# Splitting of $H\mathbb{Z}$-module spectra

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?

• Because $\tilde \alpha$ is the wedge sum of maps $\Sigma^k H(\pi_k M) \to M$ inducing isomorphism on $\pi_k$. – John Rognes Jan 6 '19 at 10:47
• Maybe the missing observation here is that $H\mathbb{Z}\wedge SG\cong HG$ for all abelian groups $G$? – Denis Nardin Jan 6 '19 at 10:48
• @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. – user09127 Jan 11 '19 at 16:30
• @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 $π_*(H\mathbb{Z}∧X)\cong H_*X$? – Denis Nardin Jan 11 '19 at 16:56
• 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 $\pi_0$...) – Dylan Wilson Jan 11 '19 at 19:50

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 $$\bigvee_k \Sigma^k H(\pi_k M) \to M$$.