## Is there an infinity × infinity lemma for abelian categories?

Many people know that there is a (3×3) nine lemma in category theory. There is also apparently a sixteen lemma, as used in a paper on the arXiv (see page 24). There might be a twenty-five lemma, as it's mentioned satirically on Wikipedia's nine lemma page.

Are the 4×4 and 5×5 lemmas true? Is there an n×n lemma? How about even more generally, if I have an infinity × infinity commutative diagram with all columns and all but one row exact, is the last row exact too? For all of these, if they are true, what are their exact statements, and if they are false, what are counterexamples?

Note: There are a few possibilities for what infinity × infinity means -- e.g., it could be Z×Z indexed or N×N indexed. Also, in the N×N case, there are some possibilities on which way arrows point and which row is concluded to be exact.

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Yes, there is an $n\times n$ lemma, and even an $\mathbb N\times\mathbb N$ lemma. The spectral sequence argument that Reid gives works. Another elementary proof uses the salamander lemma, a result of George Bergman's that I blogged about at SBS. It's exactly the same as the proof of the $3\times 3$ lemma I wrote up there.

Here's a counterexample to the $\mathbb Z\times\mathbb Z$ lemma. If you read about the salamander lemma, you'll understand how I came up with it. Here Z denotes $\mathbb Z$ and all non-zero maps are the identity

0 -> 0 -> 0 -> 0 -> 0
|    |    |    |
v    v    v    v
0 -> 0 -> 0 -> Z -> 0
|    |    |    |
v    v    v    v
0 -> 0 -> Z -> Z -> 0
|    |    |    |
v    v    v    v
0 -> Z -> Z -> 0 -> 0
.
.
.


Extend the diagram by copies of Z down and to the left, and put 0's everywhere else. All columns are exact, and all rows except one (the one with a single Z in it) are exact.

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 Thanks. Do you mind stating the N×N lemma exactly? – aorq Nov 8 2009 at 2:25 Suppose you have lower-right quadrant bicomplex (where the arrows go down and to the right; everything outside the lower right quadrant is zero) in which all the columns are exact and all the rows are exact, except possibly one. Then that row is also exact. – Anton Geraschenko♦ Nov 8 2009 at 3:15

For an n×n square, the "lemma" follows by applying the spectral sequence of a double complex in two different ways: if we first take homology along columns, then the E2 page is 0, while if we first take homology along rows, then the E2 page is 0 outside of the row in question, and there are no possible further differentials. Since both spectral sequences converge to the homology of the total complex, the unknown row must be exact also.

For Z×Z indexed double complexes, there might be convergence issues for the spectral sequences; I'm not sure.

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 Thanks for the answer, Reid. I've accepted Anton's because he answers the N×N and Z×Z questions. – aorq Nov 8 2009 at 22:18