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A hyperphantom map of spectra is a stable map $p:D\to E$ such that $pu=0$ for every $u:{\Sigma }^{-j}{\Sigma }^{\infty }X\to D$, for arbitrary based spaces $X$ and all nonnegative $j$. Equivalently it induces zero on the entire represented cohomology theory. This is a stronger condition than being a phantom map of spectra, which tests only finite spectra.

A stable map $u:D\to E$ induces a degree-preserving additive natural transformation of represented cohomology theories by postcomposition with ${\Sigma }^{q}u$ in degree $q$. This transformation commutes with the suspension isomorphisms. The resulting mapis onto, but need not be one-to-one. Its kernel consists of hyperphantom maps of spectra: maps $u$ for which $uv=0$ for every stable mapfor every based space $X$ and every $j\ge 0$. Equivalently, they induce zero in all degrees of represented cohomology theory on all based spaces. ThereforeThis condition tests arbitrary suspension spectra with negative shifts. Merely testing finite spectra defines a phantom map of spectra, which is a weaker condition. In particular the homology conclusion in Question 5 should not be substituted for the cohomology condition here.

Choose a small skeleton $\mathcal{F}$ of the finite spectra, including every integer suspension. There is such a set of representatives: a finite list of stable cells and their attaching maps specifies a finite cell model, and the possible attaching maps form sets. For every $F\in \mathcal{F}$ and every stable map $v:F\to D$, take one copy of $F$ and formDefine the evaluation map $e:X\to D$ by taking its restriction to the $(F,v)$ summand to be $v$. Complete it to the cofiber sequence of spectraThis is the universal evaluation phantom map construction.

First, $p$ is nonzero. If $p=0$, applying $[D,-{]}_{\mathrm{st}}$ to the cofiber sequence of spectra gives the exact portionSince $p_{\ast }(1_D)=p=0$, exactness supplies $s:D\to X$ with $es=1_D$. Thus $D$ is a retract of $X$. This is also a wedge summand: complete $s$ to a cofiber sequence of spectra $D\to X\to C^{\prime }\to \Sigma D$. The left inverse $e$ makes this triangle split, giving $X\cong D\vee C^{\prime }$ in the stable homotopy category. One can see the splitting directly from exactness: the connecting map is zero because $e$ extends $1_D$ across $s$, and consequently the quotient map $X\to C^{\prime }$ admits a section. This contradicts the assumption on $D$. Hence $p\ne 0$.

For any finite spectrum $F$ and any $v:F\to D$, its chosen representative has a summand inclusion ${\iota }_v:F\to X$ with $e{\iota }_v=v$, after transferring along the representative isomorphism. Consecutive maps of a cofiber sequence of spectra compose to zero, soThus $p$ is a phantom map of spectra. To establish the requested conclusion on all spaces, rather than just coefficient groups, let $K$ first be a finite based CW complex. Put $A={\Sigma }^{\infty }K$. The Spanier-Whitehead dual $A^{\vee }=F(A,\mathbb{S})$ is a finite spectrum. The evaluation and coevaluation adjunction of Spanier-Whitehead duality identifiesUnder this identification $p_{\ast }$ is postcomposition with $p$. Its source is a finite spectrum, including when $k$ is negative, so the calculation $pv=0$ proves that $p_{\ast }$ is zero on ${\tilde{D}}_k(K)$ for every $k$.

Now let $Y$ be an arbitrary based CW complex. It is the filtered union, or filtered homotopy colimit of spaces, of its finite based subcomplexes $K$. The suspension spectrum functor and smash product of spectra preserve that homotopy colimit. The shifted sphere spectrum is compact, so taking stable homotopy groups givesEvery element of the first group therefore comes from a finite $K$, where $p_{\ast }$ has just been proved zero. Consequently $p_{\ast }$ is zero on $Y$ as well. A based CW complex replacement extends the conclusion to arbitrary based spaces in the homotopy category. We have constructedThis uses continuity of represented homology theory, not a continuity assertion for represented cohomology theory; the latter need not hold.

A phantom map of spectra is a stable map $p:D\to E$ with $pu=0$ for every map $u:F\to D$ from a finite spectrum. Such a map can be nonzero even though it induces zero on all represented homology theories on based spaces.

For a phantom map of spectra $p:D\to E$ and a finite CW complex $K$, Spanier-Whitehead duality identifies $D_k(K)$ with maps from the finite spectrum ${\Sigma }^k({\Sigma }^{\infty }K{)}^{\vee }$. Hence $p_{\ast }=0$ there. An arbitrary CW complex is the filtered union of its finite subcomplexes, and represented homology theory commutes with that filtered union because spheres are compact. Thus $p$ induces zero on the entire represented homology theory.

Choose one representative of each finite spectrum and form $X={\bigvee }_{(F,u:F\to D)}F$. Let $e:X\to D$ evaluate each summand and let $p:D\to \mathrm{cofib}(e)$. Every map from a finite spectrum factors through its corresponding summand, so $p$ is a phantom map of spectra. If $p=0$, exactness supplies a section of $e$, making $D$ a retract of $X$. Consequently a spectrum that is not a retract of any wedge of finite spectra has a nonzero such phantom.