Commit beac890d authored by Leonard Guetta's avatar Leonard Guetta
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gotta go

parent 777544a7
......@@ -1168,25 +1168,48 @@ isomorphisms, which means by definition that $P$, $P'$ and $P''$ are \good{}.
Let $P$ be the free $2$\nbd{}category defined as follows:
\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 $\beta: f \Rightarrow 1_A$.
In picture, this gives
\item generating $0$\nbd{}cells: $A$ and $B$,
\item generating $1$\nbd{}cells: $f,g : A \to B$,
\item generating $2$\nbd{}cells: $\alpha,\beta,\gamma : f \to g$.
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]
\qquad \quad \text{ or }
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=E,to=F,Rightarrow,"\gamma",bend left]
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
\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]
and by reasoning as in the proof of Lemma \ref{lemma:squarenerve}, one can
show that the square induced by the nerve
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)
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{}.
\section{The ``Bubble-free'' conjecture}
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