### security commit

parent 25e266fb
 \chapter{Homollogy of ontractible $\oo$\nbd-categories its consequences} \chapter{Homology of contractible $\omega$-categories its consequences} \section{Contractible $\oo$-categories} Recall that $\sD_0$ is the terminal object of $\oo\Cat$ that for any $oo$\nbd-category, we denote by $p_X : X \to \sD_0$ the canonical morphism. \begin{definition} ... ...
 ... ... @@ -8,6 +8,8 @@ \maketitle \tableofcontents \include{omegacat} \include{homtheo} \include{hmtpy} ... ...
 ... ... @@ -1077,7 +1077,7 @@ We can now prove the expected result. \] there exists a \emph{unique} arrow $u : Y \to A$ making the whole diagram commutes. \end{paragr} \emph{In the following paragraph, we use freely the notations from \ref{paragr:defglobe} and advicee the reader to refer to it if needed.} \emph{In the two following paragraphs, we use freely the notations from \ref{paragr:defglobe} and advice the reader to refer to it if needed.} \begin{paragr} For all $0 \leq k < n$, we denote by $\sD_n \coprod_{\sD_k} \sD_n$ the $n$-category defined as the following amalgamated sum: $... ... @@ -1086,10 +1086,85 @@ We can now prove the expected result. \sD_n \ar[r] & \displaystyle\sD_n \coprod_{\sD_k} \sD_n. \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd}$ By definition, an arrow $\sD_n \coprod_{\sD_k} \sD_n \to C$ for an $\oo$-category $C$ amounts to the same data as a pair of $k$-composable $n$-cells of $X$. Hence, $\sD_n \sD_k \sD_n$ represents the functor By definition, an arrow $\sD_n \coprod_{\sD_k} \sD_n \to C$, where $C$ is an $\oo$-category, amounts to the same data as a pair of $k$-composable $n$-cells of $C$. Hence, $\sD_n \coprod_{\sD_k} \sD_n$ represents the functor \begin{align*} \oo\Cat &\to \Set \\ C &\mapsto C_n \times_{C_k} C_n . C &\mapsto C_n \times_{C_k} C_n. \end{align*} The $k$-compostion operation $(\shortminus) \comp_k (\shortminus)$ Given a pair of $k$-composable $n$-cells $(x,y)$ of an $\oo$-category $C$, we write $\langle x,y \rangle : \sD_n \coprod_{\sD_k} \sD_n \to C$ for the associated $\oo$-functor. By the Yoneda Lemma, the $k$-compostion operation $(\shortminus) \comp_k (\shortminus) : C_n \times_{C_k} C_n \to C_n,$ which is obviously natural in $C$, yields an $n$-functor $\nabla^n_k : \sD_n \to \sD_n \coprod_{\sD_k} \sD_n.$ The $n$-cell $\nabla^n_k$ is nothing the the $k$-composition of the two non-trivial $n$-cells of $\sD_n \coprod_{\sD_k} \sD_n$. In more practical terms, a commutative triangle $\begin{tikzcd} \sD_n \ar[r,"\langle x \rangle"] \ar[d,"\nabla^n_k"'] & C \\ \sD_n \coprod_{\sD_k} \sD_n \ar[ru,"{\langle x',x'' \rangle}"'] & \end{tikzcd}$ means exactly that $x = x' \comp_k x''$. \end{paragr} \begin{paragr} For all $0 \leq k < n$, we denote by $\kappa^n_k : \sD_n \to \sD_k$, the $\oo$-functor $\kappa^n_k := \langle \1^{n}_{e_k}\rangle : \sD_n \to \sD_k.$ Let $C$ be an $\oo$-category. More conceptually, $\kappa^n_k$ is induced from the Yoneda Lemma by the unit map $\1^{n}_{(\shortminus)} : C_k \to C_n,$ which is obviously natural in $C$. In practical terms, a commutative triangle $\begin{tikzcd} \sD_n \ar[r,"\langle x \rangle"] \ar[d,"\kappa^n_k"'] & C \\ \sD_k \ar[ru,"{\langle y \rangle}"'] & \end{tikzcd}$ means exactly that $x=1^{n}_y$. \end{paragr} \begin{definition}\label{def:conduche} An $\oo$-functor $F : C \to D$ is an \emph{discrete $\oo$-functor} if it is right orthogonal to the $\oo$-functors $\kappa^n_k : \sD_n \to \sD_k$ and $\nabla^n_k : \sD_n \to \sD_n \coprod_{\sD_k} \sD_n$ for all $k,n \in \mathbb{N}$ such that $0 \leq k < n$. \end{definition} \begin{paragr} Let us unfold Definition \ref{def:conduche}. The right orthogonality to $\kappa^n_k$ means that for any $n$-cell $x$ of $C$ and any $k$-cell $y$ of $D$ such that $F(x)=\1^{n}_y,$ there exists a unique\footnote{The map $z \mapsto 1^n_k(z)$ being injective, the uniqueness actually comes for free.} $k$-cell $z$ of $C$ such that $x=\1^n_z \text{ and } F(z)=y.$ Similarly, the right orthogonality to $\nabla^n_k$ means that for any $k$-cell $x$ of $C$, if $f(x) = y' \comp_k y''$ with $(y',y'')$ a pair of $k$-composable $n$-cells of $D$, then there exists a \emph{unique} pair $(x',x'')$ of $k$-composable $n$-cells of $C$ such that \begin{enumerate} \item $x=x'\comp_k x''$, \item $F(x')=y'$ and $F(x'')=y''$. \end{enumerate} \end{paragr} Since the class of discrete $\oo$-functors is a right orthogonal class of arrows, it has many good properties. For later reference, we recall one of them in the following proposition, whose proof is ommited. \begin{proposition}\label{prop:pullbackconduche} The class of Discrete Conduché $\oo$-functors is stable by pullback. This means that for any cartesian square in $\oo\Cat$ $\begin{tikzcd} C' \ar[d,"F'"'] \ar[r] & C \ar[d,"F"] \\ D' \ar[r] & D\ar[from=1-1,to=2-2,"\lrcorner",very near start,phantom] \end{tikzcd}$ if $F$ is a discrete Conduché $\oo$-functor, then so is $F'$. \end{proposition}
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