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corrected minor error in description of square category component
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Peter LeFanu Lumsdaine
Peter LeFanu Lumsdaine
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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.

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 $\beta \alpha$).

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.

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.

added 2 characters in body
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Tim Campion
Tim Campion
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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 $g,g' : \rightrightarrows 2$,$g,g' : 1 \rightrightarrows 2$; 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 $\beta \alpha$).

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.

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 $g,g' : \rightrightarrows 2$, 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 $\beta \alpha$).

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.

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 $\beta \alpha$).

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.

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Tim Campion
Tim Campion
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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 $g,g' : \rightrightarrows 2$, 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 $\beta \alpha$).

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.

Post Made Community Wiki by Tim Campion