Timeline for Is $\textbf{FHILB}$ locally regular?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2015 at 16:02 | comment | added | Chris Heunen | That's true, but finite ones do (i.e. products of finitely many objects) | |
| May 1, 2015 at 12:06 | comment | added | King Kong | Thanks Chris. I'll have a look at that paper. But one more question, I thought Cartesian products didn't always exist in $\textbf{HILB}?$ | |
| Apr 29, 2015 at 20:51 | comment | added | Chris Heunen | Yes, for $\mathbf{Hilb}$ regularity and local regularity are also equivalent. $\mathbf{Hilb}$ is still finitely complete and finitely cocomplete. I'm not sure about pullbacks of regular epis, but if you replace regular epis by so-called zero epis you do get a good factorization system and everything works beautifully, see arxiv.org/abs/0902.2355. | |
| Apr 29, 2015 at 8:08 | comment | added | King Kong | Thanks Chris. Don't know why I had it in my head that $\textbf{FHILB}$ wasn't regular. But I am right in thinking that $\textbf{HILB}$ (where you include infinite dimensional spaces) isn't regular, aren't I? If that's the case, then I guess that the same argument shows that $\textbf{HILB}$ isn't locally regular? | |
| Apr 29, 2015 at 8:03 | vote | accept | King Kong | ||
| Apr 28, 2015 at 12:42 | comment | added | Andrej Bauer | How close are we to concluding that $\mathbf{FHilb}$ is regular because it's aglebraic? (It isn't quite algebraic, but seems "mostly" so.) | |
| Apr 28, 2015 at 9:35 | history | answered | Chris Heunen | CC BY-SA 3.0 |