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Nov
5
comment Why is stable equivalence necessary in topological K-theory?
I think there's something basic I'm missing. Stable equivalence is an equivalence relation on isomorphism classes of bundles. Say $E$ and $F$ are isomorphism classes. Then it appears to me that $E = F$ iff $E + G = F + G$ in K-theory. I suppose my question is, why is stable equivalence not "lies in the same isomorphism class"? In particular, what is an example of two non-isomorphic bundles such that when a trivial bundle is added to them, they become isomorphic?
Nov
4
comment Why is stable equivalence necessary in topological K-theory?
Hi Ryan, thanks for your clarification, but it's still murky for me. I don't see why $E \cong F \not \Leftrightarrow (E \oplus G) \cong (F \oplus G)$
Nov
3
comment Why is stable equivalence necessary in topological K-theory?
Are you saying that stable equivalence endows isomorphism classes with a formal inverse? That I don't see.
Nov
3
comment Why is stable equivalence necessary in topological K-theory?
Clarified, I hope.
Nov
3
awarded  Editor
Nov
3
revised Why is stable equivalence necessary in topological K-theory?
clarified question
Nov
3
asked Why is stable equivalence necessary in topological K-theory?
Nov
3
awarded  Scholar
Nov
3
awarded  Student
Nov
3
accepted Why are holomorphic functions defined in reference to an open set?
Nov
3
asked Why are holomorphic functions defined in reference to an open set?