TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Projective comparison maps exist

Statement

Assume the Axiom of Dependent Choice.

Let $u : A \to B$ be a morphism, and let $P_{∙} \to A$ and $Q_{∙} \to B$ be projective resolutions. Then there exists an augmentation-preserving chain map $f_{∙} : P_{∙} \to Q_{∙}$ lifting $u$ .

Facts & Assumptions

Given: A morphism $u : A \to B$ and projective resolutions $P_{∙} \to A$ , $Q_{∙} \to B$ .

[L1]

Degree zero can be lifted across the target augmentation (Lifting a map through degree zero of a projective resolution).

[L2]

A partial comparison map extends one degree at a time (Extending a partial comparison map by one degree).

[L3]

The required notion is an augmentation-preserving chain map (Augmentation-preserving maps of projective resolutions).

[L4]

Dependent choice licenses the countable successor-by-successor selection of compatible lifts (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · direct

1.1L1construct

By [L1], choose $f_{0} : P_{0} \to Q_{0}$ with $ε_{Q} f_{0} = u ε_{P}$ .

2.1L2L4step 1.1choose

Starting from the degree-zero lift in step 1.1, every partial comparison map through degree $n$ extends to one through degree $n + 1$ by [L2]. The successive choices depend on the previously chosen partial map, so [L4] produces maps $f_{n}$ in every degree.

3.1L3step 1.1step 2.1∎

The family $(f_{n})$ is a chain map by construction, and step 1.1 gives the augmentation identity. Hence [L3] is satisfied, so $f_{∙}$ is a comparison map lifting $u$ .

Projective comparison maps exist

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources