Let ${\mathcal{C}}$ be a stable ${\infty}$-category. For us, this means that we have three important properties:

1. ${\mathcal{C} }$ admits finite limits and colimits.
2. ${\mathcal{C}}$ has a zero object: that is, the initial object is also final.
3. A square in ${\mathcal{C}}$ is a pull-back if and only if it is a push-out.

This is equivalent to the stability of ${\mathcal{C}}$. Actually, stability is usually defined using slightly weaker conditions, and then it takes a little work to show that these stronger ones are implied. We’ll just work with these. Stability can be thought of as a higher-categorical version of being triangulated; a general stable $\infty$-category has many of the properties (in a higher categorical sense) of the homotopy category of spectra, or the (classical) derived category.

Our goal is to show that in this case, we have an equivalence of ${\infty}$-categories

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

between simplicial objects in ${\mathcal{C}}$ and filtered (nonnegatively) objects in ${\mathcal{C}}$. The idea here is that the geometric realization of a simplicial object comes with a canonical filtration, given by geometric realizing the ${n}$-truncations for each ${n}$. This is going to give the associated filtered object. (We don’t know that the geometric realization exists, but the realizations of the truncations will.)

We will actually prove something stronger: for each ${n}$, there is an equivalence

$\displaystyle \mathrm{Fun}(\Delta^{op}_{\leq n}, \mathcal{C}) \simeq \mathrm{Fun}( [0, n], \mathcal{C}), \ \ \ \ \ (1)$

where ${[0, n] \subset \mathbb{Z}_{\geq 0}}$ is the subcategory of elements ${\leq n}$. In other words, ${n}$-truncated simplicial objects are the same as ${n}$-filtered objects of ${\mathcal{C}}$. (Note that, as a simplicial set, the nerve of ${[0, n]}$ is ${\Delta^n}$.) These equivalences will be compatible, and taking inverse limits will give the Dold-Kan correspondence.

1. Motivation

Let’s try to motivate why such a result might be reasonable. To give an ${n}$-filtered object of ${\mathcal{C}}$ (by which I simply mean an element of ${\mathrm{Fun}([0, n], \mathcal{C})}$), is the same as giving an ${n-1}$-filtered object of ${\mathcal{C}}$ together with a morphism out of the colimit of the ${n-1}$-filtered object (i.e., of the last element). That is, we have an equivalence

$\displaystyle \mathrm{Fun}( [0, n], \mathcal{C}) \simeq \mathrm{Fun}( [0, n-1], \mathcal{C}) \times_{\mathcal{C}} \mathrm{Fun}(\Delta^1, \mathcal{C}).$

Stated alternatively, to give a functor ${F: [0, n] \rightarrow \mathcal{C}}$ is the same as specifying a restricted functor ${[0, n-1] \rightarrow \mathcal{C}}$ and a colimit of ${F}$ (together with appropriate compatibility data).

If we expect an equivalence of the form (1), then we should expect that to give an ${n}$-truncated simplicial object is the same as giving an ${n-1}$-truncated object together with a specification of a colimit for the full thing. In other words, we find:

Goal: To give an ${n}$-truncated simplicial object ${F: \Delta_{\leq n}^{op} \rightarrow \mathcal{C}}$ is the same as giving an ${n-1}$-truncated simplicial object ${\Delta_{\leq n-1}^{op} \rightarrow \mathcal{C}}$ together with a desired colimit of ${F}$ (along with some compatibility data).

So, once we’re given an ${n-1}$-truncated simplicial object, there should be a unique choice of a degree ${n}$ element that will make the ${n}$-truncated colimit what we want. We will make this precise in the following manner. Let ${\Delta^+}$ be the category of finite ordered sets (so in other words, ${[-1] = \emptyset}$ is included), and let ${\Delta^{+}_{\leq n}}$ be defined to be finite ordered sets of cardinality ${\leq n}$.

Observe that ${\Delta^{+}_{\leq n-1}}$ has ${[-1]}$ as initial object, and consequently we can regard elements of

$\displaystyle \mathrm{Fun}(\Delta^{op +}_{\leq n-1}, \mathcal{C})$

as consisting of ${n-1}$-truncated simplicial objects in ${\mathcal{C}}$, together with an extra object (the image of ${-1}$). In ${\infty}$-language, we can say that

$\displaystyle \Delta^{op +}_{\leq n-1} = \Delta^{op \vartriangleright}_{\leq n-1}$

is the cone on ${\Delta^{op}_{\leq n-1}}$.

Definition 7 An augmented ${n-1}$-truncated simplicial object of ${\mathcal{C}}$ is an element of ${\mathrm{Fun}(\Delta_{\leq n-1}^{op+}, \mathcal{C})}$.

The upshot of all this is that if we expect the Dold-Kan correspondence to hold, then to give an augmented ${n-1}$-truncated simplicial object in ${\mathcal{C}}$ should be equivalent to giving an ${n}$-truncated simplicial object of ${\mathcal{C}}$. In fact, an “augmented ${n-1}$-filtered object” is precisely the same as an ${n}$-filtered object. So we should expect an equivalence

$\displaystyle \mathrm{Fun}(\Delta_{\leq n-1}^{op+}, \mathcal{C}) \simeq \mathrm{Fun}(\Delta_{\leq n}^{op}, \mathcal{C}). \ \ \ \ \ (2)$

This is specific to the stable case.

