I'm studying the construction of Adams spectral sequence in Hatcher's "Algebraic Topology" chapter five; I'm in trouble with an "algebraic statement".

Let $X$ be a connective CW-spectrum of finite type, $\mathcal A$ be the Steenrod algebra and $H^*(X)$ the graded cohomology of $X$, regarded as an $\mathcal A$-module (the cohomology will be in $\mathbb Z_p$ with $p$ a fixed prime). Choose $\alpha_i$ generators of $H^*(X)$ as $\mathcal A$-module; each $\alpha_i$ can be represented by a map $X \rightarrow K(\mathbb Z_p, m)$, so these $\alpha_i$ determine a map $X \rightarrow K_0$, where $K_0$ is a wedge of Eilenberg-Maclane spectra, and $K_0$ has finite type (since $X$ has finite type too, so each $H^k(X)$ is finitely generated). The statement I'm in trouble with is the following: $H^*(K_0)$ is a free $\mathcal A$-module.

It has surely to be related with the fact that the cohomology of a $K(\mathbb Z_p, n)$ is a free $\mathcal A$-module in dimension $<2n$, but I feel like I'm getting losing myself among the definitions and I can't prove this assertion. Hatcher claims that this is related to the natural isomorphism $[X, \bigvee K(G, n_i)] \simeq \Pi_i [X, K(G, n_i)]$, but I can't figure this out. Thanks in advance.

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    $\begingroup$ I think you might be confusing spectra with their component spaces. The EM spectra $KG$ consists of all the spaces $KG_n=K(G,n)$ for $n\geq 0$. Look at the discussion on page 592 and try to calculate the cohomology of the EM spectrum $KG$ using the inverse limit formula $H^\ast(KG)=\lim_n H^{\ast+n}(KG_n)=\lim_nH^{\ast+n}(K(G,\ast+n))$. $\endgroup$ – Tyrone Mar 10 at 12:10
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    $\begingroup$ Now apply, say, $Sq^r$ and note that you get a consistent sequence in $\prod_n H^{\ast +r+n}(K(G,n))$. Since $Sq^r$ acts freely on $H^{*+N}K(G,N)$ for large $N$, the sequence is non-zero. $\endgroup$ – Tyrone Mar 11 at 11:37
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    $\begingroup$ To make things a bit clearer, perhaps, stick to $G=\mathbb{Z}_2$ and consider $H^0K\mathbb{Z}_2$. Now choose any element $\theta$ of $\mathcal{A}$ and apply it to this group to get a non-zero element in $H^{|\theta|}K\mathbb{Z}_2$. On the other hand, if $x\in H^rK\mathbb{Z}_2$, then it corresponds to a consistence sequence which is eventually determined by a unique operation $\theta$ applied to the fundamental class $\iota_n$ for large $n$. In this way you can see easily that the action is free. $\endgroup$ – Tyrone Mar 11 at 11:40
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    $\begingroup$ Ok, the key observations I missed were that applying $Sq^r$ the sequence is still consistent, but (I suppose) it follows from the naturality of the cohomology operations (the fact that they commute with the structure maps of the spectrum), and the fact that eventually the sequence is determined by a unique operation applied to the fundamental class. Thank you very much @Tyrone. $\endgroup$ – Giuseppe Bargagnati Mar 11 at 11:43
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    $\begingroup$ The point is that a spectrum $X$ contains the data of the stabilisation maps $\Sigma X_n\rightarrow X_{n+1}$, and to determine homotopy, homology or cohomology groups you need to consider the limits over these maps. Thus, although the space $K(G,n)$ is only free over $\mathcal{A}$ in degrees $<2n$, the spectrum $KG=(K(G,n))_{n\geq 0}$ is free over $\mathcal{A}$ in all degrees. $\endgroup$ – Tyrone Mar 11 at 11:44

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