Timeline for For an additive category $\mathcal{A}$, how does one show $K_0(\mathcal{A})\cong K_0(\mathcal{K}^b(\mathcal{A}))$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| May 19, 2020 at 10:14 | comment | added | Arthur Pander Maat | Thank you! I was unsure about the connection between split exact sequences and mapping cone sequences in a general additive setting so was trying to do something using the cones directly, but this is much more straightforward. | |
| May 19, 2020 at 8:47 | vote | accept | Arthur Pander Maat | ||
| May 18, 2020 at 20:03 | comment | added | Takumi Murayama | Ah, good catch! Good thing I wrote "related." Thank you for the clarification! | |
| May 18, 2020 at 20:01 | comment | added | R. van Dobben de Bruyn | @TakumiMurayama: careful, that's a little different! As Beilinson noted, there are two definitions of $K(\mathscr A)$: one for an additive category and one for an abelian category. SGA5 only deals with the latter, but the OP is asking about the former. (Moreover, they do not clearly agree if $\mathscr A$ happens to be abelian.) | |
| May 18, 2020 at 19:01 | comment | added | Takumi Murayama | A reference for the related isomorphism $K_0(\mathscr{A}) \overset{\sim}{\to} K_0(D^b(\mathscr{A}))$ is [SGA5, Exposé VIII, no. 4], although Grothendieck does not explicitly check that $\chi$ is an inverse… | |
| May 18, 2020 at 16:49 | history | answered | R. van Dobben de Bruyn | CC BY-SA 4.0 |