LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Canonical truncation is a complex and has the claimed cohomology

Statement

Canonical truncations are functorial on complexes and on the homotopy category, with $H^{i} (τ^{\leq n} X) = H^{i} (X)$ for $i \leq n$ and zero for $i > n$ , and $H^{i} (τ^{\geq n} X) = H^{i} (X)$ for $i \geq n$ and zero for $i < n$ . They preserve quasi-isomorphisms and descend to functors on the derived category.

Facts & Assumptions

Given: Canonical truncations are functorial on complexes and on the homotopy category, with $H^{i} (τ^{\leq n} X) = H^{i} (X)$ for $i \leq n$ and zero for $i > n$ , and $H^{i} (τ^{\geq n} X) = H^{i} (X)$ for $i \geq n$ and zero for $i < n$ . They preserve quasi-isomorphisms and descend to functors on the derived category.

[F1]

Canonical upper truncation uses the kernel at its cut, and lower truncation uses the cokernel at its cut (Canonical truncation of a complex).

Proof

1.1

Since $d^{n} d^{n - 1} = 0$ , the prescribed factors through the kernel and cokernel exist. Their adjacent composites are zero by the same equation, while all other composites are unchanged or zero. Maps of complexes preserve kernels and images, hence induce the truncation maps and preserve identities and composition. This also gives zero truncations of the zero complex.

F1algebra

1.2

For the upper truncation the cycles in degree $n$ are $\ker d_{X}^{n}$ and the boundaries are $im d_{X}^{n - 1}$ ; degree $n - 1$ has the same kernel because the target inclusion is monic. Other retained degrees are unchanged. For the lower truncation the kernel at $n$ is $(\ker d_{X}^{n}) / (im d_{X}^{n - 1})$ and there are no incoming boundaries; in degree $n + 1$ the boundary image is unchanged because the map to the cokernel is epic. Deleted degrees have zero cohomology.

F1algebra

1.3

If $f - g = d h + h d$ , truncate $h$ unchanged where both adjacent terms remain. For $τ^{\leq n}$ restrict $h^{n}$ to $\ker d_{X}^{n}$ and set higher components to zero; the homotopy identity at the boundary holds because $d_{X}^{n}$ vanishes there. For $τ^{\geq n}$ compose $h^{n + 1}$ with the quotient onto the target cokernel and set lower components to zero; the degree- $n$ identity follows modulo target boundaries. Thus homotopic maps remain homotopic.

F1algebra

2.1

The cohomology formulas imply that truncating a quasi-isomorphism is a quasi-isomorphism in every retained or deleted degree. Composing truncation on $K$ with localization therefore inverts denominators, and the localization universal property descends it to $D$ . The natural truncation maps descend as well.

step 1.2step 1.3algebra∎

Depends on

Canonical truncation of a complex

Used by

Cited to discharge well-definedness by Canonical truncation of a complex.

Dependency tree · two levels

3 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

12.15, all four chain and four cochain truncations (standard reference, not scraped)