Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Equality in K⁰ is stable isomorphism over compact bases

Statement

Assume AC. For a compact Hausdorff space X,

[E][F]=[E][F] in K0(X)

if and only if there is a finite-rank bundle H such that

EFHEFH.

Equivalently, there is an N0 such that

EFεNEFεN.

In particular, [E]=[F] if and only if EεNFεN for some N.

Facts & Assumptions

Given: AC, a compact Hausdorff space X, and finite-rank complex bundles E,F,E,F over X.

[F1]

Equality in the Grothendieck group is the common-summand relation (Complex topological K⁰ by Grothendieck completion).

[F2]

Under AC, every finite-rank bundle over a compact Hausdorff base has a finite-rank complement in a trivial bundle (Finite-rank complement theorem over compact Hausdorff bases).

[A1]

AC is used only through [F2] to obtain the complement.

Proof

technique · direct
1.1

By [F1], the first displayed equality holds exactly when some bundle H satisfies the first stable-isomorphism display. This proves both directions of the first equivalence, including H=0X when no added summand is needed.

F1
2.1

Apply [F2] to H. There are a bundle H and N0 with HHεN. Adding H to both sides of the isomorphism in step 1.1 gives the trivial-stabilization display. Conversely, that display is the relation in step 1.1 with H=εN.

F2A1step 1.1algebra
3.1

Set F=F=0X in the proved equivalence. Then [E]=[F] exactly when EεNFεN for some N, including N=0.

step 2.1

Depends on

Used by

Dependency tree · two levels

8 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