Alphabeta Math
TheoremStatement: AI-adaptedProof: 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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A chosen chain of injective embeddings gives an injective resolution

Statement

Let A be an object of an abelian category. Suppose one has chosen a monomorphism A↣I0 into an injective object and, for each n≥0, a monomorphism Cn↣In+1 from the current cokernel Cn of the previous displayed map into an injective object. Then composing each quotient map with its chosen embedding produces a coaugmented complex 0→A→I0→I1→I2→⋯ that is an injective resolution of A.

Facts & Assumptions

Given: An object A of an abelian category, together with a chosen injective embedding of A and a chosen injective embedding of each successive cokernel.

[L1]

A chosen injective embedding of the current cokernel extends a partial coaugmented resolution by one exact step (One-step extension of a partial injective resolution).

[L2]

An injective resolution is an exact coaugmented complex of injective objects (Injective resolutions in an abelian category).

Proof

technique · direct
1.1L1givenconstruct

Start with the chosen monomorphism A↣I0. Applying [L1] to the chosen embedding C0↣I1 makes 0→A→I0→I1 exact, and repeating the same step with the chosen embedding of each later cokernel produces an exact coaugmented complex 0→A→I0→I1→I2→⋯ whose terms are all injective.

2.1L2step 1.1∎

By [L2], the complex assembled in step 1.1 is an injective resolution of A, including the case A=0 when the chosen initial embedding may be 0↣0.

Depends on

Used by

Dependency tree · two levels

5 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