@@ -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 $F'$.