Commit 56004c82 authored by Leonard Guetta's avatar Leonard Guetta
Browse files

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 $k<n$ (which all are the iterated units on the only object of $B^nM$). As it happens, this property is sufficient enough to characterize $n$-categories isomorphic to a $B^nM$ for some monoid $M$. This result will be a consequence of the famous ``Eckmann-Hilton argument''.
\end{paragr}
\begin{lemma}[Eckmann-Hilton argument]
Let $(M,\times,1)$ and $(M,\ast,1)$ two monoid structures on the same set $M$ and with the same unit. Suppose that for every $a,b,c,d \in M$, the exchange rule holds:
\[
(a \times b) \ast (b \times c) = (a \ast c) \times (b \ast d).
\]
Then, the two operations $\ast$ and $\times$ are equal, and furthermore they are commutative.
\end{lemma}
\begin{proof}
This follows immediatly from the sequence of equalities:
\begin{align*}
a\times b &= (a\ast 1) \times (1 \ast b)\\
&= (a \times 1) \ast ( 1 \times b)\\
&= a \ast b \\
&= (1 \times a ) \ast (b \times 1)\\
&=(1 \ast b) \times (a \ast 1)\\
&= b \times a.\qedhere
\end{align*}
\end{proof}
\begin{remark}
It is not necessary to suppose in the previous lemma that the two units of the two monoid structures coincide, as it can be shown to be also a consequence of the exchange rule. The version we gave is nevertheless sufficient for our purpose.
\end{remark}
\begin{proposition}
Let $n \geq 0$ and $C$ be a $n$-category that has exactly one $k$-cell for each $0 \leq k < n$. Then, there exists a monoid $M$ such that $C \simeq B^nM$. This monoid is unique up to a (non necessarily unique) isomorphism.
\end{proposition}
\begin{proof}
For $n=1$, the existence is obvious since the data of $C$ itself is exactly that of a monoid. For $n>1$, 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 $0\leq k <n$, we have that
\[
\sk_k(C)=(\E^{(k-1)})^*
\]
for every $k \geq 0$. Altogether, this proves that $C$ is free and its $k$-basis is the set of indeterminates of $\E^{k-1}$ for every $k \geq 0$.
The fact that a sequence of morphism of cellular extension that satisfy the hypothesis given in the statement of the proposition induces a rigid $\oo$-functor is proven in a similar fashion using, this time, the second part of Proposition \ref{prop:fromcexttocat}.
For the converse part, notice that a free $\oo$-category $C$, whose basis is denoted by $(\Sigma_k)_{k \in \mathbb{N}}$, induces a sequence of cellular extensions:
\[
\E_C^{(-1)}:=\Sigma_0
\]
and
\[
\E_C^{(n)}:=(\sk_n(C),\Sigma_{n+1},\src,\trgt) \text{ for } n\geq 0.
\]
It follows from Proposition \ref{prop:criterionnbasis} that $\sk_n(C)\simeq (\E_C^{(n)})^*$ and, then, from Lemma \ref{lemma:filtration} that $C \simeq \colim_{n \geq -1}\E_C^{(n)}$.
Finally, notice that the construction $C \mapsto (\E_C^{(n)})_{n \geq -1}$ described above is obviously functorial with respect to rigid $\oo$-functors and the isomorphism $\sk_n(C)\simeq (\E_C^{(n)})^*$ is natural with respect to rigid $\oo$-functors. Since the statement of Lemma \ref{lemma:filtration} is also natural in $C$, this easily implies that any rigid $\oo$-functor arises as the colimit of sequence of morphisms of cellular extensions as described in the statement of the proposition.
\end{proof}
\begin{remark}
The previous proposition admits an obvious truncated version for free $n$-categories with $n$ finite. In that case, we only need a finite sequence $(\E^{(k)}))_{ -1 \leq k \leq n-1}$ of cellular extensions.
\end{remark}
\begin{remark}
The data of a sequence $(\E^{(n)})_{n \geq -1}$ as in Proposition \ref{prop:freeonpolygraph} is commonly referred to in the litterature of the filed as a \emph{computad} \cite{street1976limits} or \emph{polygraph} \cite{burroni1993higher}; and consequently a $\oo$-category which is free in the sense of definition \ref{def:freeoocat} is sometimes referred to as \emph{free on a computad}-or-\emph{polygraph} in the litterature. However, the underlying polygraph of a free $\oo$-category is uniquely determined by the free $\oo$-category itself (a straightforward consequence of Proposition \ref{prop:uniquebasis}), and this is why we chose the shorter terminology \emph{free $\oo$-category}.
\end{remark}
\begin{paragr}
Concretely, Proposition \ref{prop:freeonpolygraph} gives us a recipe to construct free $\oo$-categories. It suffices to give a formal list of generating cells of the form:
\begin{itemize}[label=-]
\item generating $0$-cells : $x^0, y^0, \dots$
\item generating $1$-cells : $x^1 : \sigma(x^1) \to \tau(x^1): y^1 : \sigma(y^1) \to \tau(y^1), \dots$
\item generating $2$-cells : $x^2 : \sigma(x^2) \to \tau(x^2): y^2 : \sigma(y^2) \to \tau(y^2), \dots$
\item $\dots$,
\end{itemize}
where for a generating $k$-cell $x$ with $k>0$, $\sigma(x)$ 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 $0 <k< n$, 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}
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