I have absolutely no idea whether this is true or not but it could well be useful for me in the future if it is. If we have topological operads $\mathcal{P}$ and $\mathcal{Q}$ and we let $W$ denote Boardman-Vogt's functor for creating cofibrant resolutions of operads. Then, is this functor monoidal with respect to the Boardman-Vogt tensor product $\otimes_{BV}$ of operads, i.e. do we have $$W(\mathcal{P} \otimes_{BV} \mathcal{Q}) \cong W\mathcal{P} \otimes_{BV} W\mathcal{Q}$$ If anyone happens to know whether this is true or not, could you also either explain why or provide a reference to the result.
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$\begingroup$ I can't recall the definition of the Boardmann-Vogt tensor product, but the easiest way to check this would be with P=Q=unit of the tensor product. I suspect this would lead to a contradiction (say by counting 0-cells on either side). This still leaves the possibility of a homotopy equivalence. $\endgroup$– James GriffinMay 6, 2015 at 17:05
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$\begingroup$ I think it works in this case because I am pretty sure that the unit of the tensor product is $\mathcal{I}=(\emptyset,\ast,\emptyset,\emptyset, \ldots )$ and then $W\mathcal{I}=\mathcal{I}$, maybe I have misunderstood what you mean? $\endgroup$– Dean BarberMay 7, 2015 at 7:59
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$\begingroup$ Ah I see, it's been a while since I looked at the details of $W$, I was thinking that $W\mathcal{I}$ was a bit larger, and had forgotten you need to enforce it to be unital. How about $\mathcal{P}=\mathcal{Q}=\text{Com}$, the commutative operad? $\endgroup$– James GriffinMay 8, 2015 at 13:51
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$\begingroup$ I'd go in a different direction---how about if $\mathcal{P}$ and $\mathcal{Q}$ are concentrated in arity $1$? Then the tensor product is the Cartesian product, right? Does it work in that case? Or concentrate $\mathcal{P}$ in arity $0$ where both the tensor product and the $W$ construction are basically trivial. Does it work there? Also, what do you mean by $\cong$---homeomorphism? natural homotopy equivalence? $\endgroup$– Gabriel C. Drummond-ColeMay 10, 2015 at 18:40
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