Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was re-reading an algebraic topology book the other day, and I came across the following problem:

Suppose that $\pi$ and $\rho$ are abelian groups and $n\geq 1$. Determine $[K(\pi,n),K(\rho,n)]$, the set of (based) homotopy classes of maps between the corresponding Eilenberg-MacLane spaces.

I believe that the following is a solution: We have two functors $K(-,n)$ from (discrete) abelian groups to (the homotopy category of nice) topological spaces, and $\pi_n = [S^n,-]$ going the other direction. When we suitably restrict these functors, they appear to be inverses. Therefore $[K(\pi,n),K(\rho,n)]\cong \hom_{Ab}(\pi,\rho)$

I have two questions. Is the solution correct, or are there errors in the logic? If it does work, is there a way to make it completely transparent that the functors are inverse to each other? And if it is correct, if we suitably topologize $\pi_n(-)$, does this extend to non-discrete topological groups?

Second, is there a different way to approach the problem which better illuminates what is going on or illustrates an important point about $K(\pi,n)$?

share|cite|improve this question
@Grigory M: I mean that if we restrict to the (full sub-)category of CW complexes with homotopy groups concentrated in degree $n$ and discrete abelian groups, the two functors seem to induce inverse equivalences, if I am thinking about things correctly. – Aaron Jun 4 '11 at 7:05
It's certainly true, but this statement is exactly equivalent to the problem you quote. You can't prove it... well, without doing something :-) – Grigory M Jun 4 '11 at 7:08
I mean, it has an obvious part: $\pi_n\circ K(-;n)\cong Id$; but the part $K(-;n)\circ\pi_n\cong Id$ relies on the problem. – Grigory M Jun 4 '11 at 7:10

1 Answer 1

up vote 15 down vote accepted

Recall that $[X,K(G,n)]=H^n(X;G)$. Hence $[K(\pi,n),K(\rho,n)]=H^n(K(\pi,n);\rho)$ — which (by Hurewicz theorem + universal coefficients) is exactly $\hom(\pi,\rho)$.

share|cite|improve this answer
Thanks. I had come up with the first part of this, and then thought, "I wish I could use Hurewicz here, but coefficients and cohomology. Oh well," and promptly abandoned the line of reasoning. However, I know that isn't the only way to do the problem, because in the book it appears before they introduce cohomology at all. Is there a good way that uses less technology? – Aaron Jun 4 '11 at 16:30
I don't know. Some obstruction theory proof, perhaps?.. Anyway, let me give a warning: if you find an easy proof, check that it doesn't "prove" that, say, $[\mathbb T^2,S^2]\to\hom(\pi_1(\mathbb T^2),\pi_1(S^2))=0$ is injective. – Grigory M Jun 4 '11 at 16:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.