Let ${\mathcal{C}}$ be a stable ${\infty}$-category. In the previous post, we needed to consider cubical diagrams

$\displaystyle f: (\Delta^1)^{n+1} \rightarrow \mathcal{C}.$

These diagrams come with an initial object and a terminal object: in fact, they are the cones on smaller diagrams. For instance, ${(\Delta^1)^{n+1}}$ is the nerve of all subsets of ${[n]}$, which is the cone on the nerve of all nonempty subsets of ${[n]}$, and also the cone on the nerve of all proper subsets of ${[n]}$. So it makes sense to talk about whether ${f}$ is a limit diagram, or whether ${f}$ is a colimit diagram.

The main result is:

Proposition 11 (Cube lemma) If ${\mathcal{C}}$ is stable, then ${f: (\Delta^1)^{n+1} \rightarrow \mathcal{C}}$ is a limit diagram if and only if it is a colimit diagram.

When ${n = 0}$, this is automatic: any diagram ${\Delta^1 \rightarrow \mathcal{C}}$ is a limit diagram if and only if it is an equivalence, and ditto for colimit diagrams. When ${n = 1}$, this is particular to the stable case: a square is a push-out if and only if it is a pull-back. We took this as more or less axiomatic, though it can be deduced from much weaker axioms, as in “Higher Algebra.”

The purpose of this post is to work through the proof of the “cube lemma.” This is more or less a piece of an attempt to work through Lurie’s version of the Dold-Kan correspondence. I’ve been doing it in a fair bit of detail for my own benefit — this means that the posts are a little more detailed than usual. In any event, the present post should stand alone from the others.

In general, we will prove this by induction on ${n}$. Start with an observation which is particular to the stable case: a diagram

is a push-out (or pull-back) if the map on cofibers ${\mathrm{cof}(f) \rightarrow \mathrm{cof}(g)}$ is an equivalence. This can be verified by contemplating the diagram

Each vertical line is a cofiber sequence, and if ${\mathrm{cof}(f) \rightarrow \mathrm{cof}(g)}$ is an equivalence, a version of the “five-lemma” shows that ${B \sqcup_A C \rightarrow D}$ must be an equivalence, too. (Namely: use the stability hypothesis to realize ${B \sqcup_A C , D}$ as cofibers of ${\mathrm{cof}[-1] \rightarrow B}$, etc.)

Let’s now prove the cube lemma, by induction on ${n}$. For ${n = 1}$, as we said, it is a fact we take for granted about stable ${\infty}$-categories (or as part of the definition). Suppose given ${f: (\Delta^1)^{n+1} \rightarrow \mathcal{C}}$. Then we have a decomposition of ${Q_{n+1} = (\Delta^{1})^{n+1} \setminus \left\{\ast\right\}}$ (for ${\ast}$ the terminal object) as

$\displaystyle Q_{n+1} = (\Delta^{1})^{n+1} \setminus \left\{\ast\right\} = (\Delta^1)^{n} \sqcup_{Q_n} Q_{n} \times \Delta^1.$

Here ${(\Delta^1)^{n}}$ is the top face, and ${Q_n}$ is the top face minus the endpoint.

This means that we have an analogous decomposition

$\displaystyle \mathrm{colim}_{Q_{n+1}} f = \mathrm{colim}_{(\Delta^1)^n} f \sqcup_{\mathrm{colim}_{Q_n} f} \mathrm{colim}_{Q_n \times \Delta^1}f.$

(Ideally, I would draw this.)

Now the colimit over ${(\Delta^1)^n}$ of ${f}$ is ${f}$ evaluated at the terminal vertex. Let’s call that ${\alpha}$: so ${\alpha}$ sits right above the terminal vertex of ${(\Delta^1)^{n+1}}$. The colimit over ${Q_n \times \Delta^1}$ of ${f}$ is the same as the colimit over the bottom copy of ${Q_n}$ of ${f}$, by cofinality. (How do we prove cofinality? Any map ${A \times \left\{1\right\} \rightarrow A \times \Delta^1}$ is cofinal: more generally, any right anodyne morphism is.)

So we get

$\displaystyle \mathrm{colim}_{Q_{n+1}} f = f(\alpha) \sqcup_{\mathrm{colim}_{Q_n^t} f} \mathrm{colim}_{Q_n^b} f.$

Here ${Q_n^t}$ refers to the top copy ${Q_n \times \left\{0\right\}}$, and ${Q_n^b}$ refers to the bottom copy ${Q_n \times \left\{1\right\}}$.

Let ${\beta}$ be the terminal vertex of ${(\Delta^1)^{n+1}}$, lying right below ${\beta}$. Then to say that ${f}$ is a colimit diagram is to say that one has a pushout diagram:

However, we are working in the stable setting, so this is equivalent to the condition that we have an equivalence on cofibers:

$\displaystyle \mathrm{cof}(\mathrm{colim}_{Q_n^t} f \rightarrow \mathrm{colim}_{Q_n^b} f) \simeq \mathrm{cof}( f(\alpha) \rightarrow f(\beta)).$

We can identify the objects on both sides of the equation. If we consider ${f}$ as a natural transformation ${(\Delta^1)^{n} \times \Delta^1 \rightarrow \mathcal{C}}$ between two maps ${f_0: (\Delta^1)^n \rightarrow \mathcal{C}}$ and ${f_1: (\Delta^1)^n \rightarrow \mathcal{C}}$, then the above equivalence is saying precisely that the diagram

$\displaystyle \mathrm{cof}(f_0 \rightarrow f_1) : (\Delta^1)^n \rightarrow \mathcal{C}$

is a colimit diagram.

In a completely analogous manner, we find that to say that ${f: ( \Delta^1)^{n+1} \rightarrow \mathcal{C}}$ is a limit diagram is equivalent to saying that

$\displaystyle \mathrm{fib}(f_0 \rightarrow f_1) : (\Delta^1)^n \rightarrow \mathcal{C}$

is a limit diagram.

Now, by induction on ${n}$, to say that ${\mathrm{cof}(f_0 \rightarrow f_1)}$ is a colimit diagram is the same as saying that it is a limit diagram (since this is a cubical diagram of a smaller size). This in turn is the same as saying that ${\mathrm{fib}(f_0 \rightarrow f_1)}$ is a limit diagram since ${\mathrm{fib}(f_0 \rightarrow f_1)}$ is a shift of the cofiber. But we have just seen that this is equivalent to saying that ${f}$ is a limit diagram.