Timeline for Are sifted (2,1)-colimits of fully faithful functors again fully faithful? (And a de-categorified variant)
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2018 at 10:43 | comment | added | David Jordan | Thanks Sam, I think I was being a bit naive with what I was assuming. Back to the drawing board... | |
| Jun 14, 2018 at 10:42 | vote | accept | David Jordan | ||
| Jun 12, 2018 at 17:52 | history | edited | Sam Gunningham | CC BY-SA 4.0 |
Rewrote for clarity following comments of Denis Nardin
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| Jun 12, 2018 at 17:25 | comment | added | Sam Gunningham | Cool, that's good to know! I edited my answer accordingly. Does it work now? | |
| Jun 12, 2018 at 17:24 | history | edited | Sam Gunningham | CC BY-SA 4.0 |
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| Jun 12, 2018 at 17:15 | comment | added | Denis Nardin | Yes, I believe that will work. You can find the characterization of cofinality for (n,1)-categories worked out in the appendix to this answer of mine, and indeed the 2-truncated simplicial diagram is (2,1)-cofinal. | |
| Jun 12, 2018 at 17:05 | comment | added | Sam Gunningham | Ah, ok. To be honest, I haven't thought about what sifted would mean in a $(2,1)$-sense. But I had the impression that a simplicial diagram should be an example of a sifted diagram (I believe this is the case in the $(\infty,1)$-setting - perhaps you can truncate at the 2-simplices in the $(2,1)$-setting?). Can I just replace the groupoids in my example with the corresponding Cech simplicial diagrams to get a counterexample? | |
| Jun 12, 2018 at 16:49 | comment | added | Denis Nardin | I don't think reflexive coequalizer are sifted in the (2,1) sense (for me sifted means I→I×I is cofinal, and the meaning of cofinal is different) | |
| Jun 12, 2018 at 16:09 | history | edited | Sam Gunningham | CC BY-SA 4.0 |
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| Jun 12, 2018 at 16:01 | history | answered | Sam Gunningham | CC BY-SA 4.0 |