Why is $H_5(K(\Bbb Z_n,4))$ finite?

Question

I want to see that the cohomology $H^i(K(\Bbb Z_n,4); \Bbb Z)$ starts with $\Bbb Z_n$ in degree 5. How do we know that $\operatorname{hom}(H_5(K(\Bbb Z_n,4);\Bbb Z), \Bbb Z)$ is zero? I.e. why is $H_5(K(\Bbb Z_n,4))$ finite?

I am probably missing something obvious.

Answer 1

$K(Z_n,4)$ is the forth Eilenberg McLane space, this implies that $\pi_4(K(Z_n,4))=Z_n$ and $\pi_n(K(Z_n,4))=1$ for $n\neq 4$. Now use the Hurewicz theorem, it implies that $\pi_5(K(Z_n,4))\rightarrow H_5(K(Z_n,4),Z)$ is surjective, since $\pi_5(K(Z_n,4))=1$, you deduce the result.

https://en.wikipedia.org/wiki/Hurewicz_theorem

Answer 2

If $G$ is a finite group, then the homology of $K(G,n)$ in any degree is finite. Prove this by induction; there is the pathspace fibration $K(G,n-1) \to * \to K(G,n)$. For $K(G,1)$, note that for any finite covering map $X \to Y$, the induced map in rational homology is injective, and the universal cover of $K(G,1)$ is contractible.

Now use the Serre spectral sequence. Inductively, $K(G,n-1)$ has trivial rational homology. Because the total space does as well, and the base is simply connected, $K(G,n)$ must have trivial rational homology. Because $K(G,n)$ has a model given by a CW complex with finitely many cells in each dimension, its integral homology is finitely generated, so the universal coefficient theorem implies that its integral homology is finite.

If you don't mind localization, a perhaps cleaner way of saying the same thing is as follows: the rationalization $K(G,n-1)_{\Bbb Q}$ is contractible, and hence the long exact sequence in homotopy groups shows that $K(G,n)_{\Bbb Q}$ is contractible. Unfortunately rationalization doesn't make sense for arbitrary fundamental groups, so to bootstrap this argument you need to start by knowing that $K(G,2)$ is rationally contractible, and the best way I know of doing this is given above. (This is similar to how a Hurewicz calculation of $\pi_n S^n$ needs to start by knowing $\pi_2 S^2$.)