@@ -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 $n,m \geq0$. 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}

We are going to prove that if $n\neq0$ or $m\neq0$, 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 $n\neq0$ or $m\neq0$, 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 $n\neq0$, 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 $n\neq0$, 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 $n\neq0$) implies that $A_{(1,n)}$ has the homotopy type of a point.

Similarly, by considering the cocartesian square

\[

\tau=\begin{tikzcd}{0\to1}\ar[d,"\langle f \rangle"]\\ A \end{tikzcd}

\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 $m\neq0$, then $A_{(m,1)}$ is \good{} and has the homotopy type of a point.

Finally, for $m,n \geq0$, consider the cocartesian square

@@ -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}.