TITLE: Quasicategories for non-simplicial model categories
QUESTION [14 upvotes]: If I have a simplicially enriched model category, then I can take the coherent nerve of the full subcategory of bifibrant opjects to obtain a quasicategory.  If I have a model category that is not simplicially enriched, then I could take the hammock localisation first and then take the coherent nerve.  I have no technical objection to this, but I find it aesthetically displeasing.  I would prefer to have a construction that accepts an arbitrary model category $\mathcal{C}$ and produces a quasicategory $N'(\mathcal{C})$ with the same homotopy theory in a single step.  Ideally, all objects of $\mathcal{C}$ would appear as $0$-simplices in $N'(\mathcal{C})$, not just the bifibrant ones.  Does such a thing exist in the literature?
One idea (in the spirit of the thesis of James Cranch) is as follows.  Given a left proper model category $\mathcal{C}$, an $n$-simplex of $N'(\mathcal{C})$ would be an $n$-fold cospan diagram consisting of objects $X_{ij}$ for $0\leq i\leq j\leq n$.  There would be cofibrations  $X_{ij}\to X_{i,j+1}$ and weak equivalences $X_{i+1,j}\to X_{ij}$ and the resulting squares would be cocartesian.  I have not made any serious attempt to check whether this works.
Alternatively, it may be that an answer is essentially contained in the paper of Barwick and Kan on Relative Categories, but I have not properly digested that.

REPLY [7 votes]: You can consider the marked simplicial set $(N(\mathcal{C}),\mathcal{W})$, where $N$ is the usual nerve functor and $\mathcal{W}$ is the class of weak equivalences in your model category $\mathcal{C}$. Then take $N'(\mathcal{C})$ to be the underlying simplicial set of any fibrant replacement of $(N(\mathcal{C}),\mathcal{W})$ in the Cartesian model structure. This paper by Hinich shows that this procedure gives the "correct" $\infty$-category associated to $\mathcal{C}$. Perhaps some unwinding of definitions will give you a more explicit construction.