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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A complex is zero in the derived category exactly when it is acyclic

Statement

A complex $X$ becomes a zero object in $D (A)$ if and only if $H^{n} (X) = 0$ for every integer $n$ .

Facts & Assumptions

Given: A complex $X$ becomes a zero object in $D (A)$ if and only if $H^{n} (X) = 0$ for every integer $n$ .

[F1]

Every quasi-isomorphism becomes invertible in the derived category (The localization functor sends quasi isomorphisms to isomorphisms).

[F2]

Cohomology factors through the derived category (Cohomology factors through the derived category).

Proof

1.1F1algebra

First $Q (0)$ is a zero object: a roof $X \leftarrow U \to 0$ represents the ordinary zero map, because its numerator is the composite $U \to X \to 0$ ; dually a right roof out of $0$ is zero. Hence both Hom sets involving $Q (0)$ are singletons. If $X$ is acyclic, $X \to 0$ is a quasi-isomorphism, so $Q (X) ≅ Q (0)$ .

2.1F2step 1.1∎

Conversely, if $Q (X)$ is a zero object it is isomorphic to $Q (0)$ . The factored cohomology functors take this isomorphism to $H^{n} (X) ≅ H^{n} (0) = 0$ for every $n$ . Thus $X$ is acyclic.

Depends on

Dependency tree · 2 levels

Sources

13.11.1–13.11.6