Lemma 20.23.3. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Assume $I$ comes equipped with a total ordering. The map $c$ is a morphism of complexes. In fact it induces an isomorphism

$c : \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F})$

of complexes.

Proof. Omitted. $\square$

There are also:

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