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.

 

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