6
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Indeed, any free category is cofibrant as a (locally discrete) simplicial category. Where do you see a suggestion to the contrary? (Note also that $[2]$ is not isomorphic to $\mathfrak{C}(\Delta[2])$, since the hom-space Hom($0$,$2$) in the latter is isomorphic to $\Delta[1]$.) $\endgroup$
    – Alexander Campbell
    Commented Nov 13, 2019 at 11:01
  • $\begingroup$ Of course, my mistake, I'll edit and shift every number down by 1 $\endgroup$
    – Giulio Lo Monaco
    Commented Nov 13, 2019 at 13:03
  • $\begingroup$ It was really more of a feeling: since in almost every source where the structure of $\mathfrak{C}$ is described the authors seem to care about telling me that it provides a cofibrant replacement functor for the discrete simplicial categories of the form $[n]$, I've been naturally led to believe that we actually need to replace them in order to get something cofibrant. $\endgroup$
    – Giulio Lo Monaco
    Commented Nov 13, 2019 at 13:53
  • 4
    $\begingroup$ As pointed out in Valery's answer, both $[n]$ and $\mathfrak{C}\Delta^n$ are cofibrant simplicial categories. The key point here is that both of these actually describe functors $\Delta\to \mathrm{sCat}$, but only the second of these functors is cofibrant as an object of $Fun(\Delta, \mathrm{sCat})$ (where cofibrant means "Reedy cofibrant"). $\endgroup$
    – Charles Rezk
    Commented Nov 13, 2019 at 16:33

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .