# Lax monoidal fibrant replacement for marked simplicial sets

The category $$\mathrm{Set}_{\Delta}^{+}$$ of marked simplicial sets has a model structure (the (co)cartesian model structure) constructed by Lurie in HTT.3.1.3. I would like to know if this model structure admits a fibrant replacement functor $$R$$ that is lax monoidal with respect to the cartesian product of marked simplicial sets, that is, for which there are natural maps (necessarily equivalences) $$R(X) \times R(Y) \to R(X \times Y),$$ which commute with the fibrant replacement maps coming from $$X \times Y$$ and which are associative and unital in the usual sense.

• More generally, what's an example of a lax monoidal fibrant replacement functor other than $Ex^\infty$ or the induced fibrant replacement $Ex^\infty_\ast$ for the Bergner model structure? Saal Hardali once suggested a fibrant replacement like your $R$ might exist in the comments here. Welcome to MO, btw! Oct 1 '20 at 21:41
• Thanks Tim - yes, an analogue of $Ex^\infty$ is what I was hoping for. Oct 1 '20 at 21:59
• I think if such an $R$ exists, it will probably be quite different from $Ex^\infty$. Suppose for $X \in sSet^+$ we had some kind of $R(X)$, equivalent to $X$ and with the same 0-simplices, such that an $n$-simplex of $R(X)$ is given by some kind of map $S_{n,k} \to X$ where $S_{n,k}$ is equivalent to $\Delta[n]$. Already when $n = 1$ and $X = \Lambda^1[2]$ this fails: in $R(\Lambda^1[2])$ there should be a 1-simplex from 0 to 2. But there is no $S_{1,k} \in sSet^+$ admitting a map to $\Lambda^1[2]$ which hits both 0 and 2 and is equivalent to $\Delta[1]$. Oct 1 '20 at 22:27
• I think there's not much to be gained by loosening the assumption that $R(X)_0 = X_0$, so it seems one would have to loosen the assumption that $S_{n,k}$ is equivalent to $\Delta[n]$, at which point it seems we must be seriously diverging from the story of $Ex^\infty$. Oct 1 '20 at 22:30
• Garner's small object argument provides a sort of "canonical" and "small" fibrant replacement. I wonder if it preserves finite products in this case... Oct 2 '20 at 17:20

Not a complete answer. The following is inspired by Dmitri Pavlov's answer here.

Let's recall the work of Dugger and Spivak. Let $$\mathcal G$$ be a "category of gadgets closed under wedges" (Definition 5.4) (a full subcategory of $$sSet_{\ast,\ast}$$ satisfying certain conditions). Dugger and Spivak define a functor

$$\mathfrak C^{\mathcal G}: sSet \to sCat$$

$$Ob \mathfrak C^{\mathcal G}(S) = S_0 \qquad \qquad Hom_{\mathfrak C^{\mathcal G}(S)}(a,b) = N(\mathcal G \downarrow S)_{a,b}$$

where $$N$$ is the ordinary nerve, $$\downarrow$$ is the ordinary over-category, and the subscript means we restrict to have the correct endpoints. They show (Thm 5.2) that $$\mathfrak C^{\mathcal G}$$ is connected by a zigzag of natural DK equivalences to the usual functor $$\mathfrak C$$ (adjoint to the homotopy coherent nerve $$\mathcal N$$).

Moreover, the map $$\theta: \mathfrak C^{\mathcal G}(X) \times \mathfrak C^{\mathcal G}(Y) \to \mathfrak C^{\mathcal G}(X \times Y)$$ of Proposition 6.2 appears to make $$\mathfrak C^{\mathcal G}$$ into a lax monoidal functor! It is defined on homspaces by taking binary products (a category of gadgets is required to be closed under finite products). Be warned -- I have not checked this carefully! If there's a difficulty, then it probably lies in having to choose binary products in a coherent way...

Assuming this is true, the composite functor $$R = \mathcal N Ex^\infty_\ast \mathfrak C^{\mathcal G}$$ is a composite of lax monoidal functors and so is lax monoidal. It takes values in quasicategories, but unfortunately it is only connected by a zigzag of natural weak equivalences to the identity.

Also, this is all in the setting of the Joyal model structure -- I'm not sure about adapting it to the marked case.

EDIT: Although Dugger and Spivak don't say it, I believe there is a natural transformation $$\mathfrak C \Rightarrow \mathfrak C^{\mathcal G}$$, i.e. $$1 \Rightarrow \mathcal N \mathfrak C^{\mathcal G}$$. It sends a simplex $$\sigma \in X_n$$ to the simplicial functor $$\mathfrak C \Delta^n \to C^{\mathcal G} X$$ which does the obvious thing on objects, sends the "free" 1-morphism $$f_{ij}$$ from $$i$$ to $$j$$ to the composite $$\Delta^{j-i} \to \Delta^n \xrightarrow \sigma X$$ (where the face $$\Delta^{j-i} \subseteq \Delta^n$$ is the one whose long edge is the edge from $$i$$ to $$j$$), sends the homotopy $$f_{jk} f_{ij} \to f_{ik}$$ to the obvious map $$\Delta^{k-j} \vee \Delta^{j-i} \to \Delta^{k-i}$$, and is otherwise determined by the equality between the two composite maps $$\Delta^{l-k} \vee \Delta^{k-j} \vee \Delta^{j-i} \to \Delta^{l-i}$$ and 2-coskeletality (the homspaces of $$\mathfrak C^{\mathcal G} X$$ are nerves of categories and therefore 2-coskeletal).

With this transformation $$1 \Rightarrow \mathcal N \mathfrak C^{\mathcal G} \Rightarrow R = \mathcal N Ex^\infty_\ast \mathfrak C^\mathcal{G}$$ (assuming it's a levelwise weak equivalence), we have a fibrant replacement in the usual sense. However, the natural transformation $$1 \Rightarrow R$$ is not monoidal. (It does appear to be monoidal up to a natural homotopy defined using diagonals, which can probably be made as coherent as you want, but you'd have to be careful about what that means...)

• Thanks for this Tim. It looks very promising, though I also don’t see how to adapt to the marked case. Interesting to know though - thanks for the answer. Oct 17 '20 at 1:34