Timeline for What is the intuitive difference between these two simplicial subdivision functors?
Current License: CC BY-SA 4.0
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| Oct 18, 2023 at 19:03 | comment | added | HenrikRüping | Peters example is here slightly simpler. We have $S^1 = \Delta^1 \cup_{\partial \Delta^1} pt$. Since sd is a left adjoint, it is compatible with pushouts, so $sd(S^1) = sd(\Delta^1) \cup_{sd(\partial \Delta^1)} sd(pt)$, e.g. we take the qoutient of $sd(\delta^1)$ by the subdivision of the boundary. To see what $\pi$ does, we can use that the target is the nerve of a poset. Thus we only have to figure out what happens on zero-simplices and then a simplex $0,1,,..,n$ will be send to the chain $\pi(0),..,\pi(n)$, | |
| Oct 18, 2023 at 18:36 | comment | added | FShrike | Thanks for this example. I'm having a bit of trouble verifying the action of $\pi$ here; it seems like I would need to resort to an explicit calculation of the simplices of the colimit defining $\operatorname{sd}(X)$ which I'm too tired to accurately do right now. Is there an easier way without getting one's hands dirty? | |
| Oct 17, 2023 at 19:05 | comment | added | Peter LeFanu Lumsdaine | Excellent example; worth also noting that $S^1 = \Delta^1/\delta \Delta^1$ is already fairly illustrative, and is easier to draw. | |
| Oct 17, 2023 at 13:49 | history | answered | HenrikRüping | CC BY-SA 4.0 |