False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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The filtration on homology is h n of the filtered subcomplex

Statement

It is false that $F_{p} H_{n} (C)$ is always $H_{n} (F_{p} C)$ , or that the latter always embeds into $H_{n} (C)$ .

Facts & Assumptions

Given: The identity complex with the displayed two-step filtration.

[F1]

The filtration on H is the image of the map from filtered-piece homology (Induced filtration on homology).

[F2]

Integer complexes can be computed using ordinary subgroup kernels and quotient cokernels (Abelian-group model for spectral-sequence computations).

Refutation

technique · direct

1.1

Let $C_{1} = C_{0} = Z$ and $d_{1} = 1_{Z}$ , with all other groups zero. Set $F_{p} C = 0$ for p<0, let $F_{0} C$ be its degree-zero stalk, and set $F_{p} C = C$ for p≥1. This is a filtration by subcomplexes because the restricted degree-zero differential is zero. In $F_{0} C$ , cycles at degree zero are ℤ and boundaries are zero, so $H_{0} (F_{0} C) = Z$ .

2.1

In C the boundary image at degree zero is $i m (1_{Z}) = Z$ , hence $H_{0} (C) = Z / Z = 0$ . Consequently [F1] gives $F_{0} H_{0} (C) = im (Z \to 0) = 0$ , not ℤ. The map kills the nonzero element 1, so it is not injective either.

F1F2step 1.1∎

Source notes

Stacks §12.24, Definition 12.24.5; the identity-complex counterexample is calculated here.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Charles A. Weibel, An Introduction to Homological Algebra, Chapter 5 (standard reference, not scraped)
Romyar Sharifi, Homological Algebra (standard reference, not scraped)