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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-29
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A sufficiently long generator-extension iteration is injective

Statement

Assume the Axiom of Choice. Let A be a locally small Grothendieck category with generator U, and let (Mα) be the transfinite iteration of the one-step generator extension functor starting at an object M. Let κ bound the cardinalities of the sets of subobjects of all subobjects N⊆U. If λ is a limit ordinal with cf⁡(λ)>κ, then Mλ is injective.

Facts & Assumptions

Given: The Axiom of Choice, the transfinite tower (Mα) in a locally small Grothendieck category with generator U, the bound κ from [L2], and a limit ordinal λ with cf⁡(λ)>κ.

[A1]

AC supplies the choice hypothesis in both the transfinite factorization and the generator-subobject injectivity test (The Axiom of Choice).

[L1]

Extension from subobjects of the fixed generator detects injectivity (Extension from subobjects of a generator detects injectivity).

[L2]

If cf⁡(λ)>κ, every map from a subobject of the generator to Mλ factors through an earlier stage, and all transition maps are monic (Transfinite iteration of the generator extension preserves monomorphisms and factorizes small-source maps).

[L3]

Every map from a subobject of the generator into one stage extends across the generator at the next stage (The one-step generator map is a functorial monomorphism).

Proof

technique · direct
1.1L2L3givenconstruct

Let f:N→Mλ with N⊆U. By [L2], write f=jα,λg for some α<λ and g:N→Mα. Since λ is a limit ordinal, α+1<λ. By [L3], there is h:U→Mα+1 whose restriction to N is ηMαg. Compatibility of the transition maps gives jα+1,λh∣N=jα+1,ληMαg=jα,λg=f, so jα+1,λh extends f across N⊆U.

2.1A1L1step 1.1∎

Thus every map from every subobject of U extends to U. Under [A1], [L1] makes Mλ injective.

Depends on

Used by

Dependency tree · two levels

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Sources