Commit 7fe0121c authored by Leonard Guetta's avatar Leonard Guetta
Browse files

gotta go

parent 28ef12b9
......@@ -1307,7 +1307,13 @@ with $n$ occurences of $f$ and $m$ occurences of $g$.
\item $F(\alpha)=1_{(1,1)}$.
\end{itemize}
This last equation makes sense since $(1,1)=(0,1)+(1,0)=(1,0)+(0,1)$. For any
$1$\nbd{}cell $w$ of $P$ (encoded as a finite words on the alphabet $\{f,g\}$) 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 $P/{\star}$ (\ref{paragr:comma}) to the terminal $2$\nbd{}category
$1$\nbd{}cell $w$ of $P$ (encoded as a finite words on the alphabet $\{f,g\}$)
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 $P/{\star}$
(see \ref{paragr:comma}) to the terminal $2$\nbd{}category
\[
P/{\star} \to \sD_0
\]
......@@ -1315,9 +1321,48 @@ with $n$ occurences of $f$ and $m$ occurences of $g$.
\begin{itemize}[label=-]
\item A $0$\nbd{}cell is a $1$\nbd{}cell of $B^1(\mathbb{N}\times \mathbb{N})$.
\item For $(n,m)$ and $(n',m')$ two $0$\nbd{}cells of $P/{\ast}$, a
$1$\nbd{}cell
\item
\end{itemize}
$1$\nbd{}cell from $(n,m)$ to $(n',m')$ is a $1$\nbd{}cell $w$ of $P$ such
that the triangle
\[
\begin{tikzcd}[column sep=small,row sep=small]
\star \ar[rr,"F(w)"] \ar[rd,"{(n,m)}"']& & \star \ar[dl,"{(n',m')}"]\\
&\star&
\end{tikzcd}
\]
is commutative. More explicitely, if $F(w)=(n'',m'')$, the commutativity of
the previous triangle means
\[
n'+n''=n \text{ and } m'+m''=m.
\]
\item Given two parallel $1$\nbd{}cells $w$ and $w'$ of $P/\star$, a $2$\nbd{}cell of
$P/{\star}$ from $w$ to $w'$ is simply a $2$\nbd{}cell of $P$ from $w$ to
$w'$ seen as $1$\nbd{}cells of $P$.
\end{itemize}
From what we said earlier on the $1$\nbd{}cells and $2$\nbd{}cells of $P$, it
follows easily that for every $0$\nbd{}cell $(n,m)$ of $P/{\star}$, the
category
\[
P/{\star}((m,n),(0,0))
\]
has a terminal object, which is given by
\[
f\cdots fg\cdots g
\]
where $f$ is repeated $n$ times and $g$ is repeated $m$ times. Then, it
follows from \cite[Théorème 5.27 and Remarque 5.28]{ara2020theoreme} that $P/{\star} \to \sD_0$ is a
Thomason equivalence of $2$\nbd{}categories and this proves that $F$ is a
Thomason equivalence. Since $B^1(\mathbb{N}\times\mathbb{N})\simeq
B^1(\mathbb{N})\times B^1(\mathbb{N})$ and $B^1(\mathbb{N})$ has the homotopy
type of $\sS_1$, we conclude that $P$ has the homotopy type of $\sS_1 \times
\sS_1$, i.e.\ the homotopy type of the torus.
Consider now the commutative square
\[
\begin{tikzcd}
a
\end{tikzcd}
\]
\todo{À finir}
\end{paragr}
\section{The ``Bubble-free'' conjecture}
\begin{definition}
......
No preview for this file type
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment