For Abelian groups $\pi$ and $\rho$, what is the easiest way to see that $$[K(\pi, n), K(\rho, n)] \cong \text{Hom}(\pi, \rho)?$$My idea is the use the natural isomorphism$$[X, K(\rho, n)] \cong \tilde{H}^n(X; \rho)$$and universal coefficients, but I am stuck to that degree. Could anyone help?
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1This question is the same and has an answer, but Qiacho's answer here is more detailed... – Najib Idrissi Sep 04 '15 at 12:49
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You need one additional tool, which is the Hurewicz theorem. Let me write $B^n A$ for $K(A, n)$. Then the Hurewicz theorem tells you that
$$A \cong \pi_n B^n A \cong H_n(B^n A, \mathbb{Z})$$
and then universal coefficients tells you that
$$H^n(B^n A, C) \cong \text{Hom}(H_n(B^n A, \mathbb{Z}), C) \cong \text{Hom}(A, C)$$
as desired. (Note that all we used was that $B^n A$ has first $n$ homotopy groups $0, 0, \dots A$; we don't use anything about its homotopy groups past degree $n$.)
Qiaochu Yuan
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1@Michael: as I said, it means $K(A, n)$. This notation emphasizes that $K(A, n)$ is the $n$-fold delooping of $A$ (see, for example, http://ncatlab.org/nlab/show/delooping and https://en.wikipedia.org/wiki/Classifying_space). – Qiaochu Yuan Sep 03 '15 at 21:23