Lemma 21.12.5 (060L)—The Stacks project

Lemma 21.12.5. 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 21.7.3, there exists a covering $\mathcal{U} = \{ U_ j \to U\} $ such that $\xi |_{U_ j} = 0$ for all $j$. By Lemma 21.10.4 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 21.10.4), 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 12.32.1 we conclude that $\check\xi = 0$ as desired. $\square$

Comments (4)

Comment #568 by Nuno on April 30, 2014 at 18:45

I'm probably forgetting something about sheaf cohomology, but I didn't understand the proof of this lemma. Also, I'm not very well versed in cohomology on sites, so I'll restrict myself to the topological case. Why are the edge maps $\check H^1(\mathcal{U}, \mathcal{F}_i) \to H^1(U, \mathcal{F}_i)$ injective for all $i$ ? Since we have a canonical isomorphism $\check H^1(U, \mathcal{F}_i) \to H^1(U, \mathcal{F}_i)$ , this would mean that the canonical map $\check H^1(\mathcal{U}, \mathcal{F}_i) \to \check H^1(U, \mathcal{F}_i)$ is injective, and I believe this isn't true. As a side remark, since we already know that $H^p(U, \cdot)$ is a functor, why should we repeat the argument here to construct the canonical map?

Comment #569 by Johan on April 30, 2014 at 20:49

This horrible lemma is used only in one place and there it is used to show that if $H^1(U, \mathcal{F}_i) = 0$ for all $i$ , then $H^1(U, \mathcal{F})$ is zero. Anyway, as to your remarks.

The map $\check{H}^1(\mathcal{U}, \mathcal{G})$ to $H^1(U, \mathcal{G})$ is always injective for any abelian sheaf $\mathcal{G}$ . Namely, let $(g_{ij})$ be a cocycle. If the image is zero in cohomology, then the associated torsor is trivial. Thus the torsor has a global section. This means there are sections $g_i$ over $U_i$ such that $g_i + g_{ij} = g_j$ on the overlaps. In fact, the group $\check{H}^1(\mathcal{U}, \mathcal{G})$ classisfies exactly those torsors which are trivial over the $U_i$ .

I will explain this somewhere and improve the lemma.

OK, we do not need to construct the map: that is just nonsense!

Comment #570 by Nuno on April 30, 2014 at 21:50

I need exactly that result for formally smooth morphisms (only for schemes with Zariski topology though). I've got it now, and that is pretty clear from the proof that $H^1$ classifies torsors, which I learned not a long time ago and should not have forgotten. Thanks!

Comment #571 by Johan on May 01, 2014 at 00:04

OK, thanks for letting me know! I improved the exposition by adding a lemma on torsors and chech $H^1$ . The commit is here.

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