TITLE: Cofibrant simplicial categories
QUESTION [5 upvotes]: Reading about simplicial categories, and in particular the model structure in sCat, I found various sources, among which I can mention this one or section 16.2 in Riehl's book “Categorical homotopy theory”, which seem to imply that the trivial simplicial structures on the posets $[n]$ are not cofibrant. Now, this leaves me a bit puzzled, in that I can exhibit a very explicit description of the initial maps of these objects as compositions of pushouts of generating cofibrations. For instance, $[2]$ is the smallest of these simplicial categories which is claimed not to be cofibrant. $[1]$ is already known to be cofibrant (for one thing, it is isomorphic to $\mathfrak{C}\Delta^1$). Now consider the following two diagrams:
$\require{AMScd}$
\begin{CD}
   \emptyset @>>> [1]\\
    @VVV @VVV\\
    [0] @>>> [1] \coprod [0]
\end{CD}
$\require{AMScd}$
\begin{CD}
   [1]_{\emptyset} @>f>> [1] \coprod [0]\\
    @VVV @VVV\\
   [1]_{\Delta^0} @>>> [2]
\end{CD}
where $f$ selects the terminal point of $[1]$ and the adjoined point. It is clear that the first diagram is a pushout. It really seems to me that the second one is, too, both by explicit construction of a pushout in $\textbf{sCat}$ and by checking the universal property. The left vertical maps in both diagrams are generating cofibrations. I conclude that the inclusion $[1] \to [2]$ is also a cofibration. Is there a mistake that I don't see? What goes wrong in this argument?

REPLY [2 votes]: Simplicial categories $[n]$ are indeed cofibrant. The Simplicial category $\mathfrak{C}\Delta^n$ is also cofibrant and weakly equivalent to $[n]$, so the former is a cofibrant replacement of the latter. Note that $\mathfrak{C}\Delta^n$ is isomorphic to $[n]$ only when $n \leq 1$ (so this is not true for $n = 2$). The point is that simplicial functors $\mathfrak{C}\Delta^n \to C$ carry more information than $[n] \to C$, so the homotopy coherent nerve of $C$ remembers enough information about $C$ as opposed to the ordinary nerve.