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A map $f:X\to Y$ of CW-complexes is called a phantom if $f$ restricted to the $n$-skeleton of $X$ is contractible for all $n$. The first non-trivial example of such a map, with $X=\Sigma\mathbb{P}^\infty(\mathbb{C})$ and $Y$ an infinite wedge of 4-spheres, was constructed by J. F. Adams and G. Walker in 1964. Subsequently, it was shown that many other examples exist. In particular, in 1966 B. Gray showed that there are continuously many non-homotopy equivalent phantoms $K(\mathbb{Z},2)\to S^3$.

I would like to ask if there is a way to construct a non-trivial phantom map, or at least to prove such maps exist, by hand (say, using the material covered in Hatcher's Algebraic topology).

The motivation is this: a colleague of mine has gone away for a while and has asked me if I could replace him during his topology problem classes. I agreed but the instructions given to me were rather vague ("just show them some cool examples..."). I have enough examples to fill all the sessions (a couple of those found on MO by the way), but still I was wondering if I could construct a phantom by hand, and learn myself how to do this. The students seem to be pretty smart but haven't seen much beyond the basic cohomology and homotopy theory, not yet anyway.

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+1 for the motivation of this question! –  Sean Tilson Sep 26 '12 at 14:15
    
Can you clarify why the Adams-Walker example does not fit your requirements? –  Chris Schommer-Pries Sep 27 '12 at 15:53
    
Chris -- true, the construction by Adams and Walker is quite elementary compared to some other, but still the proof that it works uses cohomology operations modulo an arbitrary $p$ ($p=2$ does not suffice). I could explain what those are, but if there is a more elementary trick that produces phantoms, I'd be interested to know. –  algori Sep 29 '12 at 20:23
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4 Answers

Some phantoms arise from considerations in algebra.

For abelian groups $A$ and $B$ and $n > 1$, there is an isomorphism $$ [K(A,n), K(B,n+1)] \cong Ext(A,B). $$ This is a consequence of:

  • $H_{n+1} K(\mathbb{Z},n) = 0$;

  • the Kunneth formula shows $H_{n+1} K(F,n) = 0$ when $F$ is free; and

  • any resolution $0 \to R \to F \to A \to 0$ gives a fibration $K(R,n) \to K(F,n) \to K(A,n)$ and a Serre spectral sequence that determines $H^{n+1}(K(A,n);B)$.

If $K$ is a finite CW-complex and $A$ is any abelian group, any element of $H^{n}(X;A)$ lifts to an element of $H^n(X;A')$ for a finitely-generated subgroup $A' < A$. Consequently, if $K$ is any finite subcomplex of $K(A,n)$, the map $K \to K(A,n)$ factors up to homotopy through $K(A',n)$ for some finitely-generated subgroup $A' < A$.

Therefore, to produce a phantom map it suffices to find $A$, $B$, and an element of $Ext(A,B)$ which restricts to zero in $Ext(A',B)$ for any finitely-generated subgroup $A'$. For instance, basically any element of the (gigantic) group $Ext(\mathbb{Q},\mathbb{Z})$ satisfies this.

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Why do you call $\mathbb R = Ext(\mathbb Q,\mathbb Z)$ gigantic ? –  Ralph Sep 26 '12 at 6:40
    
Tyler -- thanks, but isn't this, like Oscar's example, an example of a map $f:X\to Y$ such that for any finite $K$ and any $g:K\to X$, the composition $f\circ g$ is null homotopic? The definition used in my question is a bit stronger: any phantom wrt to it has this property but not vice versa. –  algori Sep 26 '12 at 10:37
    
Algori, you're correct; I didn't see on a first look that you--and McGibbon, in his paper in the Handbook--are working with a different definition of "phantom" than I'm used to. Of course, your definition nixes anything I try to do with Eilenberg-Mac Lane spaces, and I don't have another example at hand. –  Tyler Lawson Sep 26 '12 at 13:24
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Ralph, unless I've miscalculated the group $Ext(\mathbb{Q},\mathbb{Z})$ is the quotient of the additive group $\mathbb{Q}\otimes\widehat{\mathbb{Z}}$ (the finite adeles) by the subgroup $\mathbb{Q}$ of rational numbers. That feels "large" to me, although of course abstractly both are just uncountable rational vector spaces. –  Tyler Lawson Sep 26 '12 at 13:36
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For any CW complex $X$ there is a map $\theta_X:X\to \bigvee_{n=1}^\infty \Sigma X_n$, where $X_n$ denotes the $n$-skeleton of $X$. This map can be realized as the cofiber of the folding map $\bigvee_{n=1}^\infty X_n \to X$. By construction, the map $\theta_X$ is phantom, called the Universal Phantom Map out of $X$.

