### 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 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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