LemmaStatement: AI-generatedProof: AI-generated

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.

Kernel models the homotopy fiber when the cokernel is acyclic

PDF p. 13: Supported Chow group

Statement

Let $f : A_{∙} \to B_{∙}$ be a chain map of abelian groups, $K = \ker f$ , and suppose its cokernel complex is acyclic, meaning that all its homology groups vanish. Define $F_{n} = A_{n} \oplus B_{n + 1}$ with differential $d_{F} (a, b) = (d_{A} a, f (a) - d_{B} b)$ . Then $K \to F$ , $a \mapsto (a, 0)$ , induces an isomorphism on every homology group. The complex $F$ is the standard homotopy-fiber model of $f$ .

Facts & Assumptions

Given: A chain map $f$ with acyclic cokernel, and $K, F$ as stated.

Goal: Prove that the inclusion $K \to F$ is a quasi-isomorphism, meaning a map inducing isomorphisms on all homology groups.

Proof

1.1givenalgebra

Since $f$ commutes with differentials, $d_{F}^{2} = 0$ ; the inclusion of $K$ into $F$ is a degreewise injective chain map. Its quotient is $Q_{n} = (A_{n} / K_{n}) \oplus B_{n + 1}$ with the induced differential.

2.1givenstep 1.1algebra

Identify $A / K$ with the image subcomplex $I \subset B$ . The map $Q_{n} \to (B / I)_{n + 1}$ sending $(x, b)$ to the class of $b$ is a surjective chain map to the shifted cokernel, whose differential is the negative of that of $B / I$ . Its kernel is the complex $D_{n} = I_{n} \oplus I_{n + 1}$ with differential $(x, y) \mapsto (d x, x - d y)$ .

3.1givenstep 2.1algebra∎

The maps $h_{n} : D_{n} \to D_{n + 1}$ given by $h_{n} (x, y) = (y, 0)$ satisfy $d h + h d = i d$ , so $D$ is acyclic. The shifted cokernel is acyclic by hypothesis; the long exact homology sequence of $0 \to D \to Q \to (B / I) [1] \to 0$ therefore makes $Q$ acyclic. Applying the same exact sequence to $0 \to K \to F \to Q \to 0$ proves the claimed isomorphisms.

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