TITLE: On the endofunctor $\Sigma^\infty\Omega^\infty$
QUESTION [5 upvotes]: Consider the co-monad $M:=\Sigma^\infty \Omega^\infty$ on the category of spectra.
It is clear that given a pointed space $X$,  $M\Sigma^\infty X=\Sigma^\infty E(X)$,where $E(X)$ is the free unital $E_\infty$-algebra on $X$.
As $\Sigma^\infty$ commutes with colimit,  $M\Sigma^\infty X$  is equivalent to
the free $E_\infty$-algebra on  the spectrum $\Sigma^\infty X$, that is,
to: $$
\lor_n (\wedge^n (\Sigma^\infty X)_{hS_n}).\tag{1}
$$
Is this right?

It looks like given an  arbitrary spectrum $A$, there is no equivalence between
$\Sigma^\infty\Omega^\infty A$ and $
\lor_n (\wedge^n (A)_{hS_n})
$
Is there any general condition for having such  an equivalence?

REPLY [4 votes]: For connected spectra $A$, there is an equivalence
$$\Sigma^\infty \Omega^\infty A \simeq \bigvee_{n=1}^\infty A^{\wedge n}_{h\Sigma_n}$$
if and only if $A$ is a wedge summand of a suspension spectrum.
This is Theorem 1.2 in N. Kuhn's paper Suspension Spectra and Homology Equivalences (TAMS, 1983). Kuhn calls spectra that are wedge summands of suspension spectra spacelike.
I am guessing that if $A$ is not connected then such an equivalence can not exist. It is easy to see that there can not be an equivalence of ring spectra.