Topological spaces have diagonal maps $X \rightarrow X \times X$ and $X \rightarrow X \wedge X$, and suspension spectra also have diagonal maps $\Sigma^\infty X \rightarrow \Sigma^\infty(X \wedge X) \cong (\Sigma^\infty X) \wedge (\Sigma^\infty X)$. What about general spectra? (i.e. symmetric spectra, S-modules, or any other convenient definition.) I always assumed you could, but I haven't thought through it carefully. And if not, can we still get a cup product on $E^\*(X)$ when $E$ and $X$ are spectra?



do spectra have diagonal maps?
No. Let $[X,A]$ be the set (in fact abelian group) of homotopy classes of maps from one spectrum to another. If $A$, $B$, and $C$ are spectra, any natural map of sets $[X,A]\times [X,B]\to [X,C]$ is induced by a map $A\times B\to C$. Since $A\times B=A\coprod B$, this amounts to two maps $A\to C$ and $B\to C$ inducing two homomorphisms $[X,A]\to [X,C]$ and $[X,B]\to [X,C]$ which are then added to give one homomorphism $[X,A]\times [X,B]=[X,A]\oplus [X,B] \to [X,C]$. This map cannot be distributive over addition except by being identically zero. 


EDIT Taking $A$, $B$, and $C$ to be Eilenberg-MacLane spectra, this rules out nontrivial natural bilinear products on ordinary cohomology of spectra. More generally it rules out such products on generalized cohomology of spectra. And it also rules out any nontrivial natural map $X\to X\wedge X$ because if $A\to A\wedge A$ were nontrivial then this would lead to a natural bilinear map $[X,A]\times [X,A]\to [X,A\wedge A]$ that (for example when $X=A$) is nontrivial.


