One can show that spectrifications of maps between $\Sigma$-cofibrant prespectra that are spacewise homotopy equivalences are homotopy equivalences of spectra. I'm interested in the following:

Does this result hold if only the source prespectrum is assumed to be $\Sigma$-cofibrant?

The reason for my interest is the following. The adjoint from prespectra to inclusion prespectra is difficult to deal with. I see that sometimes people replace a prespectrum by a thickened one, via a cylinder construction, as in Chapter X, Section 5 of Elmendorf-Kriz-Mandell-May's "Rings, modules and algebras in stable homotopy theory", to get a $\Sigma$-cofibrant prespectrum so that we have a nicer description of the spectrification. This is done, for example, when defining the topological Hochschild spectrum in Hesselholt-Madsen's "On the $K$-theory of finite algebras over Witt vectors of perfect fields". Say the original prespectrum is $t$, the cylinder construction yielding a $\Sigma$-cofibrant prespectrum is denoted by $K$ and spectrification is denoted by $L$. We have a map of prespectra $Kt \to t$ which is a spacewise homotopy equivalence. I would hope that $LKt$ and $Lt$ are related in a nice way so as to say that the intermediate thickening didn't essentially change anything. The "nice way" I'm hoping for is that one can deduce that the induced map $LKt \to Lt$ is a homotopy equivalence (or perhaps weak homotopy equivalence).

  • 2
    $\begingroup$ What model are you using for spectra? $\endgroup$ Jan 30, 2018 at 19:13
  • $\begingroup$ I suppose I should have said: the one in EKMM $\endgroup$
    – M.G.
    Feb 2, 2018 at 22:26

1 Answer 1


This is false. Unfortunately, lack of cofibrancy in this kind of situation can mean an almost entire loss of control over what happens to the spectrification. Let's give an example.

Here is a prespectrum $S$: it associates to an inner product space $V$ the one-point compactification $S^V = V \cup \{\infty\}$.

For every $V$, there is a subspace $D(V)$ which is the complement of the open ball of radius $1/\dim(V)$, and a quotient space $T(V) = S^V / D(V)$. This quotient is the one-point compatictification of the ball of radius $1/\dim(V)$, so $T(V)$ is a sphere where we have collapsed all the long vectors. We have natural structure maps $S^W \wedge T(V) \to T(W \oplus V)$ that are compatible across $V$ and $W$ (the right-hand term is a larger quotient of $S^{W \oplus V}$ than the left-hand one). These assemble together into a prespectrum $T$ and there's a natural map $S \to T$ which is levelwise a homotopy equivalence. Moreover, $S$ is $\Sigma$-cofibrant (it's the suspension prespectrum of $S^0$).

$S$ will spectrify to the sphere spectrum. However, $T$ spectrifies to something entirely different. Any map from $T$ to a spectrum $Y$ (whose structure maps are homeomorphisms) will factor through $$T(V) \to {\rm colim}_{W} \Omega^W T(W \oplus V) \to Y(V)$$ and any nonzero vector $v$ in $T(V)$ gets sent to a loop through vectors of length strictly greater than or equal to $||v||$, and these become equivalent to the point $\infty$ in the colimit because we are progressively identifying vectors of shorter and shorter length with $\infty$.

Long story short, the spectrification of $T$ factors through a prespectrum $W$ which sends each $V$ to the Sierpinski two-point space, and the map $S \to W$ induces zero on homotopy groups.

  • $\begingroup$ This is great, thanks a lot! Shame though then that I don't have the desired relation $LKt$ and $Lt$. My next question then is what one can say about this relation; is there anything you could offer on this? $\endgroup$
    – M.G.
    Feb 2, 2018 at 22:25

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