TITLE: Explicit expression of the unstraightening functor
QUESTION [8 upvotes]: Hard as I tried, I couldn't find a proof of Remark 2.2.2.11 in Higher Topos Theory, or prove it myself. It seems to need an explicit formulation for the unstraightening functor, so my question is: is an explicit expression known for the unstraightening? Anyway, is it possible to obtain Remark 2.2.2.11 without having one?

REPLY [4 votes]: This is an old question, but it seems worthwhile to give the full explicit description of unstraightening, for convenient reference.  I'll do this for contravariant unstraightening, and trying to match the notation used in HTT 2.2.1 (i.e., resisting the temptation to replace $\mathcal{C}$ with $\mathcal{C}^{\mathrm{op}}$ everywhere).
Fix functors $\phi\colon \mathfrak{C}[S]\to  \mathcal{C}^{\mathrm{op}}$  and $F\colon \mathcal{C}\to \mathrm{Set}_\Delta$ of simplicial categories.  I will describe the $n$-simplices of $\mathrm{Un}_\phi F$, which is a simplicial set mapping to $S$.
Given a map $s\colon \Delta^n\to S$ let $\phi_s= \phi\circ \mathfrak{C}[s]$, a functor $\mathfrak{C}[\Delta^n]\to \mathcal{C}^{\mathrm{op}}$.  Then there is a bijective correspondence  between $n$-simplices of $\mathrm{Un}_\phi F$, and pairs $(s,g)$, where  $s\colon \Delta^n\to S$ is a map of simplicial sets, and
$$
g\colon D^n\to F\circ \phi_s^{\mathrm{op}}
$$
is a map of simplicial functors $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathrm{Set}_\Delta$.
Here $D^n$ is a particular functor $\mathfrak{C}[\Delta^n]^{\mathrm{op}} \to \mathrm{Set}_\Delta$, defined as follows.  Consider the simplicial category $\mathfrak{C}[(\Delta^n)^\rhd]\approx \mathfrak{C}[\Delta^{n+1}]$, which contains $\mathfrak{C}[\Delta^n]$ as a subcategory.  Then $D^n$ is the functor represented by the object corresponding to the cone point $v$ of $(\Delta^n)^\rhd$.   When you unwind this, you get that $D^n(x)$ is isomorphic to the nerve of a poset:
$$
D^n(x) \approx N\bigl\{ S\subseteq \{x,x+1,\dots, n\} \;\bigm|\; x\in S \bigr\} \approx (\Delta^1)^{n-x}.
$$
(In fact, $D^n$ is nothing other than the straightening of $(\mathrm{id}\colon \Delta^n\to \Delta^n)$ along $\phi=\mathrm{id}\colon \mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathfrak{C}[\Delta^n]^{\mathrm{op}}$.)
In HTT 2.2.2.11, we are interested in $s\colon \Delta^n\to \{s_0\}\to S$ which factor through a single vertex in $S$, which maps under $\phi$ to some object $C$ in $\mathcal{C}$.  In this case $F\circ \phi_s^{\mathrm{op}}$ is a constant simplicial functor $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathrm{Set}_\Delta$ with value $F(C)$.  So natural transformations $D^n\to F\circ \phi_s^{\mathrm{op}}$ are the same as maps of simplicial sets $Q^n\to F(C)$, where $Q^n$ is the enriched left Kan extension of $D^n$ along $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathfrak{C}[\Delta^0]^{\mathrm{op}}=*$.  Unwinding this should yield Lurie's description of $Q^n$, which will be as a quotient of $D^n(0)\approx (\Delta^1)^n$.