3
$\begingroup$

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?

$\endgroup$
  • 6
    $\begingroup$ Because $\tilde \alpha$ is the wedge sum of maps $\Sigma^k H(\pi_k M) \to M$ inducing isomorphism on $\pi_k$. $\endgroup$ – John Rognes Jan 6 at 10:47
  • 6
    $\begingroup$ Maybe the missing observation here is that $H\mathbb{Z}\wedge SG\cong HG$ for all abelian groups $G$? $\endgroup$ – Denis Nardin Jan 6 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 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 $π_*(H\mathbb{Z}∧X)\cong H_*X$? $\endgroup$ – Denis Nardin Jan 11 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 $\pi_0$...) $\endgroup$ – Dylan Wilson Jan 11 at 19:50
4
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ Is this too `elementary' for MathOverflow? $\endgroup$ – John Rognes Jan 11 at 22:19

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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