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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Extension by zero can be strictly smaller than direct image on a punctured interval

Statement refuted

For every open immersion j:U↪X and every sheaf of abelian groups F on U, one has j!F=j∗F.

Facts & Assumptions

Given: The open immersion j:U=(−1,0)∪(0,1)↪X=(−1,1) and the constant sheaf Z‾ on U.

[F1]

For an open immersion, direct image is computed by intersection: (j∗Z‾)(V)=Z‾(V∩U) (Direct image along an open immersion is restriction-compatible intersection).

[F2]

Extension by zero consists of sections whose support is closed in the test open (Extension by zero for abelian sheaves on an open subspace).

Counterexample

technique · direct
1.1F1givenchoose

By [F1], the global section set of j∗Z‾ is (j∗Z‾)(X)=Z‾(U)≅Z×Z, because U has two connected components. Let s be the section that is 1 on both components.

2.1F2step 1.1

The germ of s is nonzero at every point of U, so its support is all of U. But U=(−1,0)∪(0,1) is not closed in X=(−1,1), since its closure contains 0. Therefore [F2] shows that s∉(j!Z‾)(X).

3.1step 1.1step 2.1∎

Thus s lies in j∗Z‾ but not in j!Z‾, so the two sheaves are not equal. This refutes the statement.

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources