1. 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?
2. 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?



On the endofunctor $\Sigma^\infty\Omega^\infty$
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.


