Let $E \to F$ be a morphism of cohomology theories defined on finite CW complexes. Then by Brown representability, $E, F$ are represented by spectra, and the map $E \to F$ comes from a map of spectra. However, it is possible that the map on cohomology theories is zero while the map of spectra is not nullhomotopic. In other words, the homotopy category of spectra does not imbed faithfully into the category of cohomology theories on finite CW complexes. This is due to the existence of phantom maps: 


Let $f: X \to Y$ be a map of spectra. It is possible that $f$ is not nullhomotopic even if for every finite spectrum $F$ and map $F \to X$, the composite $F \to X \stackrel{f}{\to} Y$ is nullhomotopic. Such maps are called phantom maps. For an explicit example, let $S^0\_{\mathbb{Q}} = H\mathbb{Q}$ be the rational sphere. This is obtained as a filtered (homotopy) colimit of copies of $S^0$ and multiplication by $m$ maps. The universal coefficient theorem shows that there are nontrivial maps $S^0\_{\mathbb{Q}} \to H \mathbb{Z}[1]$; in fact they are parametrized by $\mathrm{Ext}^1(\mathbb{Q}, \mathbb{Z}) \neq 0$. However, these restrict to zero on any of the terms in the filtered colimit (each of which is a copy of $S^0$).


In other words, the distinction between flat and projective modules is in some sense an algebraic analog of the existence of phantom maps. Given a flat non-projective module $M$ over some ring $R$, then there is a nontrivial map in the derived category $M \to N[1]$ for some module $N$. Now $M$ is a filtered colimit of finitely generated projectives -- Lazard's theorem -- and the map $M \to N[1]$ is "phantom" in that it restricts to zero on each of these finitely generated projectives (or more generally for any compact object mapping to $M$). 
So it should not be too surprising that phantom maps of spectra exist and are interesting. 


Now spectra are analogous to the derived category of $R$-modules, but spectra also come with another adjunction: 
$$ \Sigma^\infty, \Omega^\infty: \mathcal{S}\_\* \leftrightarrows \mathcal{Sp}$$
between pointed spaces and spectra. They thus come with another distinguished class of objects, the suspension spectra. (Random question: what is the analog of a suspension spectrum in algebra?) 


**Definition:** A map of spectra $X \to Y$ is **hyperphantom** if for any suspension spectrum $T$ (let's interpret that loosely to include desuspensions of suspension spectra), $T \to X \to Y$ is nullhomotopic. 


In other words, a map of spectra is hyperphantom if the induced natural transformation on cohomology theories of spaces (not necessarily finite CW ones!) is zero. 


Is it true that a hyperphantom map is nullhomotopic? Rudyak lists this as an open problem in "On Thom spectra, orientability, and cobordism." What is the state of this problem? 



Are spectra really the same as cohomology theories?
Consider the periodic complex $K$-theory spectrum $KU$. The integral homology group $H\_i(KU)$, the direct limit of 


$$\dots \to H\_{2n+i}(BU)\to H\_{2n+2+i}(BU)\to\dots,$$
is a one-dimensional rational vector space if $i$ is even and trivial if $i$ is odd. It follows that $H^1(KU)$ is nontrivial. (It's $Ext(\mathbb Q,\mathbb Z)$.) But this can't be detected in the cohomology of suspension spectra, because $H^{2n+1}(BU)$ is trivial.


So that's an example of a "hyperphantom" map from $KU$ to the Eilenberg-MacLane spectrum $\Sigma H\mathbb Z$.


