Timeline for A more natural proof of Dold-Kan?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 3, 2017 at 19:22 | history | edited | Charles Rezk | CC BY-SA 3.0 |
fix typo
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| Feb 12, 2015 at 4:06 | history | edited | Charles Rezk | CC BY-SA 3.0 |
added 2 characters in body
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| Feb 12, 2015 at 3:59 | comment | added | Charles Rezk | @KarolSzumiło I added the quickest proof I can think of. | |
| Feb 12, 2015 at 3:58 | history | edited | Charles Rezk | CC BY-SA 3.0 |
Gave a proof.
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| Feb 11, 2015 at 16:01 | comment | added | Zhen Lin | I also tried working this out a little while ago and got stuck because $\mathbb{Z} \mathbf{\Delta}$ has lots of idempotents. But it is a nice perspective, because it is formally equivalent to the standard formulation of Dold–Kan and indeed even implies Dold–Kan for all additive categories (not just $\mathbf{Ab}$). | |
| Feb 11, 2015 at 15:12 | comment | added | Karol Szumiło | I went down that road when I was trying to figure out the proof and that is a nice way of thinking. What's not clear to me though is whether the fact that "every object in $\mathcal{C}$ is a direct sum of objects $G(n)$" can be proven with a significantly different or easier argument than the ones that appear in the standard approach. Is there some nice trick for that? | |
| Feb 11, 2015 at 14:43 | history | answered | Charles Rezk | CC BY-SA 3.0 |