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Assume that $\{E^n\}_{n \in \mathbb{Z}}$ is a cohomology theory, that is, a sequence of functors from the (homotopic) category of based CW-complexes to the category of (graded) abelian groups. Then Brown's representability theorem asserts that for any such sequence, there is always a nice sequence of spaces $\{ Z^n \}_{n \in \mathbb{Z}}$, which represents the functors $E^n$. In other words we have a natural equivalence, $E^n \cong [-,Z^n]$.

My question has to do with the "dual" version of the above narrative. Assume that someone starts with a a spectrum $Z=\{Z^n\}_{n \in \mathbb{Z}}$. Then Switzer in his book "Algebraic Topology - Homotopy and Homology", makes the following constrcution: for a given spectrum $Z$, defines a cohomology theory $E$ associated to $Z$, with $E^n(X)=[\Sigma^{\infty}X, \Sigma^nZ]$, for each pointed CW-complex $X$. My question is why by setting just $E^n(X)=[X,Z^n]$ does not suffice? In his book M.R. Adhikari - Basic Algebraic Topology, seems that defines (on pg.$482$) for a given $\Omega$-spectrum $Z$, the associated (reduced) cohomology theory to be the sequence of functors $\{ [-,Z^n] \}_{n \in \mathbb{Z}}$. Why Switzer does not just take this construction and makes it more complicated?

All I can think, is that Switzer takes any (not necessarily $\Omega$) spectrum and creates a generalized cohomology theory with his construction. Though for any given spectrum, it can be turned into an $\Omega$-spectrum hence I don't get the point.

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  • $\begingroup$ This seems to be a notational issue. In the first case, he's taking the internal Homs inside the category of spectra, whereas the second case is done inside the category of topological spaces. $\endgroup$ – user40276 Jul 6 '18 at 10:13
  • $\begingroup$ Thanks for the comment; so you say that these two thinks are essentially the same with just a different dictionary in hand? $\endgroup$ – user321268 Jul 6 '18 at 10:14
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As you say Switzer defines a cohomology theory $E^\ast$ for any spectrum $E =(E_n,\epsilon_n)$. If $E$ is an $\Omega$-spectrum, then in fact $E^n(X) = [X,E_n]$ which is (in a sense) simpler than the general definition of $E^n(X)$ as the set of homotopy classes of maps of spectra from the suspension spectrum of $X$ to the spectrum $\Sigma^nE$. And yes, you may restrict to $\Omega$-spectra, but why should you reject the more universal approach?

Consider for example the sphere spectrum $S^0$. This is far from being an $\Omega$-spectrum, but the cohomology theory associated to $S^0$ is known as stable cohomotopy. This cohomology theory is well-established in algebraic topology.

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  • $\begingroup$ Thank you for your answer. I understood the same; Switzer defines the cohomology associated to any spectra since homotopy classes of morphsisms between spectra is an abelian group. On the other hand an $\Omega$-spectrum say $Z$ consists of $H$-groups, hence the homotopy classes of maps $[X,Z^n]$ are abelian groups, and in fact as user40276 commented above, this coincides with Switzer's construction just in a different category. $\endgroup$ – user321268 Jul 6 '18 at 12:45

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