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...)

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.