Commit ee8c8c77 authored by Leonard Guetta's avatar Leonard Guetta
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skate break

parent 3bd32082
......@@ -688,9 +688,9 @@ It also follows from Lemma \ref{lemma:binervthom} that the bisimplicial nerve in
\begin{proof}
This is an immediate consequence of Proposition \ref{prop:streetvsbisimplicial} and Corollary \ref{cor:bisimplicialsquare}.
\end{proof}
egin{paragr}
\section{Zoology of $2$-categories : Basic examples}
Before embarking on computations of homology and homotopy types of $2$-categories, let us recall the \todo{rappeler le corollaire 4.5.6 ?}
\begin{paragr}
Let $n,m \geq 0$. We denote by $A_{(m,n)}$ the free $2$-category with only one generating $2$-cell whose source is a chain of length $m$ and its target a chain of length $n$:
\[
......@@ -702,7 +702,7 @@ This is an immediate consequence of Proposition \ref{prop:streetvsbisimplicial}
\end{cases}$
\item generating $2$-cell: $ \alpha : f_{m}\circ \cdots \circ f_1 \Rightarrow g_n \circ \cdots \circ g_1$.
\end{itemize}
We are going to prove that if $n\neq 0$ or $m\neq 0$, then $A_{(m,n)}$ is \good{} and has the homotopy type of a point. When $m\neq0$ \emph{and} $n\neq0$, this result is not surprising, but when $n=0$ or $m=0$ (but not both), it is \emph{a priori} less clear what the homotopy type of $A_{(m,n)}$ is and whether it is \good{} or not. For example, $A_{(1,0)}$ can be pictured as follows
Notice that $A_{(1,1)}$ is nothing but $\sD_2$. We are going to prove that if $n\neq 0$ or $m\neq 0$, then $A_{(m,n)}$ is \good{} and has the homotopy type of a point. When $m\neq0$ \emph{and} $n\neq0$, this result is not surprising, but when $n=0$ or $m=0$ (but not both), it is \emph{a priori} less clear what the homotopy type of $A_{(m,n)}$ is and whether it is \good{} or not. For example, $A_{(1,0)}$ can be pictured as follows
\[
\begin{tikzcd}
A \ar[r, bend left=70, "f",""{name=A,below}] \ar[r,bend right=70,"1_A"',""{name=B,above}] & A \ar[from=A,to=B,Rightarrow,"\alpha"]
......@@ -734,7 +734,7 @@ This is an immediate consequence of Proposition \ref{prop:streetvsbisimplicial}
\todo{Est-ce qu'il ne faudra pas dire quelque part qu'une transfo naturelle donne une transfo oplax}.
\end{proof}
\begin{paragr}
In particular, it follows from Lemma \ref{lemma:pushoutstrngdefrtract} that if $n\neq0$, then $i : \Delta_1 \to \Delta_n$ is a co-universal Thomason weak equivalences. Now consider the following cocartesian square
In particular, it follows from Lemma \ref{lemma:pushoutstrngdefrtract} that if $n\neq0$, then $i : \Delta_1 \to \Delta_n$ is a co-universal Thomason weak equivalence. Now consider the following cocartesian square
\[
\begin{tikzcd}
\Delta_1 \ar[r,"i"] \ar[d,"\tau"] & \Delta_n \ar[d] \\
......@@ -742,8 +742,24 @@ This is an immediate consequence of Proposition \ref{prop:streetvsbisimplicial}
\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]
\end{tikzcd},
\]
where $\sigma : \Delta_1 \to A_{(1,1)}$ is the $2$-functor that points to the the target of the generating $2$-cell of $A_{(1,1)}$
where $\tau : \Delta_1 \to A_{(1,1)}$ is the $2$-functor that sends the unique non-trival $1$\nbd-cell of $\Delta_1$ to the target of the generating $2$-cell of $A_{(1,1)}$. If $n\neq 0$, then $i$ is a co-universal Thomason weak equivalence and thus, so is the morphism $A_{(1,1)} \to A_{(1,n)}$. In particular, the square is Thomason homotopy cocartesian. Besides, the morphism $\tau : \Delta_1 \to A_{(1,1)}$ is a folk cofibration and since $\Delta_1$, $\Delta_n$ and $A_{(1,1)}$ are \good{}, it follows from Corollary \ref{cor:usefulcriterion} that if $n\neq 0$, then $A_{(1,n)}$ is \good{}. Note also that, since $\Delta_1$, $\Delta_n$ and $A_{(1,1)}$ have the homotopy type of a point, the fact that the previous square is Thomason homotopy cocartesian (when $n\neq 0$) implies that $A_{(1,n)}$ has the homotopy type of a point.
Similarly, by considering the cocartesian square
\[
\tau = \begin{tikzcd} {0 \to 1} \ar[d,"\langle f \rangle"] \\ A \end{tikzcd}
\begin{tikzcd}
\Delta_1 \ar[r,"i"] \ar[d,"\sigma"] & \Delta_m \ar[d] \\
A_{(1,1)} \ar[r] & A_{(m,1)}
\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]
\end{tikzcd},
\]
where $\sigma : \Delta_1 \to A_{(1,1)}$ is the $2$-functor that sends the unique non trivial $1$\nbd-cell of $\Delta_1$ the source of the generating $2$\nbd-cell of $A_{(1,1)}$, we can prove that if $m\neq 0$, then $A_{(m,1)}$ is \good{} and has the homotopy type of a point.
Finally, for $m,n \geq 0$, consider the cocartesian square
\[
\begin{tikzcd}
A_{(1,1)} \ar[r] \ar[d] & A_{(1,n)} \\
A_{(m,1)} \ar[r] & A_{(m,n)}.
\end{tikzcd}
\]
If
\end{paragr}
......@@ -476,3 +476,23 @@ We shall now proceed to give an abstract criterion to find \good{} $\oo$-categor
Then the $\oo$\nbd-category $X$ is \good{}.
\end{proposition}
\todo{Preuve à écrire. Section à reprendre un peu}
The previous proposition admits the following corollary, which will be of great use in later chapters.
\begin{corollary}\label{cor:usefulcriterion}
Let
\[
\begin{tikzcd}
A \ar[r,"u"] \ar[d,"f"] & B \ar[d,"g"] \\
C \ar[r,"v"] & D
\end{tikzcd}
\]
be a cocartesian square of $\oo\Cat$ such that:
\begin{enumerate}[label=(\alph*)]
\item the $\oo$-categories $A$,$B$ and $C$ are free and \good{},
\item at least one of the morphisms $u : A \to B$ or $f : A \to C$ is a folk cofibration,
\item the square is Thomason homotopy cocartesian.
\end{enumerate}
Then, the $\oo$-category $D$ is \good{}.
\end{corollary}
\begin{proof}
The fact that $A$,$B$ and $C$ are free and one of the morphism $u$ or $f$ is a folk cofibration ensure that the square is folk homotopy cocartesian. The conclusion follows then from Proposition \ref{prop:criteriongoodcat}.
\end{proof}
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