It is classical result of Adams that every $H\mathbb{Z}$-module spectra splits as a wedge of Eilenberg-MacLane spectra. Let me briefly recall what he writes about the proof.

Let $M$ be an $H\mathbb{Z}$-module spectrum. Adams constructs a map $$\alpha:\bigvee_k\Sigma^k S(\pi_kM)\rightarrow M$$ by taking the wedge of the maps $\Sigma^kS(\pi_kM)\rightarrow M$ inducing an isomorphism on $\pi_k$, where $SA$ denote the Moore spectrum on the abelian group $A$.

The map $\alpha$ induces a map of $H\mathbb{Z}$ by taking $\tilde{\alpha} = \mu \circ (1\wedge \alpha)$.

Now, $\tilde{\alpha}$ is without doubt a map of $H\mathbb{Z}$-modules, but why is it a weak equivalence?

  • 7
    $\begingroup$ Because $\tilde \alpha$ is the wedge sum of maps $\Sigma^k H(\pi_k M) \to M$ inducing isomorphism on $\pi_k$. $\endgroup$ Jan 6, 2019 at 10:47
  • 6
    $\begingroup$ Maybe the missing observation here is that $H\mathbb{Z}\wedge SG\cong HG$ for all abelian groups $G$? $\endgroup$ Jan 6, 2019 at 10:48
  • $\begingroup$ @JohnRognes That's what I don't quite get. I assume that the way to prove this is to show that the unit map $S\rightarrow H\mathbb{Z}$ induces an isomorphism on $\pi_0$ (which has to be true), but how would your prove that? I guess a description of the action of the product of $H\mathbb{Z}$ on homotopy groups would do the trick. But, once again, that's not a precise argument. $\endgroup$
    – user09127
    Jan 11, 2019 at 16:30
  • $\begingroup$ @user09127 What are you starting from? One quick way to see that the map $\mathbb{S}→H\mathbb{Z}$ is an iso on π_0 is that it is a map of rings and both rings are $\mathbb{Z}$. Or maybe what you're missing is that $π_*(H\mathbb{Z}∧X)\cong H_*X$? $\endgroup$ Jan 11, 2019 at 16:56
  • 2
    $\begingroup$ What is your definition of $H\mathbb{Z}$ and the unit map? (It seems hard to have a definition of those two things without also having a proof that the unit map is an isomorphism on $\pi_0$...) $\endgroup$ Jan 11, 2019 at 19:50

1 Answer 1


Perhaps it helps to first think about how you can construct a map $\alpha_k : \Sigma^k S(\pi_k M) \to M$ inducing an isomorphism on $\pi_k$. Choose a free resolution $$ 0 \to \bigoplus_{j \in J} \mathbb{Z} \to \bigoplus_{i \in I} \mathbb{Z} \to \pi_k(M) \to 0 $$ and realize it in $H_k$ by a homotopy cofiber sequence $$ \bigvee_{j \in J} S^k \to \bigvee_{i \in I} S^k \to \Sigma^k S(\pi_k M) . $$ Mapping to $M$ you obtain an exact sequence $$ \dots \to [\Sigma^k S(\pi_k M), M] \to Hom(\bigoplus_{i \in I} \mathbb{Z}, \pi_k(M)) \to Hom(\bigoplus_{j \in J} \mathbb{Z}, \pi_k(M)) \to \dots $$ In particular, $$ [\Sigma^k S(\pi_k M), M] \to Hom(\pi_k(M), \pi_k(M)) $$ is surjective. Choose $\alpha_k$ so that it maps to the identity. Then $\pi_k(\alpha_k) : \pi_k(\Sigma^k S(\pi_k M)) \to \pi_k(M)$ is an isomorphism. (You should check this last claim.)

Using the $H\mathbb{Z}$-module structure on $M$, you can factor $\alpha_k$ as the Hurewicz map $$ h : \Sigma^k S(\pi_k M) \to H\mathbb{Z} \wedge \Sigma^k S(\pi_k M) \simeq \Sigma^k H(\pi_k M) $$ followed by $$ \tilde \alpha_k : \Sigma^k H(\pi_k M) \to M . $$ By the Hurewicz theorem, $\pi_k(h)$ is an isomorphism. (One way to see this is to show that $H\mathbb{Z}$ can be built as a CW spectrum from $S$ by only adding $n$-cells for $n\ge2$, which does not change $\pi_0$.) Thus $\pi_k(\tilde \alpha_k)$ is an isomorphism. Taking the wedge sum of the maps $\tilde \alpha_k$ for all integers $k$ gives the weak equivalence $\bigvee_k \Sigma^k H(\pi_k M) \to M$.

  • 4
    $\begingroup$ Is this too `elementary' for MathOverflow? $\endgroup$ Jan 11, 2019 at 22:19
  • 1
    $\begingroup$ No. It is not too elementary. $\endgroup$ Jan 31, 2019 at 0:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .