Commit 2e48a891 authored by Leonard Guetta's avatar Leonard Guetta
Browse files

that it is for today

parent de644776
......@@ -577,7 +577,7 @@ Furthermore, this function satisfies the condition
\sigma'(\varphi(x))=F(\sigma(x)) \text{ and } \tau'(\varphi(x))=F(\tau(x)).
\]
\end{definition}
\begin{paragr}
\begin{paragr}\label{paragr:freecext}
We denote by $n\Cat^{+}$ the category of $n$-cellular extensions and morphisms of $n$-cellular extensions. Every $(n+1)$-category $C$ canonically defines an $n$-cellular extension $(\tau^s_{\leq n }(C),C_{n+1},\src,\trgt)$ where $\src,\trgt : C_{n+1} \to C_n$ are the source and target maps of $C$. This defines a functor
\begin{align*}
U_n : (n+1)\Cat &\to n\Cat^+\\
......@@ -630,7 +630,7 @@ Furthermore, this function satisfies the condition
such that $w_{x}(j(x))=1$ and $w_{x}(j(y))=0$ for any $y \in \Sigma$ with $y\neq x$. In particular, this implies that $j$ is injective.
\end{proof}
\begin{paragr}
In particular, the previous lemma tells us that for any $n$-cellular extension $\E=(C,\Sigma,\sigma,\tau)$, the set of indeterminates $\Sigma$ canonically defines a subset of the $(n+1)$-cells of $\E^*$. As of now, we will consider $\Sigma$ as a subset of $(\E^{*})_{n+1}$.
In particular, the previous lemma tells us that for any $n$-cellular extension $\E=(C,\Sigma,\sigma,\tau)$, the set of indeterminates $\Sigma$ canonically defines a subset of the $(n+1)$-cells of $\E^*$. As of now, we will consider $\Sigma$ as a subset of $(\E^{*})_{n+1}$ and $j$ as the canonical inclusion.
\end{paragr}
We can now prove the following proposition, which is the key result of this section. It is slightly less trivial than it appears.
\begin{proposition}\label{prop:fromcexttocat}
......@@ -638,14 +638,7 @@ We can now prove the following proposition, which is the key result of this sect
\end{proposition}
\begin{proof}
Notice first that since the map $i_{n+1} : \sS^n \to \sD_{n+1}$ is nothing but the canonical inclusion $\sk_{n}(\sD_{n+1}) \to \sk_{n+1}(\sD_{n+1})=\sD_{n+1}$, it follows easily from square \eqref{squarefreecext} and the fact the the skeleton functors preserve colimits that $C$ is canonically isomorphic to $\sk_n(\E^*)$ and that the map $C \to \E^*$ can be identified with the canonical inclusion $\sk_n(\E^*) \to \sk_{n+1}(\E^*)=\E^*$.
Now let $\phi : \coprod_{x \in \Sigma} \sD_{n} \to \E^*$ the bottom map of the cocartesian square of the proposition and consider the map
\begin{align*}
j: \Sigma &\to (\E^*)_{n+1}\\
x &\mapsto \phi_x(e_{n+1}),
\end{align*}
where $e_{n+1}$ is the principal $(n+1)$-cell of $\sD_{n+1}$ (\ref{paragr:defglobe}). Hence, cocartesian square \eqref{squarefreecext} can be identified with
Notice first that since the map $i_{n+1} : \sS^n \to \sD_{n+1}$ is nothing but the canonical inclusion $\sk_{n}(\sD_{n+1}) \to \sk_{n+1}(\sD_{n+1})=\sD_{n+1}$, it follows easily from square \eqref{squarefreecext} and the fact the the skeleton functors preserve colimits that $C$ is canonically isomorphic to $\sk_n(\E^*)$ and that the map $C \to \E^*$ can be identified with the canonical inclusion $\sk_n(\E^*) \to \sk_{n+1}(\E^*)=\E^*$. Hence, cocartesian square \eqref{squarefreecext} can be identified with
\[
\begin{tikzcd}[column sep=huge, row sep=huge]
\displaystyle\coprod_{x \in \Sigma}\sS^n \ar[d,"\displaystyle\coprod_{x \in \Sigma}i_{n+1}"']\ar[r,"{\langle \src(x),\trgt(x)\rangle_{x \in \Sigma}}"] & \sk_{n}(\E^*) \ar[d,hook] \\
......@@ -653,7 +646,7 @@ We can now prove the following proposition, which is the key result of this sect
\ar[from=1-1,to=2-2,very near end,phantom,"\ulcorner"]
\end{tikzcd}
\]
Since from $j : \Sigma \to (\E^{*})_{n+1}$ is injective, we can consider $\Sigma$ as a subset of $(\E^{*})_{n+1})$, and then, by definition, $\E^*$ has $\Sigma$ as an $(n+1)$-basis.
Since we have identified $\Sigma$ to a subset of the $n$-cells of $\E^*$ via $j$, the above cocartesian square means exactly that $\Sigma$ is an $(n+1)$-basis of $\E^*$.
\end{proof}
\begin{paragr}
Let $C$ be an $(n+1)$-category and $E$ be a subset $E \subseteq C_{n+1}$. This defines a cellular extension
......@@ -677,7 +670,7 @@ We can now prove the following proposition, which is the key result of this sect
is an isomorphism.
\end{proposition}
\begin{proof}
It is clear that the canonical $(n+1)$-functor $\E^*_E \to C$ sends $E$, seen as a subset of $(\E^*_E)_{n+1}$, to $E$, seen as a subset of $C_{n+1}$. It follows then from Proposition \ref{prop:fromcexttocat} that if this $(n+1)$-functor is an isomorphism, then $E$ is an $(n+1)$-base of $C$.
It is clear that the canonical $(n+1)$-functor $\E^*_E \to C$ sends $E$, seen as a subset of $(\E^*_E)_{n+1}$, to $E$, seen as a subset of $C_{n+1}$. Hence, it follows from Proposition \ref{prop:fromcexttocat} that if this $(n+1)$-functor is an isomorphism, then $E$ is an $(n+1)$-base of $C$.
Conversely, if $E$ is an $(n+1)$-base of $C$, then we can define an $(n+1)$-functor $C \to \E_E^*$ that sends $E$, seen as a subset of $C_{n+1}$, to $E$, seen as a subset of $(\E^*_E)_{n+1}$ (and which is obviously the identity on cells of dimension strictly lower than $n+1$). Then, the fact that $C$ and $\E^*$ have $E$ as an $(n+1)$-base implies that this $(n+1)$-functor $C \to \E^*$ is the inverse of the canonical one $\E^* \to C$.
\end{proof}
......@@ -728,8 +721,8 @@ We can now prove the following proposition, which is the key result of this sect
\]
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}.
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.
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
......@@ -746,7 +739,7 @@ We can now prove the following proposition, which is the key result of this sect
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 field 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}.
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 field as a \emph{computad} \cite{street1976limits} or \emph{polygraph} \cite{burroni1993higher}; 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:
......@@ -844,7 +837,7 @@ In the following definition, we consider that a binary relation $\R$ on a set $E
\begin{enumerate}[label=(\alph*)]
\item $\R$ is an equivalence relation,
\item if $x\; \R\; x'$ then $x$ and $x'$ are parallel,
\item if $x \;\R\; x'$, $y \;\R\; y'$ and if $x$ and $x'$ are $k$-composable for some $0 \leq k <n$ then
\item if $x \;\R\; y$ and $x' \;\R\; y'$, and if $x$ and $x'$ are $k$-composable for some $0 \leq k <n$ then
\[
x \comp_k x' \; \R \; y\comp_k y'
\]
......@@ -857,7 +850,7 @@ In the following definition, we consider that a binary relation $\R$ on a set $E
\begin{example}
Let $F : X \to Y$ be a morphism of $n$-magmas with $n>1$. The binary relation $\R$ on $X_n$ defined as
\[
x\R y \text{ if } F(x)=F(y)
x \; \R \;y \text{ if } F(x)=F(y)
\]
is a congruence.
\end{example}
......@@ -913,7 +906,7 @@ The following lemma is trivial but nonetheless important. Its immediate proof is
\begin{lemma}
Let $X$ be an $n$-magma with $n>1$ and $\R$ a congruence on $X$. If $\tau_{\leq n}^s(X)$ is an $(n-1)$-category and $\R$ is categorical, then $X/{\R}$ is an $n$-category.
\end{lemma}
We wish now to see on which conditions there exists a ``smallest'' congruence on an $n$-magma that contains a given binary relation $\R$ on the set of $n$-cells.
We wish now to see how to prove the existence of a congruence defined with a condition such as ``the smallest congruence that contains a given binary relation on the $(n+1)$-cells''.
\begin{lemma}\label{lemma:intersectioncongruence}
Let $X$ be an $n$-magma with $n \geq 1$ and $(\R_i)_{i \in I}$ a \emph{non-empty} family of congruences on $X$ (i.e.\ $I$ is not empty). Then, the binary relation
\[
......@@ -937,7 +930,7 @@ We wish now to see on which conditions there exists a ``smallest'' congruence on
\begin{proof}
Let $I$ be the set of congruence $\mathcal{S}$ on $X$ such that for every $(x,y) \in E$, we have $x\; \mathcal{S} \; y$. All we have to prove is that $I$ is not empty, since in that case, we can apply Lemma \ref{lemma:intersectioncongruence} to the binary relation
\[
\R:=\bigcup_{\mathcal{S} \in I}\S,
\R:=\bigcup_{\mathcal{S} \in I}\mathcal{S},
\]
which will obviously be the smallest congruence satisfying the desired condition. To see that $I$ is not empty, it suffices to notice that the binary relation ``being parallel $n$-cells'' is a congruence, which obviously is in $I$.
\end{proof}
......@@ -948,5 +941,5 @@ We wish now to see on which conditions there exists a ``smallest'' congruence on
Each four axioms of Definition \ref{def:categoricalcongruence} says that some pairs of parallel $n$-cells must be equivalent under a congruence $\R$ for it to be categorical. The result follows then from Lemma \ref{lemma:congruencegenerated}.
\end{proof}
\begin{paragr}
Let $\E=(C,\Sigma,\sigma,\tau)$ be an $n$-cellular extension
Let $\E=(C,\Sigma,\sigma,\tau)$ be an $n$-cellular extension and consider the injective map $j : \Sigma \to (\E^*)_{n+1}$ constructed in Paragraph \ref{paragr:freecext}. By induction, we define a map
\end{paragr}
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