2. Comparing augmentations and truncations

Intuitively, the functor from ${\mathrm{Fun}(\Delta_{\leq n}^{op}, \mathcal{C})}$ to ${\mathrm{Fun}(\Delta_{\leq n-1}^{op+}, \mathcal{C}) }$ consists of restricting to the ${n-1}$-skeleton, plus taking the colimit (of the whole ${n}$-truncated thing) to be the augmentation in degree ${-1}$. The functor in the other direction will realize the degree ${n}$ element as an appropriate limit of the ${n-1}$-truncation plus the augmentation: the interplay between these limits and colimits is precisely where the stability condition enters.

To be very precise, we’re going to construct the equivalence (2) as a chain of equivalences. Namely, we note that augmented ${n}$-truncated simplicial objects map to both categories: we have a diagram

$\displaystyle \mathrm{Fun}(\Delta_{\leq n-1}^{op+}, \mathcal{C}) \stackrel{P}{ \leftarrow} \mathrm{Fun}(\Delta_{\leq n}^{op+}, \mathcal{C}) \stackrel{Q}{\rightarrow} \mathrm{Fun}(\Delta_{\leq n}^{op}, \mathcal{C}) .$

The main lemma is going to be that that a suitable subcategory of ${n}$-truncated augmented objects maps equivalently under ${P, Q}$ to both ends. We will show:

Proposition 8 Let ${\mathcal{C}}$ be stable. For an ${n}$-truncated augmented simplicial object ${f: \Delta^{op +}_{\leq n} \rightarrow \mathcal{C}}$, the following are equivalent:

1. ${f}$ is a left Kan extension of its restriction to ${\Delta^{op}_{\leq n} }$ (i.e., forgetting the augmentation).
2. ${f}$ is a right Kan extension of its restriction to ${\Delta^{op +}_{\leq n-1}}$ (i.e., to an ${n-1}$-truncated augmented simplicial object).

This is precisely the techincal tool which makes the proof go. The first condition (that ${f}$ is a left Kan extension) states that the augmentation comes from the colimit over the whole ${n}$-truncated thing. The second condition states that the object in degree ${n}$ is recovered from the ${n-1}$-truncation plus the augmentation. The proposition states that we can go either way.

The proposition takes some work with cofinality, and a little effort with stable ${\infty}$-categories, to prove. But once it’s proved, we can obtain the desired equivalence (2) fairly easily. Namely, we want to show that augmented ${n-1}$-truncated simplicial objects and ${n}$-truncated simplicial objects are equivalent. We consider the diagram

$\displaystyle \mathrm{Fun}(\Delta_{\leq n-1}^{op+}, \mathcal{C}) \stackrel{P}{ \leftarrow} \mathrm{Fun}^0(\Delta_{\leq n}^{op+}, \mathcal{C}) \stackrel{Q}{\rightarrow} \mathrm{Fun}(\Delta_{\leq n}^{op}, \mathcal{C}) ,$

where ${\mathrm{Fun}^0}$ means functors that satisfy either (i.e., both) of the conclusions of the proposition. For formal reasons, the two maps are equivalences: given a subcategory ${\mathcal{E} \subset \mathcal{D}}$ of an ${\infty}$-category ${\mathcal{D}}$, to give a functor out of ${\mathcal{E}}$ is the same as giving a functor out of ${\mathcal{D}}$ which is a left (or right) Kan extension of ${\mathcal{E}}$. At least, if enough limits or colimits exist in the target ${\infty}$-category, which is true in this case.

3. Proof of the proposition

Let ${\mathcal{C}}$ be stable, and fix an ${n}$-truncated augmented simplicial object ${f: \Delta_{\leq n}^{op +} \rightarrow \mathcal{C}}$. To say that ${f}$ is a left Kan extension of its restriction to ${\Delta^{op }_{\leq n}}$ is to say that ${f([-1])}$ is the colimit of the ${n}$-truncated thing ${f|_{\Delta^{op}_{\leq n}}}$. In other words, if we regard ${\Delta^{op +}_{\leq n}}$ as the cone on ${\Delta^{op}_{\leq n}}$, then the diagram

$\displaystyle f: \Delta_{\leq n}^{op +} \rightarrow \mathcal{C}$

is a colimit diagram, exhibiting ${f([-1])}$ as the colimit of ${f|_{\Delta^{op}_{\leq n}}}$. (To be precise, it is a diagram out of a conewhich is a colimit. A colimit is the whole diagram, not just the image of the cone point.)

Now in the previous post, we saw that there was another model for ${\Delta_{\leq n}^{op}}$ up to cofinality: namely, ${\Delta^{op}_{inj, [n]/}}$. This is the opposite to the category of nonempty finite ordered sets equipped with an injective map into ${[n]}$, or equivalently the opposite to the poset of nonempty subsets of ${[n]}$. (Warning: My argument was incorrect; a correct one is in Lurie’s book.)

