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Question is the following:

Is the functor $H^n_{dR}:\text{Man}\rightarrow \text{Set}$ a sheaf with respect to open cover topology on $\text{Man}$?

More generally, are cohomology functors sheaves in general (in any reasonably non trivial Grothendieck topology)?

I am also interested in cohomology functors that arise in Algebriac geometry/topology.

Is there a way of sheafification in this setup?

I have nothing much to support this question, this is completely out of curiosity.

Edit : I am also interested in answers/references related to the comment of Piotr Achinger; that reads

"in what way is cohomology a sheaf" leads one to notions like $\infty$-topoi etc.

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    $\begingroup$ No, in fact the cohomology presheaves $U\mapsto H^n_{\rm dR}(U)$ sheafify to zero for $n>0$. $\endgroup$
    – Piotr Achinger
    Commented Jul 9, 2020 at 17:52
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    $\begingroup$ No, as almost any example demonstrates (a sphere, for instance, covered by two contractible open sets); if cohomology were a sheaf the Mayer-Vietoris sequence would split as short exact sequences of $H^k$ for each $k$. $C^*(M)$ forms a sheaf of complexes, but the gluing property does not survive passing to cohomology. $\endgroup$
    – mme
    Commented Jul 9, 2020 at 17:54
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    $\begingroup$ The question is not stupid - trying to make sense of "in what way is cohomology a sheaf" leads one to notions like $\infty$-topoi etc. $\endgroup$
    – Piotr Achinger
    Commented Jul 9, 2020 at 18:01
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    $\begingroup$ @PiotrAchinger that is interesting. can you please suggest some (possibly short) reference that introduce $\infty$-topoi when trying to understand in what sense cohomology is a sheaf? $\endgroup$
    – Praphulla Koushik
    Commented Jul 9, 2020 at 18:03
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    $\begingroup$ This upvote downvote game is funny :D 3 upvotes, 2 downvotes. $\endgroup$
    – Praphulla Koushik
    Commented Jul 9, 2020 at 18:19

1 Answer 1

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Is the functor H^n_dR:Man→Set a sheaf with respect to open cover topology on Man?

As already pointed out in the comments, the answer is no for n>0, yes for n=0.

"in what way is cohomology a sheaf" leads one to notions like ∞-topoi etc.

In the context of this question, the assignment M↦Ω(M) yields a contravariant functor from smooth manifolds to cochain complexes, and this functor satisfies the homotopy descent condition.

This was first proved by Weil in Sur les théorèmes de de Rham (DigiZeitschriften, DOI, EuDML). An accessible exposition is given by Bott and Tu in §8 of Differential Forms in Algebraic Topology.

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  • $\begingroup$ The link springerlink.com/index/10.1007/BF02564296 is broken.. $\endgroup$
    – Praphulla Koushik
    Commented Jul 12, 2020 at 3:02
  • $\begingroup$ I will read the part you mentioned from Bott and Tu. I am hearing the word “homotopic descent condition” for the first time. I can guess what it is, but I will read about it and ask if I have any further questions.. $\endgroup$
    – Praphulla Koushik
    Commented Jul 12, 2020 at 3:10
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    $\begingroup$ @PraphullaKoushik Try eudml.org/doc/139040 for Weil's paper, though EuDML is down for maintenance at the moment. Here's an unofficial link: maths.ed.ac.uk/~v1ranick/papers/weil.pdf $\endgroup$
    – David Roberts ♦
    Commented Jul 12, 2020 at 3:48
  • $\begingroup$ @DavidRoberts ok $\endgroup$
    – Praphulla Koushik
    Commented Jul 12, 2020 at 4:39
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    $\begingroup$ @PraphullaKoushik: I have no links to SpringerLink in my post. $\endgroup$
    – Dmitri Pavlov
    Commented Jul 12, 2020 at 17:47

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