Timeline for Maps between K-groups induced by rings homomorphism
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2009 at 22:04 | comment | added | t3suji | As far as I understand, if you take Grothendieck group of the derived category of complexes bounded on one side only, you get zero. | |
| Dec 19, 2009 at 16:13 | comment | added | Frank Moore | Well, I was thinking about the ones that were bounded only on the left; what I wrote I guess would not even be defined unless one is in this situation. | |
| Dec 19, 2009 at 4:42 | comment | added | Hailong Dao | Preserving the relations is the tricky part (I assume you are thinking about the Grothendieck group of the bounded complexes). | |
| Dec 19, 2009 at 3:37 | history | answered | Frank Moore | CC BY-SA 2.5 |