Commit ee8c8c77 by Leonard Guetta

### 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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