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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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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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The tau level is preserved by filtered chain homotopy equivalence

Statement

Let C and D be finite integer-filtered chain complexes over F2, each with total homology isomorphic to F2. A filtered chain homotopy equivalence consists of filtration-preserving chain maps f:C→D and g:D→C such that gf and fg are chain homotopic to the respective identity maps by homotopies that preserve every filtration level. Then τ(C)=τ(D).

Facts & Assumptions

Given: C,D and filtered chain maps f,g satisfying the filtered chain homotopy equivalence hypotheses in the Statement.

[F1]

The Alexander filtration and tau level: for a finite filtered complex with one-dimensional total homology, τ(C) is the least integer s for which H(FsC)→H(C) has nonzero image.

Proof

1.1F1givenalgebra

Put t=τ(C). By [F1], some class in H(FtC) maps to a nonzero class in H(C). Since f preserves filtration, its restriction sends FtC to FtD. The square formed by these restricted maps and the inclusions into total complexes commutes, so the induced map sends that class to its image under f∗. The map f∗ is an isomorphism because g∗ is its homotopy inverse; hence the resulting class in H(D) is nonzero. Therefore H(FtD)→H(D) has nonzero image and τ(D)≤t.

2.1F1step 1.1algebra∎

The same argument using the filtration-preserving map g gives τ(C)≤τ(D). Together with step 1.1 this yields τ(C)=τ(D).

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