### gotta go

parent 777544a7
 ... ... @@ -1168,25 +1168,48 @@ isomorphisms, which means by definition that $P$, $P'$ and $P''$ are \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: $f : A \to A$, \item generating $2$\nbd{}cells: $\alpha : f \Rightarrow 1_A$ and $\beta: f \Rightarrow 1_A$. \end{itemize} In picture, this gives \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 $\begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below}]\ar[r,bend right=75,"f"',""{name=B,above}] \ar[r,"1_A" pos=1/3,""{name=C,above},""{name=D,below}]& A \ar[from=A,to=C,Rightarrow,"\alpha"] \ar[from=B,to=D,"\beta" pos=9/20,Rightarrow] \end{tikzcd} \qquad \quad \text{ or } \begin{tikzcd} A. \ar[loop,in=30,out=150,distance=3cm,"f",""{name=A,below}] \ar[from=A,to=1-1,bend right,Rightarrow,"\alpha"'] \ar[from=A,to=1-1,bend left,Rightarrow,"\beta"] \begin{tikzcd}[column sep=huge] A \ar[r,bend left=75,"f",""{name=A,below,pos=8/20},""{name=C,below,pos=1/2},""{name=E,below,pos=12/20}] \ar[r,bend right=75,"g"',""{name=B,above,pos=8/20},""{name=D,above,pos=1/2},""{name=F,above,pos=12/20}]& B. \ar[from=A,to=B,Rightarrow,"\alpha"',bend right] \ar[from=C,to=D,Rightarrow,"\beta"] \ar[from=E,to=F,Rightarrow,"\gamma",bend left] \end{tikzcd}$ 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 $\begin{tikzcd} N_{\oo}(\sD_1) \ar[r,"\langle \beta \rangle"] \ar[d,"\langle \beta \rangle"] & N_{\oo}(P') \ar[d] \\ N_{\oo}(P'') \ar[r] & N_{\oo}(P) \end{tikzcd}$ We shall now prove that $P$ is \good{} using the techniques introduced in is also cocartesian. This proves that square \ref{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} \section{The Bubble-free'' conjecture} \begin{definition} ... ...
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