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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. All non-zero maps are the identity

$$\require{AMScd} \begin{CD} 0 @>>> 0 @>>> 0 @>>> 0 @>>> 0\\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> 0 @>>> 0 @>>> \mathbb{Z} @>>> 0\\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 @>>> 0 \end{CD} $$

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.

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. All non-zero maps are the identity

$$\require{AMScd} \begin{CD} 0 @>>> 0 @>>> 0 @>>> 0 @>>> 0\\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> 0 @>>> 0 @>>> \mathbb{Z} @>>> 0\\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 @>>> 0 \end{CD} $$

Extend the diagram by copies of $\mathbb Z$ 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 $\mathbb Z$ in it) are exact.

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.

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. All non-zero maps are the identity

$$\require{AMScd} \begin{CD} 0 @>>> 0 @>>> 0 @>>> 0 @>>> 0\\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> 0 @>>> 0 @>>> \mathbb{Z} @>>> 0\\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VVV @VVV\\ 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 @>>> 0 \end{CD} $$

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.

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.

I suspect that the $\mathbb Z\times\mathbb Z$ lemma is false. If I think of a counterexample, I'll edit this post.

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.