I'm struggling to understand the explicit description of the cosimplicial simplicial set $Q^{\bullet}$ on page 76 (section 2.2.2) of Lurie's book *Higher Topos Theory*, and would be grateful if someone could elaborate on that a bit.


More concretely, we have that $$Q^n = \operatorname{Map}\_{\mathfrak{C}\left[J^n\right]}(x,y),$$ where $$J^n := (\Delta^n \star \{y\}) \coprod\_{\Delta^n} \{x\},$$ and it is claimed that $Q^n$ can be obtained as a certain quotient of the simplicial set $K\_{\left[n\right]} := N(P\_{\left[n\right]})$, where $P\_{\left[n\right]}$ is the partially ordered set of nonempty subsets of $\left[n\right]$.



explicit description of the cosimplicial simplicial set $Q^{\bullet}$
Here is how I think about this; don't know if it will help.


Start with the straightening construction: this takes a map $f\colon X\to S$ of simplicial sets to a simplicial functor $\def\St{\mathrm{St}}\def\op{\mathrm{op}}\St(f)\colon \mathfrak{C}(S)^{\op}\to s\mathrm{Set}$. 


The key example is when $f=\mathrm{id}\colon \Delta^n\to \Delta^n$. This is a simplicial functor $\St(\mathrm{id})\colon\mathfrak{C}(\Delta^n)^\op\to s\mathrm{Set}$, which I'll call $\mathfrak{D}^n$. Recall that $\mathfrak{C}(\Delta^n)$ has mapping spaces $$\mathrm{map}(x,y) =N\bigl\{ \{x,y\}\subseteq S\subseteq [x,y] \bigr\},$$ i.e., nerves of posets of subsets of the interval $[x,y]$ containing the endpoints. The functor $\mathfrak{D}^n$ is defined similarly:
$$\mathfrak{D}^n(x) = N\bigl\{\{x\}\subseteq S\subseteq [x,n]\bigr\}.$$
The action $\mathfrak{D}^n(y)\times \mathrm{map}(x,y)\to \mathfrak{D}^n(x)$ is defined by union.


In general, we have
$$ \St(\Delta^n\xrightarrow{f} S) \approx \St(\Delta^n\xrightarrow{\mathrm{id}}\Delta^n)\otimes\_{\mathfrak{C}(\Delta^n)} \mathfrak{C}(S),$$
where the tensor product really means left Kan extension along the (simplicial) functor $\mathfrak{C}(f)\colon \mathfrak{C}(\Delta^n)\to \mathfrak{C}(S)$ (or, more precisely, along its opposite).


The simplicial set 
$$Q^n= \St(\Delta^n\to \Delta^0) \approx \mathfrak{D}^n\otimes\_{\mathfrak{C}(\Delta^n)} 1,$$
where $1=\mathfrak{C}(\Delta^0)$ is the terminal simplicial category. In other words, the simplicial set $Q^n$ is the colimit of the (simplicial) functor $\mathfrak{D}^n\colon \mathfrak{C}(\Delta^n)^\op\to s\mathrm{Set}$. This is just something you have to compute: it's a coequalizer of
$$\small
\coprod\_{x=0,\dots,n} N\bigl\{ \{x\}\subseteq S\subseteq [x,n] \bigr\} \leftleftarrows \coprod\_{0\leq x\leq y\leq n} N\bigl\{ \{y\}\subseteq S\subseteq [y,n]\bigr\} \times N\bigl\{ \{x,y\}\subseteq T \subseteq [x,y] \bigr\}.
$$


This turns out to be isomorphic to Lurie's quotient of $K\_{[n]}$. Explicitly, a $k$-simplex in Lurie's quotient is a sequence $\varnothing \subsetneq S\_0\subseteq \cdots\subseteq S\_k\subseteq [n]$, subject to the relation that, for each $x\in S\_0$, we identify it with the $k$-simplex $\varnothing \subsetneq S\_0\cap[x,n]\subseteq \cdots \subseteq S\_k\cap [x,n]\subseteq [n]$.


