Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-29
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Transfinite iteration of the generator extension preserves monomorphisms and factorizes small-source maps

Statement

Assume the Axiom of Choice. In a locally small Grothendieck category, starting from an object M0=M, define a transfinite sequence by Mα+1=M(Mα) and, at limit ordinals λ, by Mλ=colim⁡α<λMα. Then every transition map Mα→Mβ is monic. Let κ bound the cardinalities of the sets of subobjects of all subobjects N⊆U. If λ has cofinality greater than κ, then every map N→Mλ with N⊆U factors through some earlier stage Mα.

Facts & Assumptions

Given: The Axiom of Choice, a locally small Grothendieck category with generator U, and the transfinite sequence defined from the one-step generator extension functor.

[L1]

The successor-stage maps Mα→Mα+1 are monic (The one-step generator map is a functorial monomorphism).

[L2]

AB5 gives the directed-subobject identity (The axioms AB5 and AB5*).

[L3]

In a locally small abelian category with a generator, each object has a set of subobjects (An AB3 locally small abelian category with a generator is well-powered).

[L4]

In a cocomplete abelian category, AB5 is equivalent to exactness of filtered colimits (AB5 is equivalent to exactness of filtered colimits).

[A1]

AC makes the set of subobjects of U well-orderable and supplies its cardinality (The Axiom of Choice).

Proof

technique · direct
1.1L1construct

For every successor ordinal, the transition map Mα→Mα+1 is monic by [L1]. By transfinite induction, any transition map whose target is a successor stage is monic.

2.1L2L4step 1.1induction

Let λ be a limit ordinal and fix α<λ. For α≤β<λ, the short exact sequences 0→Mα→Mβ→coker⁡(Mα→Mβ)→0 form a filtered system. Exactness of filtered colimits under [L2, L4] makes the colimit sequence begin 0→Mα→Mλ, so the canonical map Mα→Mλ is monic. Together with step 1.1, transfinite induction now shows that every transition map in the tower is monic.

3.1L2L3A1step 2.1givenconstruct∎

By [L3], the subobjects of U form a set; [A1] gives its cardinality κ. For each N⊆U, composition with N↣U injects its subobject classes into those of U, so κ bounds all their cardinalities. Fix f:N→Mλ with N⊆U, and regard each Mα as a subobject of Mλ by step 2.1. The preimages Nα=f−1(Mα) form an increasing family of subobjects of N, and [L2] gives ⋁α<λNα=f−1 ⁣(⋁α<λMα)=N. For each distinct Nα, take its least occurrence in the ordinal λ; these indices form a set S⊆λ of size at most κ. Since cf⁡(λ)>κ, S is bounded by some γ<λ. Then Nγ contains every Nα, so the displayed join gives Nγ=N. Equivalently, f factors through Mγ.

Depends on

Used by

Dependency tree · two levels

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Sources