### that is it for today. I have to pick from 6.4.4 and finish the 'nested' diagram of tau

parent d3221095
 ... ... @@ -262,9 +262,9 @@ From the previous proposition, we deduce the following very useful corollary. \displaystyle\coprod_{x \in E}F_x \ar[r] \ar[d] & E \ar[d]&\\ A \ar[d,"\alpha"] \ar[r] & G \ar[d] \ar[r] & B \ar[d,"\delta"]\\ C \ar[r] & H \ar[r] & D \ar[from=1-1,to=2-2,phantom,"\ulcorner" very near end,"\textcircled{\tiny \textbf{1}}" near start, description] \ar[from=2-1,to=3-2,phantom,"\ulcorner" very near end,"\textcircled{\tiny \textbf{2}}", description] \ar[from=2-2,to=3-3,phantom,"\ulcorner" very near end,"\textcircled{\tiny \textbf{3}}", description] \ar[from=1-1,to=2-2,phantom,"\ulcorner" very near end,"\text{\textcircled{\tiny \textbf{1}}}" near start, description] \ar[from=2-1,to=3-2,phantom,"\ulcorner" very near end,"\text{\textcircled{\tiny \textbf{2}}}", description] \ar[from=2-2,to=3-3,phantom,"\ulcorner" very near end,"\text{\textcircled{\tiny \textbf{3}}}", description] \end{tikzcd}. \] What we want to prove is that the image by the functor $L$ of the pasting of squares \textcircled{\tiny \textbf{2}} and \textcircled{\tiny \textbf{3}} is homotopy cocartesian. Since the morphism $G \to B$ is a monomorphism, we deduce from Corollary \ref{cor:hmtpysquaregraph} that the image by the functor $L$ of square \textcircled{\tiny \textbf{3}} is homotopy cocartesian. Hence, all we have to show is that the image by $L$ of square \textcircled{\tiny \textbf{2}} is homotopy cocartesian. \todo{Parler du pasting lemma ?} On the other hand, we know that both morphisms ... ... @@ -689,3 +689,61 @@ It also follows from Lemma \ref{lemma:binervthom} that the bisimplicial nerve in This is an immediate consequence of Proposition \ref{prop:streetvsbisimplicial} and Corollary \ref{cor:bisimplicialsquare}. \end{proof} \section{Zoology of $2$-categories : Basic examples} \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 $n$: $\underbrace{\overbrace{\begin{tikzcd}[column sep=small, ampersand replacement=\&] \&\bullet \ar[r,description,"\cdots",phantom,""{name=A,below}] \& \bullet \ar[rd] \& \\ \bullet \ar[ru] \ar[rd] \& \& \&\bullet \\ \&\bullet \ar[r,description,"\cdots",phantom,""{name=B,above}] \& \bullet \ar[ru]\ar[from=A,to=B,shorten <= 2em, shorten >= 2em,Rightarrow]\end{tikzcd}}^{m}}_{n}.$ More formally, $A_{(m,n)}$ is described in the following way: \begin{itemize}[label=-] \item generating $0$-cells: $A_0,\cdots, A_m$, $B_1,\cdots,B_{n-1}$ \item generating $1$-cells: $\begin{cases} f_{i+1} : A_i \to A_{i+1} & \text{ for } 0\leq i \leq m-1 \\ g_{j+1} : B_j \to B_{j+1} &\text{ for } 1 \leq j \leq n-2 \\g_1 : A_0 \to B_1 & \\g_{n} : B_{n-1} \to A_m & \end{cases}$ \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 $\begin{tikzcd} A \ar[r, bend left=70, "f",""{name=A,below}] \ar[r,bend right=70,"1_A"',""{name=B,above}] & A \ar[from=A,to=B,Rightarrow,"\alpha"] \end{tikzcd},$ and has many non trivial $2$-cells, such as $f\comp_0 \alpha \comp_0 f$. Note that when $m=0$ \emph{and} $n=0$, then the $2$-category $A_{(0,0)}$ is nothing but the commutative monoid $\mathbb{N}$ seen as a $2$-category (\ref{paragr:bubble}) and we have already seen that it is \emph{not} \good{}. We shall refer to this $2$-category as the \emph{bubble}. \end{paragr} \begin{paragr} Recall that for $n\geq 0$, we denote by $\Delta_n$ the linear order ${0 \leq \cdots \leq n}$ seen as a small category. Let $i : \Delta_1 \to \Delta_n$ be the unique functor such that $i(0)=0 \text{ and } i(1)=n.$ \end{paragr} \begin{lemma} For $n\neq0$, the functor $i : \Delta_1 \to \Delta_n$ is a strong deformation retract (\ref{paragr:defrtract}). \end{lemma} \begin{proof} Let $r : \Delta_n \to \Delta_1$ the unique functor such that $r(n)=1\text{ and } r(k)=1 \text{ for } k\neq n.$ By definition we have $r \circ i = 1_{\Delta_1}$. Now, the natural order on $\Delta_n$ induces a natural transformation $\alpha : i\circ r \Rightarrow \mathrm{id}_{\Delta_n},$ and it is straightforward to check that $\alpha \ast i = \mathrm{id}_i$. \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 $\begin{tikzcd} \Delta_1 \ar[r,"i"] \ar[d,"\tau"] & \Delta_n \ar[d] \\ A_{(1,1)} \ar[r] & A_{(1,n)} \ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"] \end{tikzcd},$ where $\sigma : \Delta_1 \to A_{(1,1)}$ is the $2$-functor that points to the the target of the generating $2$-cell of $A_{(1,1)}$ $\tau = \begin{tikzcd} {0 \to 1} \ar[d,"\langle f \rangle"] \\ A \end{tikzcd}$ \end{paragr}
 ... ... @@ -322,7 +322,7 @@ From now on, we will consider that the category $\Psh{\Delta}$ is equipped with \end{enumerate} A deformation retract is a particular case of homotopy equivalence and thus of Thomason weak equivalence. \end{paragr} \begin{lemma} \begin{lemma}\label{lemma:pushoutstrngdefrtract} The pushout of a strong deformation retract is a strong deformation retract. \end{lemma} \begin{proof} ... ...
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