Commit 531ccbaf by Leonard Guetta

### security commit

parent 6f5095e4
 ... ... @@ -758,13 +758,13 @@ It also follows from Lemma \ref{lemma:binervthom} that the bisimplicial nerve in From Proposition \ref{prop:streetvsbisimplicial}, we deduce the proposition below which contains two useful criteria to detect Thomason homotopy cocartesian square of $2\Cat$. \end{paragr} \begin{proposition} \begin{proposition}\label{prop:critverthorThomhmtpysquare} Let \tag{$\ast$}\label{coucou}\begin{tikzcd} A \ar[r,"u"]\ar[d,"f"] & B \ar[d,"g"] \\ C \ar[r,"v"] & D \end{tikzcd} be a square in $2\Cat$ satisfying either of the following conditions: be a square in $2\Cat$ satisfying one of the following conditions: \begin{enumerate}[label=(\alph*)] \item for every $n\geq 0$, the square $... ... @@ -1230,26 +1230,76 @@ isomorphisms, which means by definition that P, P' and P'' are \good{}. \begin{itemize}[label=-] \item generating 0\nbd{}cells: A and B, \item generating 1\nbd{}cells: f,g,h : A \to B, \item generating 2\nbd{}cells: \alpha,\beta:f \to g and \delta,\gamma:g \to h. \item generating 2\nbd{}cells: \alpha,\beta:f \Rightarrow g and \delta,\gamma:g \Rightarrow h. \end{itemize} In picture, this gives: \[ \begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below,pos=8/20},""{name=E,below,pos=12/20}] \ar[r,"g",""{name=B,above,pos=8/20},""{name=C,below,pos=8/20},""{name=F,above,pos=12/20},""{name=G,below,pos=12/20}] \ar[r,bend right=75,"h"',""{name=D,above,pos=8/20},""{name=H,above,pos=12/20}] & B. \ar[from=A,to=B,Rightarrow,"\alpha"',bend right] \ar[from=C,to=D,Rightarrow,"\delta"',bend right] \ar[from=C,to=D,Rightarrow,"\gamma"',bend right] \ar[from=E,to=F,Rightarrow,"\beta",bend left] \ar[from=G,to=H,Rightarrow,"\gamma",bend left] \ar[from=G,to=H,Rightarrow,"\delta",bend left] \end{tikzcd}$ Let us prove that this $2$\nbd{}category is \good{}. Let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $f$, $g$, $\alpha$ and $\beta$ and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and $\delta$. These two $2$\nbd{}categories are copies of $\sS_2$ and we have a cocartesian square Let us prove that this $2$\nbd{}category is \good{}. Let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $f$, $g$, $\alpha$ and $\beta$, let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and $\delta$ and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$ and $g$. These two $2$\nbd{}categories are copies of $\sS_2$ and we have a cocartesian square $\begin{tikzcd} \sD_1 \ar[d,"\langle g \rangle"] \ar[r,"\langle g \rangle"] & P' \ar[d] \\\ P'' \ar[r] & P. \end{tikzcd}$ using the second part of Corollary \end{paragr} Let us prove that this square is Thomason homotopy cocartesian using the second part of Corollary \ref{prop:critverthorThomhmtpysquare}. First, all the morphisms of the previous square are isomorphisms on objects and thus, the image by $V_0$ of the above square is obviously cocartesian. Now, notice that the categories $P(A,B)$, $P'(A,B)$ and $P''(A,B)$ are respectively free on the graphs $\begin{tikzcd} f \ar[r,shift left,"\alpha"] \ar[r, shift right, "\beta"'] & g \ar[r,shift left,"\gamma"'] \ar[r,shift right,"\delta"] & h, \end{tikzcd}$ $\begin{tikzcd} f \ar[r,shift left,"\alpha"] \ar[r, shift right, "\beta"'] & g, \end{tikzcd}$ and $\begin{tikzcd} g \ar[r,shift left,"\gamma"] \ar[r,shift right,"\delta"'] & h. \end{tikzcd}$ Besides, all the categories $P(A,A)$, $P(B,B)$, $P'(A,A)$, $P'(B,B)$, $P''(A,A)$ and $P''(B,B)$ are all isormophic to the terminal category $\sD_0$. \todo{À finir} \end{paragr} \begin{paragr} Let $P$ be the free $2$\nbd{}category defined as follows: \begin{itemize}[label=-] \item generating $0$\nbd{}cell: $A$, \item generating $1$\nbd{}cells: $f , g : A \to A$, \item generating $2$\nbd{}cells: $\alpha : f\comp_0 g \Rightarrow g \comp_0 f$. \end{itemize} In picture, this gives: $\begin{tikzcd} A \ar[r,"f"] \ar[d,"g"'] & A \ar[d,"g"] \\ A \ar[r,"f"'] & A. \ar[from=2-1,to=1-2,Rightarrow,"\alpha"] \end{tikzcd}$ Now consider the $1$\nbd{}category $B^1(\mathbb{N}\times\mathbb{N})$, that is the monoid $\mathbb{N}\times\mathbb{N}$ considered as a category with only one object, and let $F : P \to B^1(\mathbb{N}\times\mathbb{N})$ be the unique $2$\nbd{}functor such that: \begin{itemize}[label=-] \item $F(f)=(1,0)$ and $F(g)=(0,1)$, \item $F(\alpha)=1_{(1,1)}$. \end{itemize} This last equation makes sense since $(1,1)=(0,1)+(1,0)=(1,0)+(0,1)$. Notice that $1$\nbd{}cells of $P$ can be encoded in finite words on the alphabet $\{f,g\}$ (concatenation corresponding to $0$\nbd{}composition) and for a word $w$ on this alphabet such that $f$ appears $n$ times and $g$ appears $m$ times, we have $F(w)=(n,m)$. Let us now prove that $F$ is a Thomason equivalence using a dual of \cite[Corollaire 5.26]{ara2020theoreme} (see Remark 5.20 of op.\ cit.\ ). If we write $\star$ for the only object of $B^1(\mathbb{N}\times\mathbb{N})$, what we need to show is that the canonical $2$\nbd{}functor from $P/{\ast}$ (\ref{paragr:comma}) to the terminal $2$\nbd{}category $P/{\ast} \to \sD_0$ is a Thomason equivalence. The $2$\nbd{}category $P/{\ast}$ is described as follows: \begin{itemize} \item A $0$\nbd{}cells is a $1$\nbd{}cell of $B^1(\mathbb{N}\times \mathbb{N})$, \item \item \end{itemize} \end{paragr} \section{The Bubble-free'' conjecture} \begin{definition} Let $C$ be a $2$\nbd{}category. A \emph{bubble} (in $C$) is a $2$\nbd{}cell ... ...
 ... ... @@ -767,7 +767,7 @@ The nerve functor $N_{\omega} : \omega\Cat \to \Psh{\Delta}$ sends equivalences \begin{example}\label{example:slicecategories} For a small category $A$ (considered as an $\oo$\nbd{}category) and an object $a_0$ of $A$, the category $A/a_0$ in the sense of the previous paragraph is nothing but the usual slice category of $A$ over $a_0$. \end{example} \begin{paragr} \begin{paragr}\label{paragr:comma} Let $u : A \to B$ be a morphism of $\oo\Cat$ and $b_0$ an object of $B$. We define the $\oo$-category $A/b_0$ (also denoted $u\downarrow b_0$) as the following fibred product: \[ \begin{tikzcd} ... ...
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