If $C$ is a symmetric monoidal category, then $BC$ is canonically an algebra over a certain $E_\infty$ operad, but if $F: C \to D$ is a symmetric monoidal functor then (as far as I can see) $BF: BC \to BD$ is not a map of algebras over that operad (unless all the morphisms $(Fx) \otimes (Fy) \to F(x \otimes y)$ are identities).

Because of this I struggle to associate a spectrum map between the two spectra arising from $BC$ and $BD$ by infinite loop space theory, I can only see how to get a zig-zag where the wrong-way map is a weak equivalence. For most practical purposes that's just as good, but nevertheless I wonder: are any of the "well known" infinite loop space machines functorial on the nose, with respect to symmetric monoidal functors?


Here is a nice gentle old-fashioned answer. Symmetric monoidal categories are functorially equivalent as symmetric monoidal categories to permutative (symmetric strict monoidal) categories, and those are functorially equivalent (essentially the same as) algebras over a certain $E_{\infty}$ operad $\mathcal{P}$, known nowadays as the categorical Barratt-Eccles operad, in Cat. Since $B$ is product preserving it gives a functor from $\mathcal P$-algebras in Cat to $B\mathcal P$-algebras in Top. That gives $B$ as a functor from symmetric monoidal categories to algebras over an $E_\infty$ operad. That goes back, at least in outline, to my 1974 paper ``$E_{\infty}$ spaces, group completions, and permutative categories".

Again in outline, by two recent papers, the same argument works equivariantly for (genuine) symmetric monoidal $G$-categories, which give genuine $G$-spectra for finite groups $G$ via infinite loop $G$-space machines. See Equivariant iterated loop space theory and permutative $G$-categories http://www.math.uchicago.edu/~may/PAPERS/GM3.pdf and Symmetric monoidal $G$-categories and their strictification http://www.math.uchicago.edu/~may/PAPERS/AddCat1.pdf}.

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    $\begingroup$ Great, I was looking for an old-fashioned answer to my old-fashioned question! What I was missing was the first step, functoriality of the strictification. (Wordings like "there is an equivalent permutative category" made me think this was an existence result. But the proof, which I've now read, is both constructive and functorial. Just replace the object set with the free monoid on the old object set.) $\endgroup$ – user142661 Jul 4 '19 at 14:46

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