### security commit

parent 32d2ff06
 ... ... @@ -1165,51 +1165,90 @@ isomorphisms, which means by definition that $P$, $P'$ and $P''$ are \good{}. \item $G(\gamma)=\alpha\comp_1\beta$. \end{itemize} Notice that we have $F\circ G = \mathrm{id}_{P'}$, which means that $P'$ is a retract of $P$. In particular, $\sH^{\sing}(P)$ is a retract of $\sH^{\sing}(P')$ and since $P'$ has the homotopy type of $K(\mathbb{Z},2)$-space (\ref{paragr:bubble}), this proves that $P$ have non-trivial singular homology groups in all even dimension. But since it is a free $2$\nbd{}category, all its polygraphic homology groups are trivial above dimension $2$, which means that $P$ is \emph{not} \good{}. \end{paragr} \begin{paragr} Let $P$ be the free $2$\nbd{}category defined as follows: \begin{itemize}[label=-] \item generating $0$\nbd{}cells: $A$ and $B$, \item generating $1$\nbd{}cells: $f,g : A \to B$, \item generating $2$\nbd{}cells: $\alpha,\beta,\gamma : f \to g$. \end{itemize} In picture, this gives $\begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below,pos=8/20},""{name=C,below,pos=1/2},""{name=E,below,pos=12/20}] \ar[r,bend right=75,"g"',""{name=B,above,pos=8/20},""{name=D,above,pos=1/2},""{name=F,above,pos=12/20}]& B. \ar[from=A,to=B,Rightarrow,"\alpha"',bend right] \ar[from=C,to=D,Rightarrow,"\beta"] \ar[from=E,to=F,Rightarrow,"\gamma",bend left] \end{tikzcd}$ Now let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,$B$, $\alpha$ and $beta$, and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,$B$,$\beta$ and $\gamma$. These $2$\nbd{}categories are simply copies of $\sS_2$. Notice that we have a cocartesian square \begin{equation}\label{square:bouquet} \begin{tikzcd} \sD_1 \ar[r,"\langle \beta \rangle"] \ar[d,"\langle \beta \rangle"] & P' \ar[d] \\ P'' \ar[r] & P, \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \end{equation} and by reasoning as in the proof of Lemma \ref{lemma:squarenerve}, one can show that the square induced by the nerve $\begin{tikzcd} N_{\oo}(\sD_1) \ar[r,"\langle \beta \rangle"] \ar[d,"\langle \beta \rangle"] & N_{\oo}(P') \ar[d] \\ N_{\oo}(P'') \ar[r] & N_{\oo}(P) \end{tikzcd}$ is also cocartesian. This proves that square \eqref{square:bouquet} is Thomason homotopy cocartesian \todo{détailler?} and in particular that $P$ has the homotopy type of a bouqet of two $2$\nbd{}spheres. Since $\sD_1$, $P'$ and $P''$ are free and \good{} and since $\langle \beta \rangle : \sD_1 \to P'$ and $\langle \beta \rangle : \sD_1 \to P'$, this proves that $P$ is \good{}. \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{}cell: $f,g: A \to A$, \item generating $2$\nbd{}cell: $\alpha,\beta : f \to g$. \end{itemize} In picture, this gives: %% $%% \begin{tikzcd} %% \end{tikzcd} %%$ \end{paragr} \begin{paragr} Let $P$ be the free $2$\nbd{}category defined as follows: \begin{itemize}[label=-] \item generating $0$\nbd{}cells: $A$ and $B$, \item generating $1$\nbd{}cells: $f,g : A \to B$, \item generating $2$\nbd{}cells: $\alpha,\beta,\gamma : f \to g$. \end{itemize} In picture, this gives $\begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below,pos=8/20},""{name=C,below,pos=1/2},""{name=E,below,pos=12/20}] \ar[r,bend right=75,"g"',""{name=B,above,pos=8/20},""{name=D,above,pos=1/2},""{name=F,above,pos=12/20}]& B. \ar[from=A,to=B,Rightarrow,"\alpha"',bend right] \ar[from=C,to=D,Rightarrow,"\beta"] \ar[from=E,to=F,Rightarrow,"\gamma",bend left] \end{tikzcd}$ Now let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,$B$, $\alpha$ and $beta$, and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,$B$,$\beta$ and $\gamma$. These $2$\nbd{}categories are simply copies of $\sS_2$. Notice that we have a cocartesian square \begin{equation}\label{square:bouquet} \begin{tikzcd} \sD_1 \ar[r,"\langle \beta \rangle"] \ar[d,"\langle \beta \rangle"] & P' \ar[d] \\ P'' \ar[r] & P, \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \end{equation} and by reasoning as in the proof of Lemma \ref{lemma:squarenerve}, one can show that the square induced by the nerve $\begin{tikzcd} N_{\oo}(\sD_1) \ar[r,"\langle \beta \rangle"] \ar[d,"\langle \beta \rangle"] & N_{\oo}(P') \ar[d] \\ N_{\oo}(P'') \ar[r] & N_{\oo}(P) \end{tikzcd}$ is also cocartesian. Since $\langle \beta \rangle : \sD_1 \to P'$ and $\langle \beta \rangle : \sD_1 \to P''$ are monomorphisms and $N_{\oo}$ preserves monomorphisms, it follows from Lemma \ref{lemma:hmtpycocartesianreedy} that square \eqref{square:bouquet} is Thomason homotopy cocartesian and in particular that $P$ has the homotopy type of a bouquet of two $2$\nbd{}spheres. Since $\sD_1$, $P'$ and $P''$ are free and \good{} and since $\langle \beta \rangle : \sD_1 \to P'$ and $\langle \beta \rangle : \sD_1 \to P'$, this also proves that $P$ is \good{} (see \ref{paragr:criterion2cat}). \end{paragr} \begin{paragr} Let $P$ be the free $2$\nbd{}category defined as follows: \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 $\delta,\gamma:g \to h$. \end{itemize} In picture, this gives: $\begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below,pos=8/20},""{name=E,below,pos=12/20}] \ar[r,"g",""{name=B,above,pos=8/20},""{name=C,below,pos=8/20},""{name=F,above,pos=12/20},""{name=G,below,pos=12/20}] \ar[r,bend right=75,"h"',""{name=D,above,pos=8/20},""{name=H,above,pos=12/20}] & B. \ar[from=A,to=B,Rightarrow,"\alpha"',bend right] \ar[from=C,to=D,Rightarrow,"\delta"',bend right] \ar[from=E,to=F,Rightarrow,"\beta",bend left] \ar[from=G,to=H,Rightarrow,"\gamma",bend left] \end{tikzcd}$ 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 $\begin{tikzcd} \end{tikzcd}$ using the second part of Corollary \end{paragr} \section{The Bubble-free'' conjecture} \begin{definition} ... ...
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