Let $f : X \to Y$ be a nonzero phantom map between spectra. Can the cofiber of $f$ be a finite spectrum?
Recall that a map $f$ is said to be phantom if $f \circ i = 0$ whenever $i : F \to X$ is a map from a finite spectrum $F$.
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This was easier than I thought: the answer is no.
Let $f : X \to Y$ be phantom, with finite cofiber $C$. Assume without loss of generality that $C$ is connective. Then $C \to \Sigma X$ factors through $(\Sigma X)_{\geq 0}$. It follows that $\Sigma X \to (\Sigma X)_{\leq -1}$ is phantom, and in particular zero on homotopy, so that $\Sigma X$ is connective. Then it follows that $Y$ is connective as well. Then because $f$ vanishes on homology and $H_\ast C$ is finitely-generated, it follows that $H_\ast \Sigma X$ and $H_\ast Y$ are also finitely-generated. Since $\Sigma X$ and $Y$ are connective, this implies that $\Sigma X$ and $Y$ are finite. So $f = 0$.
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1$\begingroup$ You could also argue via the fiber, which would be a finite F-->X through which all maps finite-->X factor. I think that forces X to be finite already? $\endgroup$– kiranCommented Feb 12 at 15:52
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$\begingroup$ @kiran I like that argument -- if you see how to complete it, I'd be interested! $\endgroup$ Commented Feb 12 at 17:33
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$\begingroup$ I was thinking: filter X as a colimit of finite spectra X_n such that the first stage is F. Then all other stages have their identity map factoring through F so theyre summands of F, so X is also. $\endgroup$– kiranCommented Feb 13 at 3:13