Arcadian Functor

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Marni D. Sheppeard

Tuesday, September 16, 2008

M Theory Lesson 223

Recall that internal precategories in Set are formed from three sets, say $C_0$, $C_1$ and $C_2$. Now imagine these are any distinct (finite) sets. Since elements of sets are sets, one can form a three element set $\{ C_0, C_1, C_2 \}$, which is isomorphic to any three element set. This set is obtained by taking unions of the component sets, and the six ways of doing this are the paths on the parity cube. In categorical terms, this cube is canonically given by colimits (pushouts) on the three initial directions in space. In other words, the permutations of three letters exist for free in the topos Set. In M Theory, it is convenient to view all groups and groupoids as derived structures. In order to specify the maps (group operations) between paths on the cube, one must fill in the squares with higher dimensional cells, but these don't live in the one dimensional Set.

As a Chu space, a permutation on three letters should be a map $3 \times 3 \rightarrow \Omega$ into the two point set of Boolean truth values. For example, the circulant $(231)$ sends the Cartesian product elements $(C_0,C_1)$, $(C_1,C_2)$ and $(C_2,C_0)$ to the value 1 and all other pairs to the value 0. Similarly, the identity $(123)$ labels $(C_0,C_0)$, $(C_1,C_1)$ and $(C_2,C_2)$ as true. In a topos, one automatically pulls back such arrows along the arrow true $1 \rightarrow \Omega$. Since the pullback arrow into 1 is unique, this square is described by the arrow into $3 \times 3$. For the identity $(123)$, this is naturally the diagonal map $\Delta: x \mapsto (x,x)$ into $3 \times 3$. Thus the permutations generalise the diagonal map by selecting different arrows.


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