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 1


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.

  • $\begingroup$ 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? $\endgroup$ Oct 8, 2019 at 16:13
  • $\begingroup$ 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. $\endgroup$ Oct 8, 2019 at 20:37
  • $\begingroup$ Thank you for your answer. $\endgroup$ Oct 9, 2019 at 8:28
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jeff Strom
    Oct 11, 2019 at 12:56

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