So, to say that ${f}$ is a colimit diagram is to say that the diagram

$\displaystyle \Delta^{op +}_{inj, [n]/} \rightarrow \mathcal{C}$

is a colimit diagram. Here the ${+}$ comes from adding a cone point for the empty set: that’s a terminal object in ${\Delta^{op +}_{inj, [n]/}}$, and the cone point goes to ${f([-1])}$. In other words, this is the opposite to the poset of allsubsets of ${[n]}$: in other words, ${(\Delta^1)^{n+1}}$. We’re somehow going to have to turn this into a condition that will involve limits—that is, right Kan extensions.

Note now that this object ${(\Delta^1)^{n+1}}$ comes with a symmetry, and we can talk about it being either a colimit diagram (using the last vertex) or being a limit diagram (using the first vertex). When ${n=1}$, this is a square, and we can talk about a square being a pull-back or a push-out. In the stable setting, though the two are equivalent! This turns out to generalize.

Lemma 9 A diagram ${(\Delta^1)^{n +1} \rightarrow \mathcal{C}}$ is a limit diagram if and only if it is a colimit diagram (for ${\mathcal{C}}$ stable).

Let’s postpone the proof of this lemma. We saw that to say that ${f}$ was a left Kan extension (that is, that the augmentation ${f([-1])}$ came from the colimit) was equivalent to saying that the diagram

$\displaystyle F: \Delta^{op +}_{inj, [n]/} \rightarrow \mathcal{C} \ \ \ \ \ (3)$

was a colimit diagram. However, because the target category is ${(\Delta^1)^{n+1}}$, this is equivalent to (3) being a limit diagram. In other words, that’s equivalent to the image of the initial object (which is the identity map ${[n] \rightarrow [n]}$) being determined as a limit of all the other objects. That is, ${F([n]) = f([n])}$ needs to be a limit of all the other objects in some sense.

It’s now easy to believe that this condition is equivalent to ${f}$‘s being a right Kan extension. However, there is a bit of checking to do. So far, we’ve said that ${f}$ was a left Kan extension if and only if ${ F: \Delta^{op +}_{inj, [n]/} \rightarrow \mathcal{C}}$ was a limit diagram at the cone point ${[n]}$. In other words, this is saying that

$\displaystyle f([n]) = \lim_{\Delta^{op +, \leq n-1}_{inj, [n]/}} F.$

This is a complicated condition, but it is saying that ${f([n])}$ is the limit over all injections of ${[k], k \leq n-1}$ equipped with a map to ${[n]}$ (or rather, the opposite to that).

To turn this into a condition on right Kan extensions, we have to replace ${\Delta^{op +, \leq n-1}_{inj, [n]}}$ by the category

$\displaystyle \Delta^{op +, \leq n-1}_{[n]/}$

where we don’t just consider injective maps. This is what it means for ${f}$ to be a right Kan extension at ${n}$: ${f([n])}$ has to be the limit of the diagram of all things that ${[n]}$ maps to in ${\Delta^{op, \leq n-1}_{[n]/}}$.

To see this, we use a cofinality argument again:

Proposition 10 The map ${\Delta^{op +, \leq n-1}_{inj, [n]/} \rightarrow \Delta^{op +, \leq n-1}_{[n]/}}$ is right cofinal.

Here “right cofinal” is the dual to cofinal: it means that the map of opposite categories is cofinal. In other words,

$\displaystyle \Delta^{+, \leq n-1}_{inj, /[n]} \rightarrow \Delta^{\leq n-1}_{/[n]}$

is cofinal. To see this, in turn, we have to show (by Theorem A) that given an object ${[p] \rightarrow [n]}$ in the second category, the category of diagrams

is contractible. Here the map ${[m] \rightarrow [n]}$ is required to be injective. But this category has an initial object, which comes from the image of ${[p] }$ in ${[n]}$.

Let’s put this all together:

1. To say that ${f}$ is a left Kan extension is, by definition, equivalent to saying that ${f|_{\Delta^{op}_{\leq n}} }$ has colimit ${f( [-1])}$.
2. By a cofinality lemma, this is equivalent to saying that ${F: \Delta^{op +}_{inj, [n]/} \rightarrow \mathcal{C}}$ is a colimit diagram (at the cone point ${[-1] \rightarrow [n]}$). Note that this is a cubical diagram.
3. By a lemma about stable ${\infty}$-categories, this in turn is equivalent to saying that ${F}$ is a limit diagram at ${n}$. In other words, that ${F([n]) = f([n])}$ is the limit of ${\Delta^{op + \leq n-1}_{inj, [n]/}}$.
4. Using another cofinality argument, this is equivalent to saying that ${f([n])}$ is the limit of ${f|_{ \Delta^{op + \leq n-1}_{[n]/}}}$.
5. This in turn states precisely that ${f}$ is a right Kan extension at ${n}$.

This completes the proof of the (somewhat technical) lemma we need for the Dold-Kan correspondence, modulo the lemma on cubical diagrams.