The Stacks project

Lemma 21.40.1. Assumptions and notation as in Situation 21.38.1. For $\mathcal{F}$ in $\textit{PAb}(\mathcal{C})$ and $n \geq 0$ consider the abelian sheaf $L_ n(\mathcal{F})$ on $\mathcal{D}$ which is the sheaf associated to the presheaf

\[ V \longmapsto H_ n(\mathcal{C}_ V, \mathcal{F}|_{\mathcal{C}_ V}) \]

with restriction maps as indicated in the proof. Then $L_ n(\mathcal{F}) = L_ n(\mathcal{F}^\# )$.

Proof. For a morphism $h : V' \to V$ of $\mathcal{D}$ there is a pullback functor $h^* : \mathcal{C}_ V \to \mathcal{C}_{V'}$ of fibre categories (Categories, Definition 4.33.6). Moreover for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ V)$ there is a strongly cartesian morphism $h^*U \to U$ covering $h$. Restriction along these strongly cartesian morphisms defines a transformation of functors

\[ \mathcal{F}|_{\mathcal{C}_ V} \longrightarrow \mathcal{F}|_{\mathcal{C}_{V'}} \circ h^*. \]

By Example 21.39.3 we obtain the desired restriction map

\[ H_ n(\mathcal{C}_ V, \mathcal{F}|_{\mathcal{C}_ V}) \longrightarrow H_ n(\mathcal{C}_{V'}, \mathcal{F}|_{\mathcal{C}_{V'}}) \]

Let us denote $L_{n, p}(\mathcal{F})$ this presheaf, so that $L_ n(\mathcal{F}) = L_{n, p}(\mathcal{F})^\# $. The canonical map $\gamma : \mathcal{F} \to \mathcal{F}^+$ (Sites, Theorem 7.10.10) defines a canonical map $L_{n, p}(\mathcal{F}) \to L_{n, p}(\mathcal{F}^+)$. We have to prove this map becomes an isomorphism after sheafification.

Let us use the computation of homology given in Example 21.39.2. Denote $K_\bullet (\mathcal{F}|_{\mathcal{C}_ V})$ the complex associated to the restriction of $\mathcal{F}$ to the fibre category $\mathcal{C}_ V$. By the remarks above we obtain a presheaf $K_\bullet (\mathcal{F})$ of complexes

\[ V \longmapsto K_\bullet (\mathcal{F}|_{\mathcal{C}_ V}) \]

whose cohomology presheaves are the presheaves $L_{n, p}(\mathcal{F})$. Thus it suffices to show that

\[ K_\bullet (\mathcal{F}) \longrightarrow K_\bullet (\mathcal{F}^+) \]

becomes an isomorphism on sheafification.

Injectivity. Let $V$ be an object of $\mathcal{D}$ and let $\xi \in K_ n(\mathcal{F})(V)$ be an element which maps to zero in $K_ n(\mathcal{F}^+)(V)$. We have to show there exists a covering $\{ V_ j \to V\} $ such that $\xi |_{V_ j}$ is zero in $K_ n(\mathcal{F})(V_ j)$. We write

\[ \xi = \sum (U_{i, n + 1} \to \ldots \to U_{i, 0}, \sigma _ i) \]

with $\sigma _ i \in \mathcal{F}(U_{i, 0})$. We arrange it so that each sequence of morphisms $U_ n \to \ldots \to U_0$ of $\mathcal{C}_ V$ occurs are most once. Since the sums in the definition of the complex $K_\bullet $ are direct sums, the only way this can map to zero in $K_\bullet (\mathcal{F}^+)(V)$ is if all $\sigma _ i$ map to zero in $\mathcal{F}^+(U_{i, 0})$. By construction of $\mathcal{F}^+$ there exist coverings $\{ U_{i, 0, j} \to U_{i, 0}\} $ such that $\sigma _ i|_{U_{i, 0, j}}$ is zero. By our construction of the topology on $\mathcal{C}$ we can write $U_{i, 0, j} \to U_{i, 0}$ as the pullback (Categories, Definition 4.33.6) of some morphisms $V_{i, j} \to V$ and moreover each $\{ V_{i, j} \to V\} $ is a covering. Choose a covering $\{ V_ j \to V\} $ dominating each of the coverings $\{ V_{i, j} \to V\} $. Then it is clear that $\xi |_{V_ j} = 0$.

Surjectivity. Proof omitted. Hint: Argue as in the proof of injectivity. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08PI. Beware of the difference between the letter 'O' and the digit '0'.