# Higher-dimensional version of the “Magic Cube Lemma” for homotopy pushouts/pullbacks

The "Magic Cube Lemma" is a surprising (to me) relationship between (homotopy) pushouts and (homotopy) pullbacks of spaces:

Consider a cubical diagram $$I^3\to \mathcal{S}$$ in the $$\infty$$-category of spaces/homotopy types (where $$I=\Delta^1=\mathrm{N}\{0\to 1\}$$):

$$\require{AMScd}$$ $$\begin{CD} A @>>> @>>> B @. \\ @VVV @. @VVV @.\\ @. A' @>>> @>>> B'\\ @VVV @VVV @VVV @VVV\\ C @>>> @>>> D @. \\ @. @VVV @. @VVV\\ @. C' @>>> @>>> D'\\ \end{CD}$$

where there are supposed to be arrows connecting the back layer ($$A,B,C,D$$) to the front layer ($$A',B',C',D'$$). (Maybe a kind soul can explain how I draw actual cubical diagrams with AMScd)

Assume the following:

• The back and front layers are pushout squares
• The top layer ($$A,B,A',B'$$) and the left layer ($$A,C,A',C'$$) are pullback squares

Then the lemma asserts that also the right layer ($$B,D,B',D'$$) and the bottom layer ($$C,D,C',D'$$) are pullback squares.

A different way of stating the same lemma is as follows:

Let $$I^2\to \mathrm{Fun}(I,\mathcal S)$$ be a square of arrows in spaces.

$$\begin{CD} a @>f>> b\\ @VgVV (*) @VVg'V\\ c @>>f'> d\\ \end{CD}$$ where $$a\colon A\to A'$$, etc. From this perspective, $$f$$,$$f'$$,$$g$$ and $$g'$$ are natural transformations of diagrams $$I\to\mathcal S$$.

Then the lemma can be restated as follows:

Assume that the square $$(\ast)$$ is a pushout and that the natural transformations $$f$$ and $$g$$ are Cartesian. Then also $$f'$$ and $$g'$$ are Cartesian.

More generally, the following also holds (essentially by the fact that every colimit can be built from coproducts and pushouts):

Let $$U\colon K^\triangleright\to\mathrm{Fun}(I,\mathcal S)$$ be a colimit diagram of arrows. If for every edge $$e\colon I\to K$$ the natural transformation $$U\circ e\colon I\to\mathrm{Fun}(I,\mathcal S)$$ is Cartesian, then the same is true for every edge $$e\colon I\to K^\triangleright$$.

(Here $$K^\triangleright = K\star [0]$$ denotes the right cone on the category/simplicial set $$K$$)

I am interested in the following higher dimensional version:

Call a natural transformation $$f\colon a\to b$$ between cubical diagrams $$a,b\colon I^n\to\mathcal S$$ relatively Cartesian, if it is a Cartesian (i.e.\ a limit diagram) when viewed as a diagram $$I^{n+1}=I\times I^n\to\mathcal S$$.

(Equivalently: $$f$$ is a $$p$$-Cartesian edge for the canonical Cartesian fibration $$p\colon\mathrm{Fun}(I^n,\mathcal S)\to \mathrm{Fun}(I^n_\star,\mathcal S)$$, where $$I^n_\star$$ denotes the punctured cube obtained by removing the initial vertex $$(0,\dots,0)$$).

Let $$\begin{CD} a @>f>> b\\ @VgVV (**) @VVg'V\\ c @>>f'> d\\ \end{CD}$$ be a square $$I^2\to \mathrm{Fun}(I^n,\mathcal S)$$ of $$n$$-dimensional cubical diagrams.

Is it true that: if $$(\ast\ast)$$ is a pushout, and $$f$$ and $$g$$ are relatively Cartesian, then $$f'$$ and $$g'$$ are also relatively Cartesian?

Or more generally:

Given a colimit diagram $$K^\triangleright\to \mathrm{Fun}(I^n,\mathcal{S})$$ such that every edge in $$K$$ is sent to a relatively Cartesian transformation, is the same true for all edges of $$K^\triangleright$$?

If it is true: can one deduce it formally from the original magic cube lemma (I tried, but failed) or does one maybe need an additional input about the $$\infty$$-category $$\mathcal S$$? If it holds for $$\mathcal S$$, does it also hold in any $$\infty$$-topos?

## 1 Answer

I don't think so. Take $$n=2$$, and consider a map $$f\colon b\to a$$ between objects of $$\mathrm{Fun}(I^2, \mathcal{S})$$. If $$a$$ and $$b$$ are pullback squares, then any map $$f$$ between them is relatively Cartesian in your sense. Also, if $$a$$ is a pullback and $$f$$ is relatively cartesian, then $$b$$ must also be a pullback.

(Any commutative square $$a$$ has a collection of "total fibers", which are the fibers of the map $$a(0)\to a(1)\times_{a(12)} a(2)$$ over points in the target. The total fibers are all contractible iff $$a$$ is a pullback. A map $$b\to a$$ is relatively Cartesian iff it induces equivalences on all total fibers.)

So let $$a$$ be the pullback square which displays $$\Omega X$$ as a pullback of $$*\to X \leftarrow *$$, and let $$b$$ and $$c$$ be pullback squares of contractible spaces. Then the pushout $$d$$ is a square with $$\Sigma \Omega X$$ at the initial corner, $$\Sigma X$$ at the terminal corner, and $$*$$ at the other two spots. For $$d\to c$$ to be relatively Cartesian, since $$c$$ is certainly a pullback we would have to have $$d$$ be a pullback to, i.e., we would have $$\Sigma\Omega X\xrightarrow{\sim} \Omega\Sigma X$$, which almost never happens.

• Do you mean to write that $c\to d$ is relatively Cartesian (you wrote $d\to c$)? If so, how does this imply that $d$ must be Cartesian? I understand how the implication $d$ Cartesian => $c$ Cartesian works; are you saying that the converse also holds? – Tashi Walde Oct 8 at 16:13
• You're right, I have the arrows reversed somewhere. The converse almost holds: given a relative cartesian $c\to d$ with $c$ a pullback, you can conclude $d$ is a pullback if $\pi_0 (c(1)\times_{c(12)} c(2)) \to \pi_0(d(1)\times_{d(12)} d(2))$ is surjective. So in my example I should make sure that $\Omega \Sigma X$ is connected. – Charles Rezk Oct 8 at 20:37
• Thank you for your answer. – Tashi Walde Oct 9 at 8:28
• Interestingly, though, $\Omega\Sigma X$ is the suspension of $\Omega X$ in the category of topological monoids. I wonder if there is a category-switching context in which there might be a more positive answer. – Jeff Strom Oct 11 at 12:56