Is freeness of strict $n$-categories preserved by deleting a dimension?

Question

Let $C$ be an $n$-dimensional globular set. Then for $0 \leq i \leq n$ there is a globular set $C^{(i)}$ obtained from $C$ by forgetting the $i$-cells (so that for $j > i$, the $j$-cells of $C$ correspond to the $(j-1)$-cells of $C^{(i)}$). Moreover, if $C$ has the structure of a strict $n$-category, then $C^{(i)}$ has in a canonical way the structure of a strict $(n-1)$-category.

My question is about whether the functor $C \mapsto C^{(i)}$ preserves freeness. Here I say that a strict $n$-category is free if it is free on an $n$-computad (a.k.a. $n$-polygraph) $P$. (I believe that this is a mere property of $C$ -- that is, there can be at most one such $P$ for a given $C$.) That is, I'm asking the following:

Question: Let $C$ be the free strict $n$-category on an $n$-computad $P$. Then is $C^{(i)}$ likewise free on an $(n-1)$-computad $P^{(i)}$?

If the answer is yes, then I suppose that $P^{(i)}$ must be something like the following: it ought to have the same $(i-1)$-skeleton as $P$, and then in higher dimensions it should be generated by all cells of the appropriate dimensions, domains, and codomains in $C$ which are indecomposable in $C$ under $\circ_j$ for $j \neq i$.

I'd also be curious whether this works if we say "parity complex" (Street) or "torsion-free complex" (Forest) or "pasting scheme" (M. Johnson) or "loop-free unital augmented directed complex" (Steiner) in place of "computad".

Answer 1

No. Consider the 2-category $C = [2 \mid 1, 1]$, which has 3 objects $0,1,2$; 4 generating 1-morphisms $f,f' : 0 \rightrightarrows 1$ and $g,g' : 1 \rightrightarrows 2$; and 2 generating 2-cells $\alpha : f \Rightarrow f'$ and $\beta : g \Rightarrow g'$. $C$ arises from a computad.

Let $i=0$. Then $C^{(0)}$ is the disjoint union of the following 6 categories:

• 3 copies of the terminal category $[0]$ given by $id_0, id_1, id_2$;

• 2 copies of the arrow category $[1]$ given by $f \xrightarrow \alpha f'$ and $g \xrightarrow \beta g'$;

• 1 copy of the square category $[1] \times [1]$, given by $gf \overset{g\alpha, \beta f}{\rightrightarrows} gf', g'f \overset{\beta f', g' \alpha}{\rightrightarrows} g'f'$ (the diagonal is $\beta \alpha$).

• 1 copy of the square category $[1] \times [1]$, given by $gf \overset{g\alpha}{\rightarrow} gf' \overset{\beta f'}{\rightarrow} g'f'$, $gf \overset{\beta f}{\rightarrow} g'f \overset{g' \alpha}{\rightarrow} g'f'$ (the diagonal is $\beta \alpha$).

This last factor has 4 atomic 1-morphisms on the outside of the square; if $C^{(0)}$ were a computad, then we would have $\beta f' \circ g \alpha \neq g'\alpha \circ \beta f$, but in fact these composites are both equal to $\beta \alpha$. So $C^{(0)}$ is not a computad.