Gray showed in his thesis that every phantom map $X\to Y$ factors through $\theta_X$ up to homotopy, so essential phantom maps out of $X$ exist if and only if $\theta_X$ is nontrivial. Since we know, for example, that $\mathbb{C}P^\infty$ is the domain of essential phantom maps, it follows that $\theta_{\mathbb{C}P^\infty}:\mathbb{C}P^\infty\to \bigvee \Sigma \mathbb{C}P^n$ is an essential phantom map that is the homotopy cofiber of the map $\bigvee \mathbb{C}P^n \to \mathbb{C}P^\infty$.

This example is included in McGibbon's article in the Handbook, and is addressed in some detail in Modern Classical Homotopy Theory by Jeff Strom.

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I did some work a couple of years ago on phantom maps, but never wrote any of it up because I didn't prove what I wanted. I was focusing on the case of ring spectra. Since it's possible you'd introduce spectra to a class knowledgeable with homology and cohomology, I'll give you a couple of examples from that category. My favorite two papers on this are:

  • Phantom Maps and Chromatic Phantom Maps by Hovey and Christensen
  • Homological Dimensions of Ring Spectra by Hovey and Lockridge (preprint)

In the first paper they prove, as Proposition 2.5, that all maps from $\hat{S^0}$ to $H\mathbb{Z}$ are phantom. This proof only requires the Universal Coefficient Theorem, a straight-forward computation of $H^0(\hat{S^0})$ and $H^1(\hat{S^0})$, and the fact that Hom$(\hat{\mathbb{Z}}, \mathbb{Z}) = 0$. Even if you waive your hands a bit about what this $\hat{S^0}$ is and tell them it's a completion which has $H_*(\hat{S^0}) = \hat{\mathbb{Z}}$, you'd still have an example here without needing to go much into the theory of spectra.

Corollary 2.7 might also be interesting, but requires more knowledge of spectra: "let $X$ have bounded below cohomotopy groups and let $Y$ be a finite spectrum. Then all maps from $X$ to $Y$ are phantom"

In the second paper, the following wisdom is given (for $E$ an $S$-algebra, aka ring spectrum), which might help for a class seeing this for the first time:

"Recall that a phantom map is a map $f$ for which $[C, f]_∗ = 0$ for every compact $E$-module $C$. So, if we think of a map whose cofiber is a ghost as the homotopy-theoretic analogue of an epimorphism, then a map whose cofiber is a phantom is the homotopy-theoretic analogue of a pure epimorphism"

If you like these examples, I can check through my stack of papers about phantom maps more completely and try to come up with more nice examples. Also, probably Neil Strickland has written something on the subject. He's got lots of notes for various courses on his webpage and seems to like thinking about phantom maps.

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David, thanks! I've checked the first paper you mention; it seems the definition used there is similar to the one Tyler and Oscar use for spaces; the one I'm interested in is somewhat more restrictive, and I'm not sure it makes sense for spectra, at least not without some without modifications, as it involves skeleta. –  algori Sep 26 '12 at 13:30
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I seem to remember that a typical example is: let $X$ be the mapping telescope of countably many iterations of $\cdot 2 : S^2 \to S^2$. One calculates $H^3(X;\mathbb{Z})$, say with cellular cochains, and finds that it is non-zero, so there is an essential map $f : X \to K(\mathbb{Z},3)$. But any finite skeleton of $X$ deformation retracts onto an $S^2$, so $f$ restricts to a nullhomotopic map on any finite skeleton.

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Hi Oscar. Did you mean to say that the restriction to any finite subcomplex is null-homotopic? Your $X$ is three-dimensional. –  Mark Grant Sep 26 '12 at 6:21
    
That is indeed what I meant, and I now see that the question was about a slightly different notion of phantom map than I thought. –  Oscar Randal-Williams Sep 26 '12 at 6:55
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