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...)




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**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.



homology of $B S^{-1} S$ computation in the proof that $+ = Q$
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.


