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Jun 29, 2022 at 10:37 answer added John Rognes timeline score: 1
Jan 16, 2017 at 20:06 comment added ACL It looks like it is the $3\times 3$ lemma (either for complexes in abelian categories, or in triangulated categories).
Jan 4, 2013 at 17:05 comment added Mariano Suárez-Álvarez Re: your update: You can refer to this MO question!
Jan 4, 2013 at 17:04 history edited Johannes Nordström CC BY-SA 3.0
Update clarifying the question
Jan 3, 2013 at 21:21 answer added ubunke timeline score: 2
Jan 3, 2013 at 21:12 history edited Johannes Nordström CC BY-SA 3.0
Corrected subscripts in statement of lemma
Jan 3, 2013 at 20:38 answer added Eric Wofsey timeline score: 4
Jan 3, 2013 at 20:32 answer added Fernando Muro timeline score: 4
Jan 3, 2013 at 20:25 comment added Mariano Suárez-Álvarez ... we really get a lot of isomorphisms. Maybe this is what you are seeing? (I am assuming everything converges; this should follow from the fact that your $C_{i,j}$ are bounded, I think!)
Jan 3, 2013 at 20:24 comment added Mariano Suárez-Álvarez Look at your grid as a double complex, and let $C$ be the total complex. There is an action of $G=\mathbb Z$ on it by the translation you described, so we can compute hypercohomology $\mathbb H^\bullet(G,C)$. Using one of the two hypercohomology spectral sequences, we see is zero because of $C$ is exact; the other hypercohomology spectral sequence has then $E_2$ page looking like $H^\bullet(H^\bullet(\mathbb Z,C))$ and converges to zero. Since $\mathbb Z$ has global dimension $1$, this spectral sequence has only two rows (columns?) and degenerates at $E_3$; since the limit is zero, ...
Jan 3, 2013 at 20:10 comment added Mariano Suárez-Álvarez $C_{23}$ should be $C_{13}$ and $C_{32}$ should be $C_{31}$ in the statement of the lemma, no?
Jan 3, 2013 at 19:43 history asked Johannes Nordström CC BY-SA 3.0