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.

2$\begingroup$ 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

2$\begingroup$ Thanks Tim  yes, an analogue of $Ex^\infty$ is what I was hoping for. $\endgroup$ – Michael Ching Oct 1 '20 at 21:59

1$\begingroup$ 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 0simplices, 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 1simplex 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

1$\begingroup$ 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

1$\begingroup$ 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 overcategory, 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" 1morphism $f_{ij}$ from $i$ to $j$ to the composite $\Delta^{ji} \to \Delta^n \xrightarrow \sigma X$ (where the face $\Delta^{ji} \subseteq \Delta^n$ 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 $\Delta^{kj} \vee \Delta^{ji} \to \Delta^{ki}$, and is otherwise determined by the equality between the two composite maps $\Delta^{lk} \vee \Delta^{kj} \vee \Delta^{ji} \to \Delta^{li}$ and 2coskeletality (the homspaces of $\mathfrak C^{\mathcal G} X$ are nerves of categories and therefore 2coskeletal).
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