PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The degree-zero injective construction of Ext is Hom

Statement

Assume the Axiom of Dependent Choice, and let $I$ be supplied injective resolution data on the objects under consideration. For $0 \to N \overset{ι}{\to} I^{0} \overset{d^{0}}{\to} I^{1} \to \dots$ , $f \mapsto ι f$ gives a natural isomorphism $Hom (M, N) ≅ {Ext}_{I}^{0} (M, N)$ .

Facts & Assumptions

Given: The displayed injective resolution and an object $M$ .

Proof

technique · direct

1.1

Exactness gives $\ker d^{0} = im ι$ ; thus $ι f$ is a degree-zero cocycle, and $f \mapsto ι f$ is injective because $ι$ is monic.

givenconstruct

2.1

If $d^{0} g = 0$ , then $g$ factors uniquely through $ι$ . Since there are no negative-degree coboundaries, this is the required $H^{0}$ identification. Precomposition makes it natural in $M$ . For $b : N \to N^{'}$ , choose a comparison extension between the supplied resolutions; its degree-zero square with the coaugmentations commutes, so the identification is natural in $N$ , and the independence lemma makes this map independent of the chosen extension.

step 1.1algebra∎

Depends on

Used by

Ext zero as Hom in both constructions Example

Dependency tree · two levels

17 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, Chapter 2 (standard reference, not scraped)