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?

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    $\begingroup$ Nice question! Nothing substantive to say, but regarding your random side-question: wouldn't the natural analog of the suspension spectrum of a space in algebra be the free abelian group (or free R-module) on a set? $\endgroup$ Commented Dec 31, 2012 at 3:20
  • $\begingroup$ @David: I'd say it would be a bounded below chain complex. Anyway, any answer to this question is valid I guess ;-) $\endgroup$ Commented Dec 31, 2012 at 3:31
  • $\begingroup$ Perhaps, although it's not really clear to me why sets (in relation to the derived category of $R$-modules) should take the place of "spaces." The $\infty$-category of spectra can be described as the stabilization of the category of spaces, and the derived category of $R$-modules (if $R$ is commutative) can be described as the stabilization of $E_\infty$-algebras over and under $A$. That's a candidate, although $E_\infty$-algebras over and under the sphere don't seem to have much to do with spaces. $\endgroup$ Commented Dec 31, 2012 at 3:59
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    $\begingroup$ @Akhil - Sets take the place of spaces in relation to the abelian category of R-modules. For the derived category spaces (or simplicial sets) are still the analog, and your adjunction and Fernando's (the derived version of mine) are literally the same --- both special cases of an adjunction (suspension spectra for the sphere, Dold-Kan for a discrete ring) between spaces and R-module spectra for any $E_\infty$ ring spectrum. $\endgroup$ Commented Dec 31, 2012 at 16:40
  • $\begingroup$ @David: OK, agreed. $\endgroup$ Commented Dec 31, 2012 at 21:28

2 Answers 2


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 $\Sigma H\mathbb Z$.

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    $\begingroup$ This is very neat! This says that in some sense there are characteristic classes of stable vector bundles that vanish on any space but not on spectra. $\endgroup$ Commented Jan 1, 2013 at 8:55
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    $\begingroup$ Nice, I should have remembered that. Notice how classical it is. $\endgroup$
    – Peter May
    Commented Jan 1, 2013 at 13:33
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    $\begingroup$ The integral homology of $KU$ is a favorite example, but I hadn't realized it was example of this phenomenon until I thought a while. $\endgroup$ Commented Jan 1, 2013 at 15:54
  • $\begingroup$ Why $H_i(KU) \cong \mathrm{colim}_n H_{2n +i}(BU)$ and not $H_i(KU) \cong \mathrm{colim}_n H_{2n +i}(BU \times \mathbb{Z})$? Are they the same? $\endgroup$
    – CNS709
    Commented Jul 9, 2022 at 15:20
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    $\begingroup$ Yes, they're the same. The canonical map from $\Sigma^2(BU\times \mathbb Z)$ to $BU\times \mathbb Z$ factors through $BU\times 0$. $\endgroup$ Commented Jul 9, 2022 at 17:34

The answer to this question is in LMS (I.6.9 of http://www.math.uchicago.edu/~may/BOOKS/equi.pdf) and in McClure's contribution to BMMS (VII\S1 of http://www.math.uchicago.edu/~may/BOOKS/h_infty.pdf), which gives full details. Let $T = \{T_i\}$ be a prespectrum. There is a cylinder construction $ZT$ that gives a weakly equivalent $\Omega$-spectrum (I first defined it in 1968 http://www.math.uchicago.edu/~may/PAPERS/7.pdf).
It is the telescope of the desuspensions $\Sigma^{-i} \Sigma^{\infty} T_i$. There is no loss of generality in taking $X=ZT$ in your question. There results a $lim^{1}$ exact sequence of the form

$$ 0 \to lim^{1}[\Sigma^{1-i} \Sigma^{\infty} T_i,Y] \to [X,Y] \to \lim[\Sigma^{-i} \Sigma^{\infty} T_i,Y]\to 0.$$

This has nothing to do with finite CW spectra, a priori, and it can be viewed via the usual adjunctions as giving a precise measure of the difference between the stable homotopy category of spectra and the homotopy category of based spaces. McClure (VII\S4 op cit) gives a clear criterion for when the $lim^{1}$ term vanishes and examples where the criterion holds. It is obviously not to be expected that the $lim^{1}$ term vanishes in general. One way to construct counter-examples is to relate this $lim^{1}$ exact sequence with the one given by approximating $X$ by a CW-spectrum, but I'll leave that to the interested reader. Of course, the elements of this $lim^{1}$ term are your hyperphantom maps.

Added as an edit: In answer to Tom Goodwillie's comment, the adjunctions I referred to give that if $Y$ is an $\Omega$-spectrum with $i$th space $Y_i$, then $$[\Sigma^{-i} \Sigma^{\infty} T_i,Y] \cong [T_i,Y_i]. $$ The brackets refer to spectra on left and based spaces on the right. Therefore the original $lim^{1}$ exact sequence can be rewritten as $$ 0 \to lim^{1}[T_i,Y_{i-1}] \to [X,Y] \to \lim[T_i,Y_i]\to 0.$$ The $lim$ and $lim^1$ terms are computed in terms of homotopy classes of maps of based spaces. That is what I had in mind with my sloppy statement about comparing homotopy categories. This is really a comparison between the stable homotopy category and the category of cohomology theories on based spaces, answering the original question.

Everyone go have fun: it's New Year's Eve (with a whole new meaning to the countdown to midnight).

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    $\begingroup$ (Presumably "homotopy category of based spaces" should be "stable homotopy category of based spaces".) $\endgroup$ Commented Dec 31, 2012 at 17:44
  • $\begingroup$ Tom, I edited in an elaboration in response to your comment. $\endgroup$
    – Peter May
    Commented Dec 31, 2012 at 18:36
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    $\begingroup$ Interesting. To elaborate for those who were initially confused (such as myself), to show that a map $X \to Y$ is hyperphantom, it suffices to show that $\Sigma^\infty \Omega^\infty X \to X \to Y$ is nullhomotopic (and the same with $X \to Y$ replaced by any (de)suspension) because the universal map from a suspension spectrum into $X$ is $\Sigma^\infty \Omega^\infty X \to X$. $\endgroup$ Commented Dec 31, 2012 at 21:17
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    $\begingroup$ However, while I may be misunderstanding something, it is still not clear to me how to construct an explicit example of a map $X \to Y$ satisfying this condition (that is, the above description of $X$ as a filtered colimit of suspension spectra is sufficiently special that one might imagine the $\lim^1$ terms would tend to vanish). $\endgroup$ Commented Dec 31, 2012 at 21:27
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    $\begingroup$ Akhil, that goes back before my time. One of the motivations for constructing the stable homotopy category back in early 1960's was the understanding that the category of cohomology theories cannot be triangulated. I can't remember counterexamples or even if I ever knew them. I'm pretty sure there was some discussion in Boardman's (unpublished) work. In any case, the non-triangulability has to be explained by non-vanishing lim^1 terms. $\endgroup$
    – Peter May
    Commented Jan 1, 2013 at 0:45

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