n \leq k}$ , we get categories ${\mathcal{I}^0_{\leq k}, \Delta^{op}_{\leq k}}$ , and the associated map

$\displaystyle \mathcal{I}^0_{\leq k} \rightarrow \Delta^{op}_{\leq k}$

does admit a section (given by sending ${[m] \mapsto ([m], k)}$ , and so we can actually regard ${\Delta^{op}_{\leq k}}$ as a (full) subcategory of ${\mathcal{I}^0_{\leq k}}$ .

It will suffice to show that

$\displaystyle \mathrm{Fun}(\Delta^{op}_{\leq k}, \mathcal{C}) \simeq \mathrm{Fun}'(\mathcal{I}^0_{\leq k}, \mathcal{C})$

for each ${k}$ . We can do this by observing that the “prime” condition on ${ \mathrm{Fun}'(\mathcal{I}^0_{\leq k}, \mathcal{C})}$ is precisely saying that the functor ${\mathcal{I}^0_{\leq k} \rightarrow \mathcal{C}}$ is a right Kan extension of its restriction to ${\Delta^{op}_{\leq k}}$ (regarded as a subcategory), and the functor in this direction can be identified with right Kan extension. This establishes the equivalence above, and shows (by taking homotopy inverse limits) that ${f_1}$ is an equivalence as desired.

This completes the proof of the Dold-Kan correspondence, and shows that there is a functorial (for exact functors) chain of three equivalences between simplicial and filtered objects in any stable $\infty$ -category.

This is one of the first genuinely non-formal results I’ve learned in higher category theory — it’s one that is simply different from the 1-categorical case, and which uses the inherently higher-categorical notion of stability. There is the Dold-Kan correspondence for abelian categories, but it relies on somewhat hard-to-motivate constructions, and takes values in chain complexes. This version of the Dold-Kan correspondence is based on an extremely natural functor, familiar once one knows that a geometric realization admits a skeletal filtration. What’s not clear to me yet is whether one can (in some manner) derive the classical version from this version.

Climbing Mount Bourbaki