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I was skimming though Goerss' Topological Modular Forms (after Hopkins, Miller and Lurie) to try to find some motivation for wanting a smash product of spectra and on page 1005-08 they say that if you start from a cohomology theory $h^*$ (represented by a spectrum $K_*$, which means $h^n(X)=[X,K_n]$ for any (pointed) space $X$) together with a product $h^p \otimes h^q \to h^{p+q}$ then `by taking universal examples' you get a map $K_p \wedge K_q \to K_{p+q}$.

I don't really see how the argument follows. If I'm given a map $h^p \times h^q \to h^{p+q}$ I can easily construct $K_p \times K_q \to K_{p+q}$, using the canonical projections (in other words $K_p \times K_q$ is the product in Top).

But, as there are no obvious (at least to me) morphisms $K_p \wedge K_q \to K_p, K_q$ I wouldn't know how to obtain the claimed morphism.

Might it be true that $[-,E\wedge K] = [-,E] \otimes [-,K]$ for (infinite) loopspaces $E$ and $K$?

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I believe they are referring to the specific process by which we get spectra for coh. theories, i.e. Brown Representability. In the proof of Brown representability, for a functor $h$ you find an object $X$ such that $T:h(-)\to[-,X]$ is an iso, and you use a special element in $h(X)$, I believe called a universal element, such that $T(Y)(f)=h(f)(u)$, and so perhaps by the process you get the necessary map. –  Jon Beardsley Nov 17 '11 at 21:55
    
Where $f:Y\to X$ and $T$ is a natural transformation. –  Jon Beardsley Nov 17 '11 at 21:57
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specifying a universal element u is equivalent to an isomorphism of functors $h \simeq [-,K]$, I'm not sure I understand what your argument would be though. –  Jacob Bell Nov 17 '11 at 22:05
    
Yeah you're right, perhaps not such a great idea. Maybe by universal examples they mean what you plug into $h$. Yeah, I don't know. Sorry! –  Jon Beardsley Nov 17 '11 at 22:10
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I think I will commit the sin of accepting my own answer: may god have mercy on my soul. –  Jacob Bell Nov 19 '11 at 22:16

2 Answers 2

In multilinear algebra a bilinear map $M\times N\to P$, where $M,N,P$ are modules over a ring $R$, is represented by a linear map $M\otimes_R N\to P$: we need the tensor product to talk about bilinearity without leaving the category of modules and linear maps. When $A=M=N=P$, this fits $A$ with a product structure $\mu$, i.e. the structure of an $R$-algebra.

In the case of a cohomology theory $E^\cdot$ with values in a category of $R$-modules, we know that the cohomology module of a space comes with a natural grading. If we write $F\otimes G$ for the functor $X\mapsto F(X)\otimes G(X)$, where $F,G$ takes values in a category with tensor products, a product structure* on a cohomology theory is really a natural transformation $E^\cdot\otimes E^\cdot\to E^\cdot$ of $R$-linear maps of graded modules, which decomposes into morphisms $E^k\otimes E^l\to E^{k+l}$.

We want these morphisms to be represented on the level of spectra, and as in the case of multilinear maps we need the smash product $\wedge$ to talk about "bilinear maps" $E^\cdot\wedge F^\cdot\to G^\cdot$ without leaving the category of spectra. The definition itself is straightforward: the smash product functor is left-adjoint to $\hom$.

The proof of its existence, however, requires a construction, which turns out to be really difficult to provide: there are now many of them, and as far as I know they all have both strengths and weaknesses. Some, for instance, are only well behaved in "the" homotopy category of spectra, not on the actual level of spectra. So to answer the question on the last line to the best of my knowledge: yes, this should at the very least be what we want.

*For $E^\cdot$ ordinary singular cohomology with values in an $R$-algebra $A$, the product structure on $E^\cdot$ is induced by the coproduct on chain complexes $C_\cdot=C_\cdot(X,Y;A)$, the tensor product of cochains $\xi,\eta$, and the $R$-algebra structure $\mu$ on $A$, i.e. the composition $C_\cdot\to C_\cdot\otimes_R C_\cdot\stackrel{\xi\otimes\eta}{\to} A\otimes_R A \stackrel{\mu}{\to} A.$

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thank you for your answer. I can see that all this \emph{motivates} $\wedge$ of spectra, but I still don't understand how to construct the map Goerss had in mind. –  Jacob Bell Nov 19 '11 at 0:06
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I think a way to make sense of all this is as follows.

The `correct' way to define a (reduced) cohomology theory with products is in fact by using the smash product (cf the relative version of the kunneth formula [Hatcher, Theorem 3.21]). To be a little bit more precise we require the existence of a morphism $$ h^p(X) \otimes h^q(Y) \to h^{p+q}(X\wedge Y)$$

so if $h^*$ is represented by a spectrum $\{K_n\}$, by considering the cohomology classes corresponding to the identity maps of $K_p$ and $K_q$, this extra structure gives in fact a morphism $$ K_p \wedge K_q \to K_{p+q}$$

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