TheoremStatement: 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.

Two out of three for quasi isomorphisms

Statement

For composable complex maps $f : X \to Y$ and $g : Y \to Z$ , if any two of $f, g, g f$ are quasi-isomorphisms, then so is the third.

Facts & Assumptions

Given: For composable complex maps $f : X \to Y$ and $g : Y \to Z$ , if any two of $f, g, g f$ are quasi-isomorphisms, then so is the third.

[F1]

Identities and composites of quasi-isomorphisms are quasi-isomorphisms (Quasi isomorphisms contain identities and are closed under composition).

Proof

1.1F1algebra

Fix any degree $n$ and write $a = H^{n} (f)$ and $b = H^{n} (g)$ . Functoriality gives $H^{n} (g f) = b a$ . If $a, b$ are invertible then $b a$ is invertible, also when any cohomology object is zero.

2.1step 1.1algebra∎

If $a, b a$ are invertible, then $b = (b a) a^{- 1}$ is invertible. If $b, b a$ are invertible, then $a = b^{- 1} (b a)$ is invertible. These are the other two possible pairs, and the argument holds in every degree.

Two out of three for quasi isomorphisms

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources