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Trying to learn some basics of (topological) K-theory and came up with the following question:

From what I can gather, we can define (complex, topological) K-theory as [itex] K^n(X) = [X, B^n Gr^\infty(m)] [/itex] with m going to infinity (indeed for m large enough, the answer is independent of it) and where B is taking the classifying space (this means: for any topological space X, we define BX such that [itex] \Omega (B X) = X[/itex] where [itex]\Omega[/itex] takes the loop space).

Let me know if so far I have made a mistake. As an illustration this tells us [itex]K^0(X) = [X,Gr^\infty] [/itex] (where I have implicitly taken the limit [itex]m \to \infty[/itex]) which indeed classifies vector bundles on X up to (stable) isomorphism/equivalence.

So Bott periodicity tells us [itex]K^n(X) \cong K^{n+2}(X)[/itex]. In other words it tells us [itex] [X, B^n Gr^\infty] \cong [X, B^{n+2} Gr^\infty] [/itex]. My question is: have I made a mistake, ordoes Bott periodicity imply

[tex]\boxed{ B^n Gr^\infty \simeq B^{n+2} Gr^\infty} \;?[/tex]

To focus on a specific example, let's take n = -1. Then this would tell us thatthe classifying space of the infinite Grassmannian is homotopic to U(n)(for n large enough). Is this true?...

EDIT:It seems wikipedia agrees with the above bold statements. However, not completely. For example it implies that in fact [itex] B^2 Gr^\infty \simeq \mathbb Z \times Gr^\infty [/itex]. This seems weird, as I wouldn't expect [itex] [X, Gr^\infty ] \cong [X, \mathbb Z \times Gr^\infty ][/itex] ... Do we have to take reduced cohomology/mod out by something to make it all work, or does it work like this anyway and do I just fail to see it?

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# Does Bott periodicity imply homotopy equivalences?

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