As connective spectra are equivalent to group-like $E_{\infty}$ algebras in spaces, the $\infty$-category of connective spectra is monadic over the $\infty$-category of spaces though the usual $\Sigma^{\infty} \dashv \Omega^{\infty}$ adjunction.

Is it also the case that the category of spaces is comonadic over the category of spectra (or maybe connective spectra) ?

If so do we have a nice description of "what is a (unstable) space" in terms of stable homotopy theory, i.e. can we say something about what are the $\Sigma^{\infty} \Omega^{\infty}$-coalgebra in spectra in terms of more familiar structure (in the same way that a $\Omega^{\infty} \Sigma^{\infty}$-algebra is the same as a group-like $E_{\infty}$-algebra).

For example $\Sigma^{\infty} X$ naturally has the structure of a cocomutative (in the $E_{\infty}$-sense) co-algebras (coming from the diagonal map of $X$). Can we characterize $\Sigma^{\infty} \Omega^{\infty}$-coalgebra, as such cocommutative coalgebra satisfying some additional conditions ?

I'm interested in how to deduce some properties of a higher toposes from properties of its category of spectra, and more generally, how much of a higher topos can be understood from its category spectra. so this sort of explicit description could be useful and any description of the co-monad only involving more classical structure on the category of spectra (like the smash product and the subcategory of connective spectra) might be useful.

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    $\begingroup$ Something like this is true rationally, since unstable rational homotopy types are some extra structure on a rational vector space. More generally, there is an analogous statement "v_n-periodicially". Here are two references: www3.nd.edu/~mbehren1/papers/BKTAQ8.pdf, arxiv.org/pdf/1803.06325v1.pdf $\endgroup$ – Dylan Wilson Apr 18 '18 at 18:49
  • $\begingroup$ whoops- made a silly comment then edited it, but: You need to at least restrict to something like simply connected spaces to have hope, otherwise equivalences are not detected stably $\endgroup$ – Dylan Wilson Apr 18 '18 at 18:51
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    $\begingroup$ With the 1-connected restriction, I think comonadicity is a result of Blomquist and Harper. $\endgroup$ – Tim Campion Apr 18 '18 at 19:10
  • $\begingroup$ Right, it cannot be comonadic simply because $\Sigma^{\infty}$ is not conservative. so together with Blomquist & Harper results I think this answer my question. thanks $\endgroup$ – Simon Henry Apr 19 '18 at 2:28

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