PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

Left derived maps preserve identities

Statement

Assume the Axiom of Dependent Choice.

Let $P$ be a supplied projective resolution datum on a class $D$ and $F : A \to B$ an additive functor between abelian categories. For every object $A \in D$ and every $n \in Z$ , $L_{n}^{P} F (1_{A}) = 1_{L_{n}^{P} F (A)} .$

Facts & Assumptions

Given: An object $A \in D$ and an integer $n$ .

[L1]

The map $L_{n}^{P} F (1_{A})$ is defined from any comparison lift of the identity on the chosen resolution of $A$ (The left derived map relative to supplied resolution data).

[L2]

Any comparison map lifting $1_{A}$ is homotopic to the identity chain map (Comparison of the identity is homotopic to the identity).

[L3]

Homology sends the identity chain map to the identity and respects composition (Homology respects identities and composition).

Proof

technique · direct

1.1L1L2given

Let ${\tilde{1}}_{∙}$ be any comparison lift of $1_{A}$ used in [L1]. By [L2], ${\tilde{1}}_{∙}$ is homotopic to the identity chain map on the chosen projective resolution of $A$ .

2.1L3step 1.1algebra∎

Applying $F$ preserves that homotopy relation as in the construction of the left derived map, so the induced map on homology agrees with the map from the identity chain map. By [L3], that latter map is $1_{L_{n}^{P} F (A)}$ . Therefore $L_{n}^{P} F (1_{A}) = 1_{L_{n}^{P} F (A)}$ .

Depends on

Dependency tree · 2 levels

Sources

Joseph J. Rotman, An Introduction to Homological Algebra, 2nd ed.