Timeline for Is the category of vector bundles over a topological space abelian?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2021 at 0:21 | comment | added | Todd Trimble | Nico "Taking stalks" (not fibers). In brief: taking stalks of sheaves $F$ at points $x$ is a colimit of functors $F \mapsto F(U)$ where $U$ ranges over open neighborhoods of $x$ (more precisely, over the system of natural transformations $F(U) \to F(V)$ where $V \subseteq U$ are open neighborhoods). All these functors preserve colimits (and also finite limits), so a colimit of them will also preserve colimits. But the crucial fact is that the colimit over the system is a filtered colimit, and that a filtered colimit of functors to sets preserving finite limits again preserves finite limits. | |
| Oct 11, 2021 at 15:49 | comment | added | Nico | Why is "taking fibres at x" an exact functor? | |
| Jul 1, 2013 at 14:57 | comment | added | Todd Trimble | No, it doesn't change. | |
| Jun 30, 2013 at 20:50 | comment | added | Tobias Diez | Does this answer change if one looks at smooth vector bundles over a fixed manifold? I doubt it, but one often reads 'exact sequence of (smooth) vector bundles' and I'm only used to exact sequences in Abelian categories. | |
| Sep 13, 2012 at 17:07 | comment | added | Qiaochu Yuan | Re: the first paragraph, when the space $X$ is compact Hausdorff, the Serre-Swan theorem asserts that the category of vector bundles over $X$ is precisely the category of finitely-generated projective modules over $C(X)$ (no sheaves required). | |
| Sep 13, 2012 at 16:49 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 399 characters in body
|
| Sep 13, 2012 at 16:44 | history | answered | Todd Trimble | CC BY-SA 3.0 |