LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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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.

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Extending a partial comparison map by one degree

Statement

Let $u : A \to B$ be a morphism, and let $P_{∙} \to A$ and $Q_{∙} \to B$ be projective resolutions. Suppose morphisms $f_{0}, \dots, f_{n}$ have already been chosen so that the augmentation condition holds and the chain-map squares commute through degree $n$ . Then there exists $f_{n + 1} : P_{n + 1} \to Q_{n + 1}$ making the next square commute as well.

Facts & Assumptions

Given: Projective resolutions $P_{∙} \to A$ and $Q_{∙} \to B$ , a morphism $u : A \to B$ , and a partial comparison map through degree $n$ .

[L1]

A projective resolution is exact, so at degree $n$ its cycle object is the image of the next differential (Projective resolutions in an abelian category).

[L2]

The $n$ th cycle object is the kernel of the degree- $n$ differential (Cycle and boundary subobjects of a complex).

[L3]

Projective objects lift across epimorphisms (Projective object).

Proof

technique · direct

1.1L1L2givenalgebra

If $n = 0$ , the augmentation identity gives $ε_{Q} f_{0} d_{1}^{P} = u ε_{P} d_{1}^{P} = 0,$ so $f_{0} d_{1}^{P}$ lands in $\ker (ε_{Q}) = Z_{0} (Q)$ . If $n > 0$ , the previous squares commute and $d_{n}^{Q} f_{n} d_{n + 1}^{P} = f_{n - 1} d_{n}^{P} d_{n + 1}^{P} = 0,$ so again $f_{n} d_{n + 1}^{P}$ lands in $Z_{n} (Q) = \ker (d_{n}^{Q})$ by [L2]. Exactness of $Q_{∙}$ at degree $n$ makes the canonical map $Q_{n + 1} ↠ Z_{n} (Q)$ epic by [L1].

2.1L3step 1.1construct∎

The object $P_{n + 1}$ is projective, so [L3] lifts $f_{n} d_{n + 1}^{P} : P_{n + 1} \to Z_{n} (Q)$ across the epimorphism $Q_{n + 1} ↠ Z_{n} (Q)$ . Writing the lift as $f_{n + 1}$ gives $d_{n + 1}^{Q} f_{n + 1} = f_{n} d_{n + 1}^{P},$ so the partial comparison map extends by one degree.

Depends on

Used by

Projective comparison maps exist Theorem

Dependency tree · two levels

12 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

Charles A. Weibel, An Introduction to Homological Algebra (standard reference, not scraped)
Romyar Sharifi, Homological Algebra (standard reference, not scraped)