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 $X$ ,

$[E] - [F] = [E^{'}] - [F^{'}] in K^{0} (X)$

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

$E \oplus F^{'} \oplus H ≅ E^{'} \oplus F \oplus H .$

Equivalently, there is an $N \geq 0$ such that

$E \oplus F^{'} \oplus ε^{N} ≅ E^{'} \oplus F \oplus ε^{N} .$

In particular, $[E] = [F]$ if and only if $E \oplus ε^{N} ≅ F \oplus ε^{N}$ 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

1.1F1

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 $H = 0_{X}$ when no added summand is needed.

2.1F2A1step 1.1algebra

Apply [F2] to $H$ . There are a bundle $H^{'}$ and $N \geq 0$ with $H \oplus H^{'} ≅ ε^{N}$ . 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 $H = ε^{N}$ .

3.1step 2.1∎

Set $F = F^{'} = 0_{X}$ in the proved equivalence. Then $[E] = [F]$ exactly when $E \oplus ε^{N} ≅ F \oplus ε^{N}$ for some $N$ , including $N = 0$ .

Equality in K⁰ is stable isomorphism over compact bases

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources