@@ -1077,7 +1077,7 @@ We can now prove the expected result.
\]
there exists a \emph{unique} arrow $u : Y \to A$ making the whole diagram commutes.
\end{paragr}
\emph{In the following paragraph, we use freely the notations from \ref{paragr:defglobe} and advicee the reader to refer to it if needed.}
\emph{In the two following paragraphs, we use freely the notations from \ref{paragr:defglobe} and advice the reader to refer to it if needed.}
\begin{paragr}
For all $0\leq k < n$, we denote by $\sD_n \coprod_{\sD_k}\sD_n$ the $n$-category defined as the following amalgamated sum:
\[
...
...
@@ -1086,10 +1086,85 @@ We can now prove the expected result.
\sD_n \ar[r]&\displaystyle\sD_n \coprod_{\sD_k}\sD_n. \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end]
\end{tikzcd}
\]
By definition, an arrow $\sD_n \coprod_{\sD_k}\sD_n \to C$ for an $\oo$-category$C$ amounts to the same data as a pair of $k$-composable $n$-cells of $X$. Hence, $\sD_n \sD_k \sD_n$ represents the functor
By definition, an arrow $\sD_n \coprod_{\sD_k}\sD_n \to C$, where $C$ is an $\oo$-category, amounts to the same data as a pair of $k$-composable $n$-cells of $C$. Hence, $\sD_n \coprod_{\sD_k}\sD_n$ represents the functor
\begin{align*}
\oo\Cat&\to\Set\\
C &\mapsto C_n \times_{C_k} C_n.
C &\mapsto C_n \times_{C_k} C_n.
\end{align*}
The $k$-compostion operation $(\shortminus)\comp_k (\shortminus)$
Given a pair of $k$-composable $n$-cells $(x,y)$ of an $\oo$-category $C$, we write
\[
\langle x,y \rangle : \sD_n \coprod_{\sD_k}\sD_n \to C
\]
for the associated $\oo$-functor.
By the Yoneda Lemma, the $k$-compostion operation
\[(\shortminus)\comp_k (\shortminus) : C_n \times_{C_k} C_n \to C_n,\] which is obviously natural in $C$, yields an $n$-functor
\[
\nabla^n_k : \sD_n \to\sD_n \coprod_{\sD_k}\sD_n.
\]
The $n$-cell $\nabla^n_k$ is nothing the the $k$-composition of the two non-trivial $n$-cells of $\sD_n \coprod_{\sD_k}\sD_n$. In more practical terms, a commutative triangle
\[
\begin{tikzcd}
\sD_n \ar[r,"\langle x \rangle"]\ar[d,"\nabla^n_k"']& C \\
Let $C$ be an $\oo$-category. More conceptually, $\kappa^n_k$ is induced from the Yoneda Lemma by the unit map
\[\1^{n}_{(\shortminus)} : C_k \to C_n,\]
which is obviously natural in $C$. In practical terms, a commutative triangle
\[
\begin{tikzcd}
\sD_n \ar[r,"\langle x \rangle"]\ar[d,"\kappa^n_k"']& C \\
\sD_k \ar[ru,"{\langle y \rangle}"']&
\end{tikzcd}
\]
means exactly that $x=1^{n}_y$.
\end{paragr}
\begin{definition}\label{def:conduche}
An $\oo$-functor $F : C \to D$ is an \emph{discrete $\oo$-functor} if it is right orthogonal to the $\oo$-functors
\[
\kappa^n_k : \sD_n \to\sD_k
\]
and
\[
\nabla^n_k : \sD_n \to\sD_n \coprod_{\sD_k}\sD_n
\]
for all $k,n \in\mathbb{N}$ such that $0\leq k < n$.
\end{definition}
\begin{paragr}
Let us unfold Definition \ref{def:conduche}. The right orthogonality to $\kappa^n_k$ means that for any $n$-cell $x$ of $C$ and any $k$-cell $y$ of $D$ such that
\[
F(x)=\1^{n}_y,
\]
there exists a unique\footnote{The map $z \mapsto1^n_k(z)$ being injective, the uniqueness actually comes for free.}$k$-cell $z$ of $C$ such that
\[
x=\1^n_z \text{ and } F(z)=y.
\]
Similarly, the right orthogonality to $\nabla^n_k$ means that for any $k$-cell $x$ of $C$, if
\[
f(x)= y' \comp_k y''
\]
with $(y',y'')$ a pair of $k$-composable $n$-cells of $D$, then there exists a \emph{unique} pair $(x',x'')$ of $k$-composable $n$-cells of $C$ such that
\begin{enumerate}
\item$x=x'\comp_k x''$,
\item$F(x')=y'$ and $F(x'')=y''$.
\end{enumerate}
\end{paragr}
Since the class of discrete $\oo$-functors is a right orthogonal class of arrows, it has many good properties. For later reference, we recall one of them in the following proposition, whose proof is ommited.
\begin{proposition}\label{prop:pullbackconduche}
The class of Discrete Conduché $\oo$-functors is stable by pullback. This means that for any cartesian square in $\oo\Cat$
\[
\begin{tikzcd}
C' \ar[d,"F'"']\ar[r]& C \ar[d,"F"]\\
D' \ar[r]& D\ar[from=1-1,to=2-2,"\lrcorner",very near start,phantom]
\end{tikzcd}
\]
if $F$ is a discrete Conduché $\oo$-functor, then so is $F'$.
