# Are cohomology functors sheaves?

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.

• No, in fact the cohomology presheaves $U\mapsto H^n_{\rm dR}(U)$ sheafify to zero for $n>0$. Jul 9, 2020 at 17:52
• 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.
– mme
Jul 9, 2020 at 17:54
• 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. Jul 9, 2020 at 18:01
• @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? Jul 9, 2020 at 18:03
• This upvote downvote game is funny :D 3 upvotes, 2 downvotes. Jul 9, 2020 at 18:19

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