@@ -777,7 +777,7 @@ We can now prove the following proposition, which is the key result of this sect

\]

is not free on a $2$-graph. The reason is that the source (resp. the target) of $\alpha$ is $g \comp_0 f$ (resp. $i \comp_0 h$) which are not generating $1$-cells.

\end{example}

\begin{example}

\begin{example}\label{example:freemonoid}

Let $n \geq1$ and $M$ be a monoid (commutative if $n>1$). The $n$-category $B^nM$ is free if and only if $M$ is a free monoid (free commutative monoid if $n>1$). If so, it has exactly one generating cell of dimension $0$, no generating cells of dimension $0 <k< n$, and the set of generators of the monoid (which is unique) as generating $n$-cells.

\end{example}

\section{Cells of free $\oo$-categories as words}

...

...

@@ -1021,20 +1021,20 @@ We can now prove the expected result.

Let us start with the surjectivity. Let $H : \E^+/{\equiv}\to\D$ be a $(n+1)$-functor. We define a map $\varphi : \Sigma\to D_{n+1}$ as

\begin{align*}

\varphi : \Sigma&\to D_{n+1}\\

\alpha&\mapsto H_{n+1}([\cc_{\alpha}]),

\alpha&\mapsto H_{n+1}([(\cc_{\alpha})]),

\end{align*}

where $[w]$ is the equivalence class under $\equiv$ of an element $w \in\T[\E]$. All we need to show is that

\[

\widehat{\varphi}=H_{n+1}.

\]

Let $z$ be an element of $\T[\E]/{\equiv}$ and let us choose $w \in\T[\E]$ such that $z=[w]$. We proceed to show that $\widehat{\varphi}(z)=H_{n+1}(z)$ by induction on the size of $w$ (Definition \ref{def:sizeword}).

If $|w|=0$, then either $w=\cc_{\alpha}$ for some $\alpha\in\Sigma$ or $w =\ii_x$ for some $x \in C_n$. In the first case, we have

If $|w|=0$, then either $w=(\cc_{\alpha})$ for some $\alpha\in\Sigma$ or $w =(\ii_x)$ for some $x \in C_n$. In the first case, we have

In particular, the previous proposition tells us that $n$-cells in a free $\oo$-category can be represented as a (well formed) words made up of the generating $n$-cells and units on lower dimensionan cells.

In other words, the previous proposition tells us that for an $n$-cellular extension $\E$, the $(n+1)$-cells of $\E^*$ can be represented as equivalence classes of (well formed) words made up of the indeterminates of $\E$ and units on $n$-cells. Note, however, that the equivalences classes are enormous. For example, for an indeterminate $\alpha$, all three following words are in the equivalence class of $(\cc_{\alpha})$:

% Besides, the way we have proven the existence of the smallest congruence on an $n$-magma was abstract and will turn out to be unsufficient

Note that when $n=0$, we have already said that a $0$-cellular extension $\E$ is nothing but a graph and that $\E^*$ is the free category on the graph. In particular, the $1$-cells of $\E^*$ can simply be encoded in strings of composable arrows of the graph, which is a much simpler description that the one obtained from the previous proposition since no equivalence relation is involved. A natural question is whether it would be possible to obtain a similar simple description for $n>0$. While it is certainly possible to have a simpler description of the $(n+1)$-cells of $\E^*$ than the one obtained in the present section (for example by getting rid of some ``pathological'' well formed words and reduce the size of the equivalence classes), it seems not possible to completely avoid the use of an equivalence relation, at least not in a canonical way. Indeed, we have already seen in Example \ref{example:freemonoid} that a free \emph{commutative} monoid $M$ gives rise to a free $n$-category $B^nM$ whose generating $n$-cells are in bijection with the generators of the monoid. And, as soon as there are at least two generators of the monoid, let say $a$ and $b$, there is no canonical way of representing elements of the monoid by a unique word on the generators, for we have

\[

ab = ba.

\]

\todo{À finir ? Il faudrait peut-être dire qu'en rajoutant des hypothèses on doit pouvoir quand même avoir une forme normale. Dans le cas n=2, quand on fait l'hypothèse que le 2-polygraphe est positif, je sais que c'est vrai.}

\end{paragr}

\section{Discrete Conduché $\oo$-functors}

\begin{paragr}

Recall that given a category $\C$ and a class $M$ of arrows of $\C$, we say that an arrow $f : X \to Y$ of $\C$ is \emph{left orthogonal} to $M$ if for every $m : A \to B$ in $M$ and every solid arrow square

\[

\begin{tikzcd}

X \ar[d,"f"']\ar[r]& A \ar[d,"m"]\\

Y \ar[r]& B,

\ar[from=2-1,to=1-2,dashed,"u"]

\end{tikzcd}

\]

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.}

\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:

\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

\begin{align*}

\oo\Cat&\to\Set\\

C &\mapsto C_n \times_{C_k} C_n .

\end{align*}

The $k$-compostion operation $(\shortminus)\comp_k (\shortminus)$