Lemma 20.25.5. Let $X$ be a topological space. Let $\mathcal{O}' \to \mathcal{O}$ be a surjection of sheaves of rings whose kernel $\mathcal{I} \subset \mathcal{O}'$ has square zero. Then $M = H^1(X, \mathcal{I})$ is a $R = H^0(X, \mathcal{O})$-module and the boundary map $\partial : R \to M$ associated to the short exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0 \]

is a derivation (Algebra, Definition 10.130.1).

**Proof.**
The map $\mathcal{O}' \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{I}, \mathcal{I})$ factors through $\mathcal{O}$ as $\mathcal{I} \cdot \mathcal{I} = 0$ by assumption. Hence $\mathcal{I}$ is a sheaf of $\mathcal{O}$-modules and this defines the $R$-module structure on $M$. The boundary map is additive hence it suffices to prove the Leibniz rule. Let $f \in R$. Choose an open covering $\mathcal{U} : X = \bigcup U_ i$ such that there exist $f_ i \in \mathcal{O}'(U_ i)$ lifting $f|_{U_ i} \in \mathcal{O}(U_ i)$. Observe that $f_ i - f_ j$ is an element of $\mathcal{I}(U_ i \cap U_ j)$. Then $\partial (f)$ corresponds to the Čech cohomology class of the $1$-cocycle $\alpha $ with $\alpha _{i_0i_1} = f_{i_0} - f_{i_1}$. (Observe that by Lemma 20.11.3 the first Čech cohomology group with respect to $\mathcal{U}$ is a submodule of $M$.) Next, let $g \in R$ be a second element and assume (after possibly refining the open covering) that $g_ i \in \mathcal{O}'(U_ i)$ lifts $g|_{U_ i} \in \mathcal{O}(U_ i)$. Then we see that $\partial (g)$ is given by the cocycle $\beta $ with $\beta _{i_0i_1} = g_{i_0} - g_{i_1}$. Since $f_ ig_ i \in \mathcal{O}'(U_ i)$ lifts $fg|_{U_ i}$ we see that $\partial (fg)$ is given by the cocycle $\gamma $ with

\[ \gamma _{i_0i_1} = f_{i_0}g_{i_0} - f_{i_1}g_{i_1} = (f_{i_0} - f_{i_1})g_{i_0} + f_{i_1}(g_{i_0} - g_{i_1}) = \alpha _{i_0i_1}g + f\beta _{i_0i_1} \]

by our definition of the $\mathcal{O}$-module structure on $\mathcal{I}$. This proves the Leibniz rule and the proof is complete.
$\square$

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