gotta go

2cat.tex

@@ -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}