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.

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    $\begingroup$ 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! $\endgroup$ – Tim Campion Oct 1 '20 at 21:41
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    $\begingroup$ Thanks Tim - yes, an analogue of $Ex^\infty$ is what I was hoping for. $\endgroup$ – Michael Ching Oct 1 '20 at 21:59
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    $\begingroup$ 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 $\Delta[1]$. $\endgroup$ – Tim Campion Oct 1 '20 at 22:27
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    $\begingroup$ 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 $Ex^\infty$. $\endgroup$ – Tim Campion Oct 1 '20 at 22:30
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    $\begingroup$ Garner's small object argument provides a sort of "canonical" and "small" fibrant replacement. I wonder if it preserves finite products in this case... $\endgroup$ – Tim Campion 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...)

  • $\begingroup$ 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. $\endgroup$ – Michael Ching Oct 17 '20 at 1:34

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