Timeline for Smooth manifolds as idempotent splitting completion
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2017 at 23:45 | comment | added | ಠ_ಠ | I don't really know anything about this, but it sounds like this the theorem states that the category of smooth manifolds is the Karoubi envelope of the category of smooth Cartesian spaces. Maybe looking into the Karoubi envelope will be helpful? | |
| Nov 28, 2015 at 21:07 | vote | accept | Arrow | ||
| Nov 26, 2015 at 9:50 | answer | added | Peter Michor | timeline score: 12 | |
| Nov 26, 2015 at 5:25 | comment | added | Ryan Budney | It sounds like this is a categorical encoding of the uniqueness of tubular neighbourhoods theorem in a few different contexts. I think that answers your first two bullet-point questions. I suppose if you are concerned with the category of smooth manifolds this could be a useful result. | |
| Nov 26, 2015 at 2:35 | history | migrated | from math.stackexchange.com (revisions) | ||
| Nov 5, 2015 at 10:32 | comment | added | Pierre Cagne | Here's the article. I access it from my university, I don't know if it is accessible from anywhere. You can always find a way to contact me if needed... | |
| Nov 5, 2015 at 8:26 | comment | added | Qiaochu Yuan | (For the example of cohomology this isn't particularly compelling because it's already clear that you can reduce to the case of open subsets of $\mathbb{R}^n$ by embedding a smooth manifold into a large enough $\mathbb{R}^n$ and taking a small enough open neighborhood of it, by the tubular neighborhood theorem.) | |
| Nov 5, 2015 at 8:18 | comment | added | Qiaochu Yuan | For starters, splitting idempotents is an absolute colimit: it is preserved by any functor whatsoever. So a functor out of a category $C$ into an idempotent complete category $D$ is the same as a functor out of the idempotent completion of $C$ into $D$. Hence if you want to know how any functor whatsoever behaves on smooth manifolds (e.g. cohomology) it suffices to know how it behaves on open subsets of $\mathbb{R}^n$. | |
| Nov 5, 2015 at 8:05 | history | asked | Arrow | CC BY-SA 3.0 |