Dear all,Please re-open this question which is put on hold:

https://math.stackexchange.com/questions/2887114/understanding-de-suspension-sigma-1-sigmax-neq-x

It was said that "unclear what you're asking" and people do not know the def of de-suspension.

However, the suspension is introduced earlier in the cited question:

https://math.stackexchange.com/questions/2884901/the-suspension-topology-and-elementary-examples

While the desuspension is also quoted/linked to the Wikipedia (withe refs given by Wiki). I also include a new note: "The desuspension is arguably firstly introduced in the …

Please re-open this question which is put on hold:

https://math.stackexchange.com/questions/2887114/understanding-de-suspension-sigma-1-sigmax-neq-x



Follow people's comments, I modify to also ask the basic definitions:


> Question: How do we define desuspension exactly? (Please see the comments below, people complain about the meanings of desuspension in Wikipedia is useless).

> Are we able to have the desuspension acting on the topological space as the suspension does? Or do we only have the desuspension act on the spectra but not the space?

@TedShifrin Proof by contradiction that $s$ is rational infinitely often:

Assume $s=\frac{ln(1/2)}{lnx} \notin \Bbb Q$.

For a given $n$, select the $n$ curves in the family
$x^s+\phi^s=1$;
$x,\phi \in \Bbb Q (0,1), s\notin \Bbb Q,$
where the $k$-th curve passes through $\left(\frac k{n+1}, \frac k{n+1}\right)$.

If $s$ is not rational, then that would mean as $n$ tends to infinity, there would be no rational solutions of $x^s+\phi^s=1$ intersecting with $\phi=x$.

This is equivalent to saying $s=\frac{ln(1/2)}{lnx}$ is never rational.