Lemma 20.25.4. In the situation above, assume Čech cohomology agrees with cohomology for the sheaves $\mathcal{F}_ i^ p$ and $\mathcal{G}_ j^ q$. Let $a_3 \in H^ n(X, \mathcal{F}_3^\bullet )$ and $b_1 \in H^ m(X, \mathcal{G}_1^\bullet )$. Then we have
\[ \gamma _1( \partial a_3 \cup b_1) = (-1)^{n + 1} \gamma _3( a_3 \cup \partial b_1) \]
in $H^{n + m}(X, \mathcal{H}^\bullet )$ where $\partial $ indicates the boundary map on cohomology associated to the short exact sequences of complexes above.
Proof.
We will use the following conventions and notation. We think of ${\mathcal F}_1^ p$ as a subsheaf of ${\mathcal F}_2^ p$ and we think of ${\mathcal G}_3^ q$ as a subsheaf of ${\mathcal G}_2^ q$. Hence if $s$ is a local section of ${\mathcal F}_1^ p$ we use $s$ to denote the corresponding section of ${\mathcal F}_2^ p$ as well. Similarly for local sections of ${\mathcal G}_3^ q$. Furthermore, if $s$ is a local section of ${\mathcal F}_2^ p$ then we denote $\bar s$ its image in ${\mathcal F}_3^ p$. Similarly for the map ${\mathcal G}_2^ q \to {\mathcal G}^ q_1$. In particular if $s$ is a local section of ${\mathcal F}_2^ p$ and $\bar s = 0$ then $s$ is a local section of ${\mathcal F}_1^ p$. The commutativity of the diagrams above implies, for local sections $s$ of ${\mathcal F}_2^ p$ and $t$ of ${\mathcal G}_3^ q$ that $\gamma _2(s \otimes t) = \gamma _3(\bar s \otimes t)$ as sections of ${\mathcal H}^{p + q}$.
Let ${\mathcal U} : X = \bigcup _{i \in I} U_ i$ be an open covering of $X$. Suppose that $\alpha _3$, resp. $\beta _1$ is a degree $n$, resp. $m$ cocycle of $\text{Tot}( \check{\mathcal{C}}^\bullet ({\mathcal U}, {\mathcal F}_3^\bullet ))$, resp. $\text{Tot}( \check{\mathcal{C}}^\bullet ({\mathcal U}, {\mathcal G}_1^\bullet ))$ representing $a_3$, resp. $b_1$. After refining $\mathcal{U}$ if necessary, we can find cochains $\alpha _2$, resp. $\beta _2$ of degree $n$, resp. $m$ in $\text{Tot}( \check{\mathcal{C}}^\bullet ({\mathcal U}, {\mathcal F}_2^\bullet ))$, resp. $\text{Tot}( \check{\mathcal{C}}^\bullet ({\mathcal U}, {\mathcal G}_2^\bullet ))$ mapping to $\alpha _3$, resp. $\beta _1$. Then we see that
\[ \overline{d(\alpha _2)} = d(\bar\alpha _2) = 0 \quad \text{and}\quad \overline{d(\beta _2)} = d(\bar\beta _2) = 0. \]
This means that $\alpha _1 = d(\alpha _2)$ is a degree $n + 1$ cocycle in $\text{Tot}(\check{\mathcal{C}}^\bullet ({\mathcal U}, {\mathcal F}_1^\bullet ))$ representing $\partial a_3$. Similarly, $\beta _3 = d(\beta _2)$ is a degree $m + 1$ cocycle in $\text{Tot}(\check{\mathcal{C}}^\bullet ({\mathcal U}, {\mathcal G}_3^\bullet ))$ representing $\partial b_1$. Thus we may compute
\begin{align*} d(\gamma _2(\alpha _2 \cup \beta _2)) & = \gamma _2(d(\alpha _2 \cup \beta _2)) \\ & = \gamma _2(d(\alpha _2) \cup \beta _2 + (-1)^ n \alpha _2 \cup d(\beta _2) ) \\ & = \gamma _2( \alpha _1 \cup \beta _2) + (-1)^ n \gamma _2( \alpha _2 \cup \beta _3) \\ & = \gamma _1(\alpha _1 \cup \beta _1) + (-1)^ n \gamma _3(\alpha _3 \cup \beta _3) \end{align*}
So this even tells us that the sign is $(-1)^{n + 1}$ as indicated in the lemma1.
$\square$
Comments (1)
Comment #8985 by Devang Agarwal on
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