Timeline for Are subfunctors of left exact functors also left exact?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 5, 2013 at 21:16 | vote | accept | Louis A | ||
| Apr 2, 2013 at 18:53 | comment | added | Eric Wofsey | It turns out that $f_!$ is still left exact. This follows from the fact that $f_*$ is left exact and for any subsheaf $\mathcal{F}\subset\mathcal{G}$, $f_!\mathcal{G}\cap f_*\mathcal{F}=f_!\mathcal{F}$ as subsheaves of $f_*\mathcal{G}$. | |
| Apr 2, 2013 at 5:21 | vote | accept | Louis A | ||
| Apr 2, 2013 at 5:21 | |||||
| Apr 2, 2013 at 5:20 | comment | added | Louis A | Thanks, does that mean that $f_!$ might not be left exact then? | |
| Apr 1, 2013 at 17:51 | comment | added | Steven Landsburg | Eric: Aagh. You're right. | |
| Apr 1, 2013 at 17:38 | comment | added | Eric Wofsey | It is obvious that a subfunctor of a left exact functor sends short exact sequences to sequences that are exact on the left (i.e., it sends injections to injections), but it is neither obvious nor true that it preserves exactness in the middle. | |
| Apr 1, 2013 at 17:32 | answer | added | Eric Wofsey | timeline score: 21 | |
| Apr 1, 2013 at 17:05 | answer | added | Omar Antolín-Camarena | timeline score: 6 | |
| Apr 1, 2013 at 16:26 | comment | added | Steven Landsburg | Yes, obviously. | |
| Apr 1, 2013 at 15:50 | history | asked | Louis A | CC BY-SA 3.0 |