Consider the following two ways of getting the zeroth space in the $K$-theory spectrum $BU \times \mathbb{Z}$:

1) Take the groupoid of finite dimensional complex inner product spaces with isometries as morphisms and apply Quillen $S^{-1}S$-construction to it. This amounts to forming a new category, whose objects are pairs $(V_+, V_-)$. A morphism $(V_+, V_-) \to (W_+,W_-)$ is an equivalence class of triples $(A, f_+, f_-)$, where $A$ is another finite dimensional inner product space and $f_{\pm} \colon V_{\pm} \oplus A \to W_{\pm}$ is an isometric isomorphism (i.e. a morphism in the former category). The equivalence relation identifies isomorphic objects $A$ and $B$ and the corresponding maps $f_{\pm}^A$ and $f_{\pm}^B$. All this can be found in a paper by Daniel Grayson and is also sketched in section 7 of this paper. Take the nerve of this new category to get the first model.

2) According to Segal we could also take the category of finite length chain complexes of finite dimensional vector spaces together with chain homotopy equivalences as morphisms and take the nerve of that to get $BU \times \mathbb{Z}$. (This is part of the $\Gamma$-space construction of $BU_{\otimes}$.)

How are the two models related?

I tried to construct a functor from the category in 2) to the one in 1), that has the chance of being an equivalence, but failed so far. Taking the homology of the chain complexes in 2) yields a functor that ends up in 1), but only "sees" the morphisms where $A = 0$, since chain homotopy equivalences always provide isomorphisms when taking homology. Nevertheless: Is this the right thing to consider?

lotsof ways of getting the zeroth space of BU x Z and these are two of them. Your first sentence implies (for me, at any rate) that these are theonlytwo. – Andrew Stacey Dec 14 '11 at 8:07