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Working in compactly generated weak Hausdorff spaces, is the category of inclusion prespectra bicomplete?

I should probably specify that by inclusion prespectra, I mean prespectra such that the adjoint structure maps are closed inclusions. It is probably complete, as close inclusions of CGWH spaces are characterized by a limit condition. But is it cocomplete? It would be enough to define a left-adjoint to the forgetful functor to prespectra (or even injection prespectra), but I was not able to construct one.

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  • $\begingroup$ I would say that presumably the answer is no (simply because the "closed inclusion" condition seems too delicate to be preserved by various quotients), but I don't have a counterexample. $\endgroup$
    – Tim Campion ♦
    Oct 20, 2018 at 19:43
  • $\begingroup$ That is my feeling as well, but I still want to make sure! $\endgroup$
    – user09127
    Oct 21, 2018 at 21:47

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No, inclusion spectra are not closed under colimits.

Let $S$ be the suspension spectrum of $S^0$, let $S'$ be like $S$ except the zeroth space is a point, and let $0$ be the spectrum which is constantly a point. These are all inclusion spectra. The pushout of $0 \leftarrow S' \to S$ fails to be an inclusion spectrum.

Hopefully something similar works with fancy categories of spectra.

Note I haven't ruled out that colimits exist in the category of inclusion spectra -- I've just shown that if they exist, they are not preserved by the inclusion functor to all spectra.

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