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Suppose you have a complex $0 \to C^1 \to C^2 \to ... \to C^n \to 0$ which can be included in a commutative diagram of the following type (see image) where all rows and columns (except the complex itself) are exact and all rows sufficiently high and sufficiently low consist only of zeros.

enter image description here

The question is, for which $i$ the cohomology group $H^i(C)$ can be non-zero.

This was an exam question I was unable to solve.

The only thing that came to my mind while trying to solve this is the following: if there is an example for small $n$, say $n=3$, then we can construct an example for big $n$, just adding zeros to the complex from the example. That would give the answer for almost all $i$.

But I am not able to find such an example.

kvardekkvar
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  • @kamills I cannot see a short exact sequence of complexes here. – kvardekkvar Dec 28 '19 at 11:21
  • I'm sorry. I misinterpreted the "all rows sufficiently large consist of zeros" since I didn't see the diagonal ellipses coming off the diagram and assumed all of the stars were zero. I'll give it another look in a bit, but I can get you started on $H^1$: Note that $H^1$ is the kernel of $C^1 \to C^2$. An element $x$ in that kernel goes to zero in the down-right composition in the square with upper-left corner at $C^1$. Then the downward image of $x$ is in the kernel of the lower map, which is injective, so the image is 0. The column is exact, so $x$ came from something above $C^1$... – kamills Dec 29 '19 at 00:03

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