### end of the day

parent 180e3bf6
 ... ... @@ -138,6 +138,16 @@ year={2020} publisher={ScienceDirect}, year={1995} } @article{freyd1972categories, title={Categories of continuous functors, I}, author={Freyd, Peter J and Kelly, G Max}, journal={Journal of pure and applied algebra}, volume={2}, number={3}, pages={169--191}, year={1972}, publisher={North-Holland} } @book{gabriel1967calculus, title={Calculus of fractions and homotopy theory}, author={Gabriel, Peter and Zisman, Michel}, ... ...
 ... ... @@ -344,7 +344,7 @@ A \emph{morphism of $\oo$-magmas} $f : X \to Y$ is a morphism of underlying $\oo By considering$n\Cat$as a subcategory of$\oo\Cat$, the previous definition also works for$n$-categories. It follows from Example \ref{dummyexample} that an$n$-category is free if and only if it has a$k$-basis for every$0 \leq k \leq n$. \end{paragr} We wish now to recall an important result due to Makkai concerning the uniqueness of$n$-basis for a free$\oo$-category. First we need the following definition. \begin{definition} \begin{definition}\label{def:indecomposable} Let$C$be an$\oo$-category. For$n>0$, an$n$-cell$x$of$C$is \emph{indecomposable} if both following conditions are satisfied: \begin{enumerate}[label=(\alph*)] \item$x$is not a unit on a lower dimensional cell, ... ... @@ -1128,7 +1128,7 @@ We can now prove the expected result. 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 An$\oo$-functor$F : C \to D$is an \emph{discrete Conduché$\oo$-functor} if it is right orthogonal to the$\oo$-functors $\kappa^n_k : \sD_n \to \sD_k$ ... ... @@ -1138,8 +1138,22 @@ We can now prove the expected result. \] for all$k,n \in \mathbb{N}$such that$0 \leq k < n$. \end{definition} Before unfolding the previous definition, let us put here for later reference the immediate following lemma. \begin{lemma}\label{lemma: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{lemma} \begin{proof} This is standard fact about right orthogonal classes in a category that admits pullbacks. See for example the dual of \cite[Proposition 2.1.1(e)]{freyd1972categories}. \end{proof} \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 Let us now give a practical version of 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 $F(x)=\1^{n}_y,$ ... ... @@ -1157,14 +1171,85 @@ We can now prove the expected result. \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$\oo\Cat$As it turns out, the definition we gave of discrete Conduché$\oo$-functor is highly redundant. \begin{lemma}\label{lemma:redundencies} Let$F : C \to D$be an$\oo$-functor and let$k n$. This follows easily from the fact that for any$m$-cell$x$in an$n$-category (seen as an$\oo$-category) with$m > n$, there exists a unique$n$-cell$x'$such that$x = \1^m_{x'}$. Details are left to the reader. \end{proof} As we shall now see, discrete Conduché$\oo$-functors have a deep connection with free$\oo$-categories. We begin by an easy property. \begin{lemma} Let$F : C \to D$be a discrete Conduché$\oo$-functor and$n \in \mathbb{N}$. An$n$-cell$x$of$C$is indecomposable (Definition \ref{def:indecomposable}) if and only if$F(x)$is indecomposable. \end{lemma} \begin{proof} When$n=0$, there is nothing to prove since every$0$-cell is indecomposable. We suppose now that$n>0$. Suppose that$x$is indecomposable. Since$x$is not a unit on a lower dimenionsal well, the right orthogonality to$\kappa^n_k$for every$0 \leq k
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