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Let $S$ denote the category of projective (left) $R$-modules with isomorphisms for arrows. We have that

$BS^{-1}S \sim B \text{GL}(R)^+ \times K_0(R)$

In proving this, in Srinivas' algebraic K-theory text, the following is casually written:

$H_p(BS^{-1}S) \cong H_P(BS^{-1}S^0) \times K_0(R)$

where $BS^{-1}S^0$ is the component containing $(0,0)$. How is this?

(It is easy to show that $BS^{-1}S \cong BS^{-1}S^0 \times K_0(R)$, and from there Kunneth theorem gives $H_p(BS^{-1}S) = H_p(BS^{-1}S^0) \otimes_{\mathbb{Z}} K_0(R) \oplus \text{Tor}(H_{p-1}(BS^{-1}S), K_0(R))$, but that doesn't seem to get us anywhere...)

Addendum: Silly (overtired) oversight of mine. $H_0(K_0(R)) = \mathbb{Z}[K_0(R)]$ and the Tor part of the Kunneth exact sequence vanishes. So we are left with $H_p(B S^{-1}S) = H_p(BS^{-1}S^0) \otimes_{\mathbb{Z}} \mathbb{Z}[K_0(R)] = H_p(BS^{-1}S^0) \oplus K_0(R)$. As Dan pointed out below though, we can better make use of the relation between the homology of a space and that of its components.

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up vote 2 down vote accepted

If you want to apply Kunneth, then you should take into account that $H_0 X$ is the free abelian group on $\pi_0 X$, so you have to replace your use of $K_0R$ by the free abelian group on $K_0R$. Then the Tor vanishes, and we see that Srinivas' formula is incorrect.

Simpler than applying Kunneth would be to think about how the homology groups of a space are related to the homology groups of its connected components: it's the direct sum. So if a space $X$ has all its components homotopy equivalent to each other, as in this case, then $H_p X$ is $\coprod_{\pi_0 X} H_p X^0 $, where $X^0$ is one of the components.

One further issue to think about is naturality of the isomorphism, since the homotopy equivalence between two components of $X$ depends on a choice.

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@Dan: Thank you. – Joshua Seaton Mar 30 '13 at 12:20

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