# Is the category of inclusion prespectra bicomplete?

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.

• 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. – Tim Campion Oct 20 '18 at 19:43
• That is my feeling as well, but I still want to make sure! – user09127 Oct 21 '18 at 21:47

## 1 Answer

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.