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.




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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 $K$)




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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?



Higher-dimensional version of the "Magic Cube Lemma" for homotopy pushouts/pullbacks
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.


