The Stacks project

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. $\square$

Comments (1)

Comment #9952 by on

I have difficulties to believe this Lemma. Let me try to explain my problem in a situation contravariant to this one, eg homology of simplicial complexes. Let c<d be two vertices connected with a line. The boundary of the ordered chain (c,d) is c-d. Alternating just produces c-d. Alternating (c,d) makes it to (c,d)-(d,c) and the bounday is 2c-2d. It doesn't fit.

There are also:

  • 2 comment(s) on Section 20.23: The alternating Čech complex

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