Lemma 21.13.5. Let $\mathcal{C}$ be a site. Let $I$ be a set. For $i \in I$ let $\mathcal{F}_ i$ be an abelian sheaf on $\mathcal{C}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. 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 Sites, Lemma 7.10.1. 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 21.8.3, there exists a covering $\mathcal{U} = \{ U_ j \to U\} $ such that $\xi |_{U_ j} = 0$ for all $j$. By Lemma 21.11.4 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 21.11.4), 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 12.29.1 we conclude that $\check\xi = 0$ as desired.
$\square$

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