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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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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→N→ιI0→d0I1→⋯, f↦ιf gives a natural isomorphism Hom⁡(M,N)≅Ext⁡I0(M,N).

Facts & Assumptions

Given: The displayed injective resolution and an object M.

Proof

technique · direct
1.1givenconstruct

Exactness gives ker⁡d0=im⁡ι; thus ιf is a degree-zero cocycle, and f↦ιf is injective because ι is monic.

2.1step 1.1algebra∎

If d0g=0, then g factors uniquely through ι. Since there are no negative-degree coboundaries, this is the required H0 identification. Precomposition makes it natural in M. For b:N→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.

Depends on

Used by

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