The Stacks project

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

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

So this even tells us that the sign is $(-1)^{n + 1}$ as indicated in the lemma¹. $\square$