Of course, before group completion, topological commutative monoids do not model all connected $E_\infty$ spaces -- among the grouplike ones, they model only products of Eilenberg-Mac Lane spaces. But after group completion, perhaps they do. Similarly, I wonder: Do strictly symmetric monoidal (i.e. the symmetry isomorphisms $a \otimes b \cong b \otimes a$ are identities) topologically-enriched categories model all connective spectra?

Maybe something even stronger is true: Do strictly symmetric monoidal groupoids (or equivalently, 1-types with a commutative monoid structure) model all connective spectra after group completion? Do strictly symmetric monoidal groups (equivalently, $K(G,1)$'s with a commutative monoid structure) model all 0-connective spectra after group completion?

If the answer to some version of this question is affirmative, I'd liken it to the fact that every space is the realization of a category -- you can get some extra strictness in a model at the expense of some noninvertibility.

This question obviously has some affinities to another question I asked a few days ago.

Of course, before group completion, topological commutative monoids do not model all connected $E_\infty$ spaces -- among the grouplike ones, they model only products of Eilenberg-Mac Lane spaces. But after group completion, perhaps they do. Similarly, I wonder: Do strictly symmetric monoidal (i.e. the symmetry isomorphisms $a \otimes b \cong b \otimes a$ are identities) topologically-enriched categories model all connective spectra?

Maybe something even stronger is true: Do strictly symmetric monoidal groupoids (or equivalently, 1-types with a commutative monoid structure) model all connective spectra after group completion? Do strictly symmetric monoidal groups (equivalently, $K(G,1)$'s with a commutative monoid structure) model all 0-connective spectra after group completion?

If the answer to some version of this question is affirmative, I'd liken it to the fact that every space is the realization of a category -- you can get some extra strictness in a model at the expense of some noninvertibility.

This question obviously has some affinities to another question I asked a few days ago.