Given a covering $\mathcal{U}$ of manifold $M$ and a sheaf $F$, $H^n(\mathcal{U},F)\to H^n(M,F)$ is injection?

Question

Given a covering $\mathcal{U}$ of manifold $M$ and a sheaf $F$, the book shows $H^1(\mathcal{U},F)\to H^1(M,F)$ is injection where $H^1(M,F)$ is the direct limit over all refinements.

The book also has shown if $\forall U\in\mathcal{U}, H^1(U,F)=0$, then $H^2(\mathcal{U},F)\to H^2(M,F)$ is injection.

$\textbf{Q:}$ Is it true that if $\forall U\in\mathcal{U}, H^i(U,F)=0$ for all $i<n$, then $H^n(\mathcal{U},F)\to H^n(M,F)$ is injection?

Ref: Kodaira, Complex Manifolds and Deformation of Complex Structures Chpt 3, Sec 3.

Answer 1

This is probably overkill, but a quick way to see it. Denote $\check{H}$ for Cech cohomology and $H$ for ordinary cohomology to avoid confusion. If $F$ is a sheaf, denote $\underline{H}^q(F)$ the presheaf $U\mapsto H^q(U,F)$. In particular $\underline{H}^0(F)=F$.

Then the Cech to derived functor spectral sequence is a spectral sequence :$$E_2^{pq}=\check{H}^p(\mathcal{U},\underline{H}^q(F))\Rightarrow H^{p+q}(M,F)$$ Using the assumption that $H^q(U,F)=0$ for any $U\in\mathcal{U}$ (and the comparison theorem), the $E_2$ page looks like :

$$ \require{AMScd} \begin{CD} \vdots@.\vdots@.\vdots@.\cdots@.\vdots@.\\ @.@.@.@.@.\\ \check{H}^0(\mathcal{U},\underline{H}^n(F))@.\check{H}^1(\mathcal{U},\underline{H}^n(F))@.\check{H}^2(\mathcal{U},\underline{H}^n(F))@.\cdots@.\check{H}^n(\mathcal{U},\underline{H}^n(F))@.\\ @.@.@.@.@.\\ [email protected]@.0@.\[email protected]\\ @.@.@.@.@.\\ \vdots@.\vdots@.\vdots@.\cdots@.\vdots@.\\ @.@.@.@.@.\\ [email protected]@.0@.\[email protected]\\ @.@.@.@.@.\\ \check{H}^0(\mathcal{U},\underline{H}^0(F))@.\check{H}^1(\mathcal{U},\underline{H}^0(F))@.\check{H}^2(\mathcal{U},\underline{H}^0(F))@.\cdots@.\check{H}^n(\mathcal{U},\underline{H}^0(F))@.\\ @.@.@.@.@. \end{CD} $$ (there might be non-zero term above the highest row but they do not matters here).

Obviously there is no non-zero differentials out of $\check{H}^n(\mathcal{U},F)$ since such a differential would leave the first quadrant. Then the differentials with target $\check{H}^n(\mathcal{U},F)$ have source (a subquotient of) $\check{H}^{n-r}(\mathcal{U},\underline{H}^{r-1}(F))=0$. It follows that $\check{H}^n(\mathcal{U},F)$ survives the spectral sequence and the boundary map $\check{H}^n(\mathcal{U},F)\to H^n(M,F)$ is an injective. By comparison again, you get the desired result.