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Equality in K⁰ is stable isomorphism over compact bases
Statement
Assume AC. For a compact Hausdorff space ,
if and only if there is a finite-rank bundle such that
Equivalently, there is an such that
In particular, if and only if for some .
Facts & Assumptions
Given: AC, a compact Hausdorff space , and finite-rank complex bundles over .
Equality in the Grothendieck group is the common-summand relation (Complex topological K⁰ by Grothendieck completion).
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).
AC is used only through [F2] to obtain the complement.
Proof
By [F1], the first displayed equality holds exactly when some bundle satisfies the first stable-isomorphism display. This proves both directions of the first equivalence, including when no added summand is needed.
Apply [F2] to . There are a bundle and with . Adding 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 .
Set in the proved equivalence. Then exactly when for some , including .
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
- Hatcher, Vector Bundles & K-Theory, Proposition 2.1 and following construction (standard reference, not scraped)
- May, A Concise Course in Algebraic Topology, Chapter 24 §1 (standard reference, not scraped)