# $\mathcal{H}^q$ is a locally constant presheaf

Let $\pi: E \rightarrow M$ be a fiber bundle with fiber $F$ and define the presheaf $\mathcal{H}^q(U)=H^q(\pi^{-1}(U))$, for every open subset $U \subset M$, where $H^q$ denotes the De Rham cohomology.

I want to show that $\mathcal{H}^q$ is a locally constant presheaf. If $U$ is contractible, it's easy to show that $H^q(\pi^{-1}(U))=H^q(F)$, but I need help to show that the restriction $p^U_V:H^q(\pi^{-1}(U)) \rightarrow H^q(\pi^{-1}(V))$ is the identity map, for every $V \subset U$ with $V$ connected.

Any help is appreciated

• What is your definition of "locally constant presheaf"? I would think that (1) is not necessary for that (it would only be necessary for a locally constant sheaf). – Eric Wofsey Jun 26 '18 at 4:03
• You're right. I will edit, it's only necessary (2). – Renné Silva Jun 26 '18 at 10:53

It's not clear to me what you want "locally constant presheaf" to mean, but it sounds like you mean "locally isomorphic to a constant sheaf". In that case, the statement you are trying to prove is very false. For instance, let $E=M=\mathbb{R}^2$ with $\pi$ the identity map. Then $F$ is a point so $H^1(F)=0$, but every nonempty open set $U$ contains a connected open set $V$ such that $H^1(\pi^{-1}(V))=H^1(V)$ is nontrivial (for instance, take $V$ to be a small open annulus).
The only sense in which $\mathcal{H}^q$ is a "locally constant presheaf" is that its sheafification is a locally constant sheaf. This is because the sheafification can be computed using only a basis of open sets, and the contractible open sets form a basis. Choosing a trivialization of the bundle over a contractible open set $U$, the restriction map $H^q(\pi^{-1}(U))\to H^q(\pi^{-1}(V))$ is an isomorphism for any contractible $V\subseteq U$ (since our trivialization gives a natural identification of both of these groups with $H^q(F)$). So, the restriction of $\mathcal{H}^q$ to contractible open subsets of $U$ is constant, and so the sheafification of $\mathcal{H}^q$ is constant when restricted to $U$.