Let $M$ be a differential manifold and $\mathcal H^k$ the presheaf of real vector spaces associating to the open subset $U\subset M$ the $k$-th de Rham cohomology vector space: $\mathcal H^k(U)=H^k\_{DR}(U)$. Is this presheaf a sheaf?  

Of course not! Indeed, given *any* non-zero cohomology class $0\neq[\omega]\in \mathcal H^k(U)$ represented by the closed $k$-form $\omega\in \Omega^k\_M(U)$ there exists (by Poincaré's Lemma) a covering $(U\_i)\_{i\in I}$ of $U$ by open subsets $U\_i\subset U$ such that $[\omega]\vert U\_i=[\omega\vert U\_i]=0\in \mathcal H^k(U\_i)$, and thus the first axiom for a presheaf to be a sheaf is violated.  

But what about the second axiom?  

**My question:**  

Suppose we are given a differential manifold M, a covering $(U\_\lambda)\_{\lambda \in \Lambda}$of $M$ by open subsets $U\_\lambda \subset M$, closed differential $k-$forms $\omega\_\lambda \in \Omega^k\_M(U\_\lambda)$ satisfying $[\omega\_\lambda]\vert U\_\lambda \cap U\_\mu=[\omega\_\mu]\vert U\_\lambda \cap U\_\mu\in \mathcal H^k(U\_\lambda\cap U\mu)$ for all $\lambda,\mu \in \Lambda$.  

Does there then exist a closed differential form $\omega\in \Omega^k(M)$ such that we have for the restrictions in cohomology: $[\omega]\vert U\_\lambda=[\omega \_\lambda]\in \mathcal H^k(U\_\lambda)$ for all $\lambda\in \Lambda$ ?


**Remarks**


1. This is an extremely naïve question which, to my embarrassment, I cannot solve.  

I have extensively browsed the literature and consulted some of my friends, all brilliant geometers (albeit not differential topologists), but they didn't know the answer offhand.
For what it's worth, I would guess (but not conjecture!) that such glueing is impossible.
2. If the covering of $X$ has only two opens then we can glue.  

This follows immediately from Mayer-Vietoris's long exact sequence
$$\cdots \to \mathcal H^k(M) \to \mathcal H^k(U\_1) \oplus \mathcal H^k(U\_2) \to \mathcal H^k(U\_1\cap U\_2)\to \cdots$$


**Update**  

My brilliant friends didn't answer offhand but a few hours later, unsurprisingly, they came back to me with splendid counterexamples! See below.



Can one glue De Rham cohomology classes on a differential manifolds?
No.


Make $M$ by gluing three strips to two discs to form a thrice-punctured sphere. Take three open sets $U\_\lambda$, each made by both discs and two of the strips. Then each $U\_\lambda$ is homeomorphic to annulus and thus has $1$-dimensional $H^1$.


The pairwise intersections, made from one strip connecting two discs, are contractible and so their $H^1$ vanishes. Thus, for $k=1$, the agreement condition on the pairwise intersections is vacuous.


If your claim held, then we could choose a $1$-form on $M$ that restricts to an arbitrary cohomology class on each of the three $U\_\lambda$, making the first de Rham cohomology of $M$ at least three-dimensional. But in fact it is only two-dimensional. Instead, there is a relation where the integrals around three clockwise loops around the three punctures sum to $0$, because these loops form the boundary of a particular subset of $M$.


It's true if the intersections $U\_\lambda \cap U\_\kappa \cap U\_\mu$ are empty for all distinct $\lambda,\kappa,\mu$, by iteratively applying the Mayer-Vietoris sequence or applying a single exact sequence in sheaf cohomology.


