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

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?

• I think for (1) you should assume that $X$ is a pointed connected space, right? Jan 3, 2022 at 4:26

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.
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.