Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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  • 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.
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Hom into an injective makes projective-resolution rows exact

Statement

Let ⋯→P1→P0→M→0 be a projective resolution and N→I∙ an injective resolution. For every q≥0, the augmented row 0⟶Hom⁡(M,Iq)⟶Hom⁡(P0,Iq)⟶Hom⁡(P1,Iq)⟶⋯ is exact. These augmentations commute with postcomposition by dI, so their edge complex is Hom⁡(M,I∙) and maps naturally to the total Hom complex.

Facts & Assumptions

Given: The two resolutions in the statement and the Hom double complex Kp,q=Hom⁡(Pp,Iq).

Proof

technique · direct
1.1givenalgebra

Each Iq is injective, so Hom⁡(−,Iq) is exact. Applying it to the augmented projective resolution gives the displayed exact row. Naturality of Hom shows that these augmentations commute with postcomposition by dI.

2.1step 1.1algebra∎

The objects in horizontal degree −1 are Hom⁡(M,Iq), with vertical differential given by postcomposition by dI. They therefore form Hom⁡(M,I∙) and give the asserted natural edge map to the total complex.

Depends on

Used by

Dependency tree · two levels

7 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