3
$\begingroup$

In the web page http://www.encyclopediaofmath.org/index.php/Moore_space it can be found the following statement:

If $K(\mathbb Z,n)$ is the Eilenberg–MacLane space of the group of integers $\mathbb Z$ and $M_k(G)$ is the Moore space with $\tilde{H}_k(M_k(G))=G$, then

$$lim_{N\rightarrow\infty}[\Sigma^{N+k}X,K(\mathbb Z, N+n)\bigwedge M_k(G)]\cong H^n(X,G)$$,

that is, $\{K(\mathbb Z,n)\wedge M_k(G)\}$ is the spectrum of the cohomology theory $H^*(\ ,G)$.

The reference of the webpage is the articule of Moore ´´On homotopy groups of spaces with single non-vanishing homotopy group´´ but I don´t find anything like that on the article.

It would be very helpful if you could give a reference where I can see the explanation or you can explain me that.

$\endgroup$

2 Answers 2

16
$\begingroup$

Let $H\mathbb{Z}$ denote the spectrum for cohomology with coefficients in $\mathbb{Z}$, so your spectrum is $H\mathbb{Z}\wedge M_k(G)$. Let $$0\to\bigoplus_I\mathbb{Z}\stackrel{f}{\to}\bigoplus_J\mathbb{Z}\to G\to 0$$ be a presentation of $G$. Then we can explicitly construct a Moore space $M_k(G)$ as the cofiber of a map $\bigvee_I S^k\to \bigvee_J S^k$ whose induced map on homology is $f$. In spectra, we then have a cofiber sequence $$\bigvee_I \Sigma^k H\mathbb{Z}\to\bigvee_J\Sigma^k H\mathbb{Z}\to H\mathbb{Z}\wedge M_k(G).$$

For any pointed finite CW-complex $X$, we then have a long exact sequence in cohomology of the form $$\dots\to\bigoplus_IH^{n+k}(X,\mathbb{Z})\to \bigoplus_J H^{n+k}(X,\mathbb{Z})\to H\mathbb{Z}\wedge M_k(G)^n(X)\to\dots$$

Now consider $X=S^0$. The long exact sequence above computes that $H\mathbb{Z}\wedge M_k(G)^{-k}(S^0)=G$ and $H\mathbb{Z}\wedge M_k(G)^n(S^0)=0$ for $n\neq -k$. Thus up to a degree shift, the cohomology theory $H\mathbb{Z}\wedge M_k(G)$ satisfies the Eilenberg-Steenrod dimension axiom, so it must coincide with $H^{*+k}(X,G)$.

$\endgroup$
2
  • $\begingroup$ Thank you, I would like to ask you something if that is ok: Are Moore spaces the spectrum of a homology theory? I was wondering that because their duals Eilenberg-McLane are. $\endgroup$ May 27, 2015 at 5:09
  • $\begingroup$ @PaulaCartagenaAtará I wouldn't say that Moore spaces are the dual of EM spaces, although I see what you are getting at. Every spectrum gives a homology theory, the sphere spectrum gives stable homotopy theory for example. So what do you mean by "are Moore spaces the spectrum of a homology theory?" What spectrum are you wanting to build from them? $\endgroup$ May 27, 2015 at 8:50
10
$\begingroup$

In case you are curious, here is a more concise but higher level argument.

The homotopy groups of the spectrum $H\mathbb Z \wedge M_k(G)$ agree with the homology groups of $M_k(G)$, which is just a $G$ in degree $k$. So the spectrum $H\mathbb Z \wedge M_k(G)$ is equivalent a shift of $HG$.

These are all standard facts in stable homotopy theory; Eric's answer above goes into more detail as to why they are true in this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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