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Is there an endofunctor $U: \mathrm{Top} \to \mathrm{Top}$ (or from some good subcategory) such that $H_n(UX) = H_{n+1}(X)$ for any $n \geq 1$

Arshak Aivazian
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  • Yes: https://math.stackexchange.com/questions/1322170/relationship-between-homology-of-suspension-of-x-and-x – Sergey Guminov Jan 21 '22 at 06:38
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    @SergeyGuminov You should look more carefully at the indexing. – Connor Malin Jan 21 '22 at 07:00
  • @SergeyGuminov That's a different question, yes, but I didn't regret reading that fact, thanks :) – Arshak Aivazian Jan 21 '22 at 09:10
  • @Aivazian Arshak. Yeah, I didn't notice the indexing. But as gor your question, I don't think such a thing exists, because by iterating $U$ you would get that all your homology vanishes, since $H^i$ vanishes for $i<0$. – Sergey Guminov Jan 21 '22 at 10:12
  • @SergeyGuminov Why not an official answer? – Paul Frost Jan 21 '22 at 10:33
  • @Paul Frost because I recently heard the term "desuspension" and was interested how it relates to this question. But I guess it's a construction for something much more abstract than topological spaces. – Sergey Guminov Jan 21 '22 at 10:55
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    Interesting question, even though the answer is that most likely no such functor $U$ exists the proof of that would be interesting. – Noel Lundström Jan 21 '22 at 22:01

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I don't believe such a functor exists, unless you restrict your attention to contractible spaces, which is uninteristing.

The reason is that then you would have $H_n(X)=H_{n-1}(UX)=\ldots=H_{-1}(U^{n+1}X)=0$.

Sergey Guminov
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