Timeline for When an exact embedding of abelian categories induces a full embedding of their derived categories?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 17, 2013 at 12:18 | answer | added | Ralf Meyer | timeline score: 3 | |
| Jun 21, 2013 at 10:57 | vote | accept | Mikhail Bondarko | ||
| Jun 19, 2013 at 19:24 | comment | added | Alexei Pirkovskii | Sorry, of course I meant that $\mathrm{Tor}_0^A(B,B)\cong B$, not $A$. | |
| Jun 19, 2013 at 19:13 | comment | added | Alexei Pirkovskii | When the categories are module categories, and the functor is restriction along a map $A\to B$ of rings, $D^b(B)\to D^b(A)$ is a full embedding iff [ \mathrm{Tor}_i^A(B,B)\cong \begin{cases} A, & i=0,\\ 0, & i>0. \end{cases} ] See W. Geigle and H. Lenzing, "Perpendicular categories with applications to representations and sheaves", J. Algebra 144 (1991), no. 2, 273--343. | |
| Jun 19, 2013 at 17:25 | answer | added | Leonid Positselski | timeline score: 6 | |
| Jun 19, 2013 at 7:43 | comment | added | Mariano Suárez-Álvarez | When the categories are module categories, and the functor is restriction along a map of rings, such a functor $F$ is called an homological epimorphism. That keyword should produce useful information. | |
| Jun 19, 2013 at 7:21 | answer | added | Sasha | timeline score: 2 | |
| Jun 18, 2013 at 17:11 | comment | added | Benjamin Antieau | It seems like this would happen when $A$ is thick in $A'$ and projectives in $A$ are projective in $A'$. Then, I think $D^b(A)$ can be identified with $D_A^b(A')$, the derived category of complexes of objects in $A'$ with bounded homology contained in $A$, which is a full subcategory of $D(A')$. See Weibel, Exercise 10.4.3. | |
| Jun 18, 2013 at 14:20 | comment | added | Mikhail Bondarko | This is true; yet when does this happen? | |
| Jun 18, 2013 at 14:06 | comment | added | anon | I'd guess: if and only if the Exts in $A$ and $A'$ agree, | |
| Jun 18, 2013 at 13:34 | history | asked | Mikhail Bondarko | CC BY-SA 3.0 |