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Čech complex for a two-open cover
Statement
Let be a topological space, let be a sheaf of abelian groups on and let be open subsets with (A sheaf on a topological space). Index the two-member family by with and put , , so that is an open cover of indexed by a linearly ordered set with ordered Čech cochains and differential (Ordered Čech cochain complex of a cover). Then the only possibly nonzero component of the differential is and . Consequently, with the Čech cohomology of the cover (Fixed-cover Čech cohomology), The differential is the difference of the two restrictions, in the order of the chosen linear order ; exchanging the roles of the two members changes its sign and leaves , and all three cohomology groups unchanged.
Facts & Assumptions
The ordered Čech -cochains are the product over increasing tuples, with the convention that an empty product is the zero group (Ordered Čech cochain complex of a cover).
The Čech differential is , a sum of restrictions inside one section group (Ordered Čech cochain complex of a cover).
The cohomology of the fixed cover is , the cohomology of the cochain complex in degree (Fixed-cover Čech cohomology).
Proof
Given: A topological space , a sheaf of abelian groups on and open subsets with , indexed as .
The increasing tuples of the two-element linearly ordered set are: the two singletons and in degree , the single pair in degree , and none at all in degrees . Evaluating the product formula of [F1] at these tuples gives , then , and then for because the product over the empty set of tuples is the zero group by [F1]; the direct product of two groups is their direct sum.
For the differential formula of [F2] at the unique increasing pair reads , the two restrictions being taken along and ; there are no other components in degree because has target , which has the single component . Hence .
Since by [step 1.1], the map is the zero map, and the complex is . By the description of the cohomology of a fixed cover in [F3] this gives , and for . Reversing the linear order interchanges the two summands of and multiplies by in the sense that the new differential is the negative of the old one composed with the swap of the summands, which leaves kernel, image and cohomology unchanged. ∎
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Sources
- The Stacks Project, Cohomology of Sheaves (standard reference, not scraped)