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Consider the (reduced) homology functor $H_*$ from the category of spectra to the category of graded Abelian groups. I wanted to know whether there is a "section" of this functor, i.e., a functor $F$ from graded Abelian groups to spectra so that $H_*\circ F=Id$.

(I have very little experience in Algebraic Topology, so I am guessing the answer is "obviously yes" or "obviously no", but I don't know which, and cannot find any reference. One can define the functor $F$ on objects to be a wedge sum of Moore spaces, but I do not know if this can be made functorial.)

Clarification: We may make impose additional restrictions if it suits our interest. In my situation, I am only interested in (possibly shifted) finite suspension spectra (i.e., suspension spectra of finite CW complexes), and finitely generated graded Abelian groups.

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up vote 8 down vote accepted

I think that the answer to your question is no, but not for the reason you may suspect. Let us try to do the "stupid" case first: fix an abelian group $A$ and you want a spectrum $X$ such that $H_0(X) =A$ and $H_i(X)=0$ for $i\neq 0$. Such a spectrum is called Moore spectrum and it always exists. However it is not functorial in $A$, that is every map between abelian groups may be realized as a map between the corresponding Moore spectra but not uniquely so. However you can realize Moore spectra functorially if you consider only 2-divisible groups.

Finally, this case is in fact the only problem you have. Supposing you were able to realize Moore spectra functorially, you could realize graded abelian groups easily enough by taking wedges of suspensions.

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Is there a good reference on that there is no Moore space functor? – Achim Krause Apr 15 '14 at 18:03
I was trying to think of one for including it in the answer but I cannot think of one. There is something in Schwede's unfinished book on symmetric spectra, but not on the version on his webpage. The idea should be to try to compute the endomorphism of $\mathbb{S}/2$ and see that the map to the endomorphism of $\mathbb{Z}/2$ doesn't split. – Denis Nardin Apr 15 '14 at 18:09
The endomorphisms of $\mathbb{S}/2$ have an additive structure (i.e. it is a ring) and the identity has no additive order 2, because otherwise $\pi_2(\mathbb{S}/2)$ would be 2-torsion (which it isn't). So you cannot send the identity to the identity because the functor must send biproducts to biproducts (and so be additive) – Denis Nardin Apr 15 '14 at 18:25
Ah, I see: There is a map $\mathbb{S}/2 \rightarrow \mathbb{S}/2 \vee \mathbb{S}/2$, uniquely up to homotopy characterized by inducing the diagonal on homology, and similarly, there is a map $\mathbb{S}/2\vee\mathbb{S}/2\rightarrow \mathbb{S}/2$ inducing addition on homology, and their composition is zero in homology, but nonzero in homotopy. – Achim Krause Apr 15 '14 at 18:41
Yes exactly. Careful that the correct symbol for the wedge of two spectra is $\vee$ (backslash vee). I know, it's annoying that LaTeX (and all the rest of mathematics) disagree with us about how a wedge must be written... – Denis Nardin Apr 15 '14 at 18:43

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