While dealing with $BO(n)$, $BSO(n)$ and $BSpin(n)$ with the universal coefficient theorem and Künneth formula, I came to have the following question:


The universal coefficient says $H^n(X;M)\cong \hom(H\_{n}(X;\mathbb{Z}),M)\oplus {\rm Ext}^{1} (H\_{n-1}(X;\mathbb{Z},M))$ for a $\mathbb{Z}$-module $M$.


When $X=BSpin(n)$, we know that $H^4(BSpin(n);\mathbb{Z})\cong \mathbb{Z}$ and it seems likely that once we know what $H\_p(BSpin(n);\mathbb{Z})$ would be for $p=3,4$ we might be able to retrieve this isomorphism with the aid of universal coefficient theorem.


So what would be $H\_p(BSpin(n);\mathbb{Z})$ at least for $p=0,1,2,3,4$? 


(I have to say that the question is not about how to prove the isomorphism $H^4(BSpin(4);\mathbb{Z})\cong \mathbb{Z}$.)


Thanks!



Homology of classifying space of spin group BSpin(n)
Consider the classifying space $EG$ of a given group $G$. In your case, $G={Spin}(n)$. We have a fibration $G\to EG\to BG$ where $EG$ is contractible and $G$ acts freely on it. Therefore, the long exact sequence in homotopy groups tells you that $\pi\_j(BG)\cong \pi\_{j-1}(G)$. But $G={Spin}(n)$ is a Lie group which is simply connected. It is also classically known that $\pi\_2(G)=0$ for finite dimensional Lie groups $G$. And, we also know that $\pi\_3({Spin}(n))=\mathbb{Z}$. This implies that $\pi\_j(B {Spin}(n))=0$ for $j=0,1,2,3$. Moreover, $\pi\_4(B {Spin}(n))\cong H\_4(B{Spin}(n);\mathbb{Z})$ by Hurewicz theorem. This is also isomorphic to $\pi\_3({Spin}(n))$. The universal coefficient theorem now tells you that $H^4(B{Spin}(n);\mathbb{Z})=\mathbb{Z}$.


