I am trying to show, or find a reference, for the following fact:
"Given O,P two infinity operads [in the sense of Lurie, HA, Definition 2.1.1.10], there always exist a tensor product".
In other words, this means the following. One can consider an infinity operad as a simplicial set over $K=N(Fin_*)$ with some marked edges, namely the inert maps. We call the ($\infty$) category of pairs $(X,\Gamma)$ where $X$ is a simplicial set and $\Gamma$ is a subset of edges that contains degenerate edges a marked simplicial set, denoted by $sSet^+_{/K}$.
Then, by the very definition, a map of infinity operads is a morphism as marked simplicial sets. Note that the product of two infinity operads, as marked simplicial set, is not necessarily an infinity operad. Indeed, one can show that there exist a model category on $sSet^+_{/K}$ such that infinity operads are exactly the fibrant objects, and every object is cofibrant.
Define bifunctors from $O,P$ to $Q$ (everything being an infinity operad) as $$Map_{sSet^+_{/K}}(O\times P, Q) $$ We say that $\alpha: O \times P \to Q$ exhibit Q as the tensor product of O,P if, for any other ifninity operad R, the composition map
$$ \alpha^*: Map_{sSet^+_{/K}}(Q,R) \to Map_{sSet^+_{/K}}(O \times P,R) $$ is an equivalence of sSets.
In literature is often said that it is enough to take a fibrant replacement of $O\times P$ in the category of marked sSets over K. I am trying to show, and here there are my attempts:
- Note that $\alpha^*$ is surjective in every degree. Indeed the diagram determined by $$O \times P \times (\Delta^n)^{\#} \to R$$ $$ O \times P \times (\Delta^n)^{\#} \to Q \times (\Delta^n)^{\#}$$ $$Q\times (\Delta^n)^{\#} \to 0,R \to 0$$
admits a lifting, because $R \to 0$ is a fibration and $O \times P \to Q$ is an acyclic cofibration. Recall that $(\Delta^n)^{\#}$ is the (left) canonical marked simplicial set associated to the n-cell, with just degenerate edges marked. This is the cosimplicial object we use for maps in degree n.
- For every marked S, the sequence
$$ S \times (\Delta^n)^{\#} \coprod S \times (\Delta^n)^{\#} \to S \times (\Delta^{n+1})^{\#} \to S \times (\Delta^n)^{\#} $$
exhibits $ S \times (\Delta^{n+1})^{\#}$ as a canonical cylinder object of $S \times (\Delta^n)^{\#}$ (canonical meaning that the last is a fibration). This is true because it is at the level of $\Delta^n$, and multiplying by $Q$ preserve at least cofibrations (~injectives) and equivalences.
I would hope that, with some being fibrant argument, one can show that every left homotopy can be done via tha cylinder object above.
As the map $O\times P \to Q$ is a weak equivalence, both are cofibrant, and $R$ is fibrant, we have that $\alpha^*$ is an equivalence between $Map(O \times P, R)_n/ homotopy$ and $Map(Q, R)_n /homotopy$.
Even if everything goes in the good direction, it seems that these facts point at $\alpha_*$ being a weak equivalence, and not an equivalence. In other words, it hints at the fact that it may be a simplicial model category (see https://ncatlab.org/nlab/show/simplicial+model+category, definition 2.1, point 3). But how the hell showing that fibers are contractible and not weakly contractible?
