Timeline for Is singular barycentric subdivision injective?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2017 at 19:29 | comment | added | Greg Friedman | And I'm still pretty sure the cancellation happens as I originally said. Look, suppose $\sigma_1$ has its endpoints at $(1,0)$, embedding the circle in the plane in the standard way. Then the barycentric subdivision of $\sigma_1$ is the sum of the two CCW-oriented arcs between $(1,0)$ and $(-1,0)$. Similarly, $\sigma_2$ has its endpoints at $(-1,0)$, and so is not $-\sigma_1$, but its subdivision is the sum of the two arcs oriented clockwise. So everything cancels as singular chains. | |
| Mar 28, 2017 at 19:27 | comment | added | Greg Friedman | If you don't reverse the orientation, nothing can cancel. | |
| Mar 27, 2017 at 13:14 | comment | added | C. Dubussy | But if we don't reverse the orientation and only do a rotation of 180 degrees, then $\sigma_1 \neq \sigma_2$ and $b(\sigma_1)=b(\sigma_2).$ | |
| Mar 27, 2017 at 11:59 | comment | added | C. Dubussy | If I'm not mistaken, the cancellation in pairs occurs in homology but not in $\Delta_p$. | |
| Mar 27, 2017 at 7:46 | history | answered | Greg Friedman | CC BY-SA 3.0 |