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Feb 25, 2020 at 0:22 comment added Skies burn Perhaps I would rephrase the first dot point as: For $m=3$ the 3-simplices of the 2-coskeleton correspond to all possible choices of four triangles in $X$ which can be arranged into a tetrahedral shape.$^\dagger$ and finish at the end with: $\dagger$ I.e. to a map on the $2$-truncations $(\Delta^3)^{(2)}\to X_\bullet^{(2)}$, which is determined by where the $4$ non-degenerate $2$-simplicies are mapped, such that these form a tetrahedron.
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Jan 13, 2017 at 13:36
Jan 13, 2017 at 13:04 history answered MetalMathician CC BY-SA 3.0