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Does the Čech cohomology always give rise to a long exact sequences given a short exact sequence of sheaves?

Clearly that cannot occur for sheaves on a paracomact (perhaps also Hausdorff, I'm not sure about that) space, but the argument one uses there relies on the topological assumptions in a crucial way and cannot be generalized to an arbitrary space, not in a straight-forward manner anyway.

Thank you for shedding any light on this issue!

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Searching for "Čehov cohomology" turns up this question and nothing else. This is not my field so it is hard to judge but perhaps some clarification what this is could be appropriate. (Or do you mean "Čech cohomology"?) –  quid Feb 25 '13 at 20:11
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He means Čech cohomology, I am quite sure as this is a student who was in my office today. I suggested he asks here. I edited the question to fix the terminology. –  Andrej Bauer Feb 25 '13 at 20:24
    
Thank you both for your hasty corrections - indeed it was Čech cohomology that I had in mind and it was little but stupidity that led me to write otherwise. –  HeWhoHungers Feb 25 '13 at 20:32
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"Čehov cohomology" sounded interesting too. –  Włodzimierz Holsztyński Feb 26 '13 at 4:15
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Cech cohomology yields a 'nice' cohomology theory if all the (multiple) intersections of the components of your covering have trivial cohomology (this is also true if you consider a limit of coverings). If you have a weird topological space, then you cannot achieve this condition; then you need hypercoverings. –  Mikhail Bondarko Feb 26 '13 at 8:25

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up vote 7 down vote accepted

It should be no. Sorry, I don't have the time to write anything detailed, so I'll just leave assembly instructions. Take a look at my answer here: equivalence of Grothendieck-style versus Cech-style sheaf cohomology describing an example due to Grothendieck where Čech differs from derived functor cohomology. If you write out what you get from the sequence $$0\to K\to \mathbb{Z}_X\to \mathbb{Z}_Y\to 0$$ described there, the Čech sequence should fail to be exact at $\check{H}^2(K)$.


Here are some details. $X$ is the affine plane over a field with the Zariski topology. $Y\subset X$ is the union of two irreducible curves $Y_i$ meeting at two points. If you are not familiar with algebraic geometry, the key point is that the sheaves $\mathbb{Z}_X$, $\mathbb{Z}_{Y_i}$ are flasque hence acylic. From the long exact sequence for sheaf cohomology (= derived functors of global sections), we have an exact sequence $$ H^1(X,\mathbb{Z}_X)=0\to H^1(Y,\mathbb{Z}_Y)\to H^2(X, K)\to H^2(X,\mathbb{Z}_X)=0$$ Using Mayer-Vietor, the second, and therefore third group, is just $\mathbb{Z}$. Now consider the sequence (I'm not saying exact sequence), $$ \check{H}^1(X,\mathbb{Z}_X)\to \check{H}^1(Y,\mathbb{Z}_Y)\to \check{H}^2(X, K)\to $$ It is known that $\check{H}^1$ coincides with $H^1$, but that $\check{H}^2(K)=0$ [cf Grothendieck "Sur quelques points...", sect 3.8]. So exactness of the second sequence fails.

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If someone can flesh this out, it would be really helpful, since it is exactly at this point that help was needed earlier today in my office. –  Andrej Bauer Feb 25 '13 at 21:49
    
Sure, I can try to do this in few days, when I finish some other things. –  Donu Arapura Feb 25 '13 at 22:02
    
Thanks! And since I am asking for favors, I vaguely remember that derived functors are Kan extensions, so cohomology should be a Kan extension. Where can I read about that? (Sorry, this is off topic, but it does not warrant a whole new question.) –  Andrej Bauer Feb 26 '13 at 0:36
    
I added some details. (Don't know much about Kan extensions, sorry.) –  Donu Arapura Feb 26 '13 at 1:47
    
The nlab page for derived functors nlab.mathforge.org/nlab/show/derived+functor talks about Kan extensions, though I'm not competent enough to decide if that helps you. –  Ketil Tveiten Feb 26 '13 at 8:29

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