Lemma 20.11.12. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $I$ be a set. For $i \in I$ let $\mathcal{F}_ i$ be an $\mathcal{O}_ X$-module. Let $U \subset X$ be open. The canonical map

$H^ p(U, \prod \nolimits _{i \in I} \mathcal{F}_ i) \longrightarrow \prod \nolimits _{i \in I} H^ p(U, \mathcal{F}_ i)$

is an isomorphism for $p = 0$ and injective for $p = 1$.

Proof. The statement for $p = 0$ is true because the product of sheaves is equal to the product of the underlying presheaves, see Sheaves, Section 6.29. Proof for $p = 1$. Set $\mathcal{F} = \prod \mathcal{F}_ i$. Let $\xi \in H^1(U, \mathcal{F})$ map to zero in $\prod H^1(U, \mathcal{F}_ i)$. By locality of cohomology, see Lemma 20.7.2, there exists an open covering $\mathcal{U} : U = \bigcup U_ j$ such that $\xi |_{U_ j} = 0$ for all $j$. By Lemma 20.11.3 this means $\xi$ comes from an element $\check\xi \in \check H^1(\mathcal{U}, \mathcal{F})$. Since the maps $\check H^1(\mathcal{U}, \mathcal{F}_ i) \to H^1(U, \mathcal{F}_ i)$ are injective for all $i$ (by Lemma 20.11.3), and since the image of $\xi$ is zero in $\prod H^1(U, \mathcal{F}_ i)$ we see that the image $\check\xi _ i = 0$ in $\check H^1(\mathcal{U}, \mathcal{F}_ i)$. However, since $\mathcal{F} = \prod \mathcal{F}_ i$ we see that $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is the product of the complexes $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}_ i)$, hence by Homology, Lemma 12.32.1 we conclude that $\check\xi = 0$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).