### end of the weekend

parent a65e77ed
 ... ... @@ -96,7 +96,8 @@ \newcommand{\PCat}{\mathbf{PCat}} \newcommand{\CellExt}{\mathbf{CellExt}} \newcommand{\Rgrph}{\mathbf{Rgrph}} \newcommand{\Mon}{\mathbf{Mon}} \newcommand{\CMon}{\mathbf{CMon}} \newcommand{\CCat}{\underline{\mathbf{Cat}}} %2-category of small categories \newcommand{\CCAT}{\underline{\mathbf{CAT}}} %2-category of big categories \newcommand{\PPder}{\underline{\mathbf{Pder}}} %2-category of prederivators ... ...
 ... ... @@ -413,6 +413,49 @@ So far, we have not yet seen examples of free $\oo$-categories. In order to do t \item if $n>1$, the $n$-magma $B^nM$ is an $n$-category if and only $M$ is commutative. \end{itemize} \end{lemma} \todo{Parler de Eckmann-Hilton ? Cf fichier source} \iffalse\begin{paragr} For $n=1$, the correspondance $M \mapsto B^nM$ obviously defines a functor $B^n : \Mon \to n\Cat,$ where $\Mon$ is the category of monoids, sand for $n>1$, a functor $B^n : \CMon \to n\Cat,$ where $\CMon$ is the category of commutative monoids. This functor is fully faithful and \end{paragr} \begin{paragr} For an integer $n\geq 1$ and a monoid $M$ (commutative if $n>1$), the $n$-category $B^nM$ has the particular property of having exactly one $k$-cell for each $k1$, this follows from the Eckmann-Hilton argument. The unicity comes from the fact that the functor \end{proof} \fi This construction will turn out to be of great use many times in this dissertation and we now explore a few of its properties. \begin{lemma}\label{lemma:nfunctortomonoid} Let $C$ be an $n$-category with $n\geq 1$ and $M=(M,\ast,1)$ a monoid (commutative if $n>1$). The map ... ... @@ -636,8 +679,8 @@ The following proposition is the key result of this section. It is slightly less \] to be both the identity functors on $\Set$. \end{paragr} \begin{proposition} Let $(\E^{(n)})_{n \geq -1}$ be sequence where: \begin{proposition}\label{prop:freeonpolygraph} Let $(\E^{(n)})_{n \geq -1}$ be a sequence where: \begin{itemize} \item $\E^{(-1)}$ is a $(-1)$-cellular extension, \item for every $n\geq 0$, $\E^{(n)}$ is an cellular extension of the $n$-category $(\E^{(n-1)})^*$. ... ... @@ -665,15 +708,54 @@ The following proposition is the key result of this section. It is slightly less Conversely, any free $\oo$-category and any rigid $\oo$-functor arise this way. \end{proposition} \begin{proof} From Proposition \ref{prop:fromcexttocat} we know that every $(\E^{(n)})^*$ has an $(n+1)$-basis, which is canonically isomorphic to the set of indeterminate of $\E^{(n)}$. Now, since for every $n\geq 0$, each $\E^{(n)}$ is a cellular extension of $(\E^{(n-1)})^*$, it follows that $\sk_{n-1}((\E^{(n)})*)=(\E^{(n-1)})^*$. Hence, by an straightforward induction, each $\E^{(n)}$ is a free $(n+1)$-category and its $k$-basis for $0 \leq k \leq n+1$ is (canonically isomorphic to) the set of indeterminates of $\E^{(k-1)}$. From the first part Proposition \ref{prop:fromcexttocat} we know that every $(\E^{(n)})^*$ has an $(n+1)$-basis, which is canonically isomorphic to the set of indeterminate of $\E^{(n)}$. Besides, since for every $n\geq 0$, $\E^{(n)}$ is a cellular extension of $(\E^{(n-1)})^*$, it follows that $\sk_{n-1}((\E^{(n)})^*)=(\E^{(n-1)})^*$. Hence, by a straightforward induction, each $\E^{(n)}$ is a free $(n+1)$-category and its $k$-basis for $0 \leq k \leq n+1$ is (canonically isomorphic to) the set of indeterminates of $\E^{(k-1)}$. Now let $C :=\colim_{n \geq -1}(\E^{(n)})^*$. Since for every $k\geq 0$, $\sk_k$ preserves colimits and since $\sk_k((\E^{(n)})^*)=(\E^{(k-1)})^*$ for all $0\leq k 0$, $\sigma(x)$ and $\tau(x)$ are $(k-1)$-cells of the free $(k-1)$-category recursively generated by the generating cells of dimension strictly lower than $k$. \end{paragr} \begin{example} Recall that for a graph $G$ (or $1$-graph in the terminology of \ref{paragr:defncat}), the free category on $G$ is the category whose objects are those of $G$ and whose arrows are strings of composable arrows of $G$; the composition being given by concatenation of strings. A $1$-category $C$ is free in the sense of Definition \ref{def:freeoocat} if and only if there exists a graph $G$ such that $C$ is (isomorphic) to the free category on $G$. The generating $1$-cells of $G$ are given by the strings of length $1$. The data of $1$-cellular extension $\E$ is nothing but the data of a graph $G$ (or $1$-graph in the terminology of of \ref{paragr:defncat}), and it is not hard to see that, in that case, $\E^*$ is nothing but the free category on $G$. That is to say, the category whose objects are those of $G$ and whose arrows are strings of composable arrows of $G$; the composition being given by concatenation of strings. Hence, from Proposition \ref{prop:freeonpolygraph}, a ($1$-)category is free in the sense of Definition \ref{def:freeoocat} if and only if it is (isomorphic to) a free category on a graph. \end{example} \begin{example} The notion of free category on a graph is easily generalized to the notion of free $n$-category on a $n$-graph (with $n \in \mathbb{N}\cup\{\infty\}$). As in the previous example, any free $n$-category on an $n$-graph is free in the sense of Definition \ref{def:freeoocat}. \emph{However}, the converse is not true for $n>1$. The point is that in a free $n$-category on an $n$-graph, the source and target of a $k$\nbd-generating cell must be $(k-1)$-generating cells; whereas for a free $n$\nbd-category, they can be any $(k-1)$-cells (not necessarily generating). For example, the free $2$-category pictured as The notion of free category on a graph is easily generalized to the notion of free $n$-category on a $n$-graph (with $n \in \mathbb{N}\cup\{\infty\}$). As in the previous example, any free $n$-category on an $n$-graph is free in the sense of Definition \ref{def:freeoocat}. \emph{However}, the converse is not true for $n>1$. The point is that in a free $n$-category on an $n$-graph, the source and target of a $k$\nbd-generating cell must be $(k-1)$-generating cells; whereas for a free $n$\nbd-category, they can be any $(k-1)$-cells (not necessarily generating). For example, the free $2$-category described as \begin{itemize}[label=-] \item generating $0$-cells : $A,B,C,D$ \item generating $1$-cells : $f:A \to B$, $g : B \to C$, $h : A \to D$, $i : D \to C$ \item generating $2$-cells : $\alpha : i\comp_0 h \Rightarrow g \comp_0 f$, \end{itemize} which can be pictured as $\begin{tikzcd} & B\ar[rd,"g"] & \\ ... ... @@ -684,3 +766,8 @@ The following proposition is the key result of this section. It is slightly less$ 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} Let $n \geq 1$ 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
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