Suppose that, given an abelian group $G$, there is a multiplication map $\mu:K(G,n)\times K(G,n) \to K(G,n)$ defined such that the induced map on the homotopy group $\mu_*:\pi_n(K(G,n) \times K(G,n)) \to \pi_n(K(G,n))$ takes $(g_1,g_2)$ to $g_1 + g_2$, where $+$ is the operation on $G$.

Does it follow that this multiplication is homotopy-commutative; that is, if $t:K(G,n) \times K(G,n) \to K(G,n) \times K(G,n)$ switches coordinates, does it follow that $\mu t$ is homotopic to $\mu$? Since $G$ is commutative, it seems that $\mu$ should be, but I'm having a hard time coming up with the actual homotopy. I know that is NOT true in general that if two maps induce the same homomorphisms on homotopy groups, then they are homotopic.

One could also ask if the fact that the operation on $G$ is associative implies that $\mu$ is homotopy-associative.


Yes, it follows that $\mu t$ is homotopic to $t$.

In general we have the following result:

Lemma: Suppose that $G$ and $H$ are abelian groups and $f,g: K(G,n) \to K(H,n)$ are maps. Then $f_* = g_*: \pi_n(K(G,n))=G \to \pi_n(K(H,n))$ if and only if $f$ is homotopic to $g$.

This follows easily from the following:

Lemma: Let $X$ be an $(n-1)$-connected CW-complex with $\pi_1(X)$ abelian then the map $\eta: H^n(X;G) \to \text{Hom}(\pi_n(X),G)$ given by $\eta[f] = f_*: \pi_n(X) \to \pi_n(K(G,n)) = G$ is an isomorphism.

Here we use that $H^n(X;G) \cong [X, K(G,n)]$ where $[ , ]$ denotes homotopy classes of maps.

A reference for this last lemma is Arkowitz - Introduction to Homotopy Theory. It is Lemma 2.5.13

  • $\begingroup$ To apply your lemma are you asserting that $K(G,n)\times K(G,n)$ is a $K(H,n)$? $\endgroup$ – Kevin Carlson Feb 17 '14 at 21:33
  • $\begingroup$ @Kevin Carlson: In general $\pi_n(X \times Y) \cong \pi_n(X) \times \pi_n(Y)$. In particular, for Eilenberg-Mac Lane spaces we get that $K(G,n) \times K(H,n)$ has the homotopy type of a $K(G \times H,n)$. $\endgroup$ – R. Frankhuizen Feb 17 '14 at 21:41
  • $\begingroup$ Of course. Thanks. $\endgroup$ – Kevin Carlson Feb 17 '14 at 21:42

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