Does Bott periodicity imply homotopy equivalences?

[nonequilibrium]

Hello!

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 K^n(X) = [X, B^n Gr^\infty(m)] 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 \Omega (B X) = X where \Omega takes the loop space).

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

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

\boxed{ B^n Gr^\infty \simeq B^{n+2} Gr^\infty} \;?

To focus on a specific example, let's take n = -1. Then this would tell us that the 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 B^2 Gr^\infty \simeq \mathbb Z \times Gr^\infty. This seems weird, as I wouldn't expect [X, Gr^\infty ] \cong [X, \mathbb Z \times Gr^\infty ] ... 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?

 

[nonequilibrium]

Aha, I have realized the answer. The key point is that [X,Y] \cong [X,Y \times \mathbb Z] since in these contexts we are looking at *base point preserving* maps, which are insensitive to the number of disconnected components.