Commit 535b2c1a authored by Leonard Guetta's avatar Leonard Guetta
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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 $k<n \in \mathbb{N}$. If $F$ is right orthogonal to $\nabla^n_k$ then it is right orthogonal to $\kappa^n_k$.
\end{lemma}
\begin{proof}
Let $x$ be an $n$-cell of $C$ and suppose that $F(x)=\1^n_y$ with $y$ a $k$-cell of $D$. Notice that
\[
\begin{tikzcd}
C' \ar[d,"F'"'] \ar[r] & C \ar[d,"F"] \\
D' \ar[r] & D\ar[from=1-1,to=2-2,"\lrcorner",very near start,phantom]
\end{tikzcd}
F(x)=\1^n_y \comp_k \1^n_y
\]
if $F$ is a discrete Conduché $\oo$-functor, then so is $F'$.
and
\[
x=x\comp_k \1^n_{\src_k(x)}=\1^n_{\trgt_k(x)}\comp_k x
\]
and
\[
F(\1^n_{\src_k(x)})=\1^n_{\src_k(F(x))}=\1^n_y=\1^n_{\trgt_k(F(x))}=F(\1^n_{\trgt_k(x)}).
\]
Using the uniqueness part of the right orthogonality to $\nabla^n_k$, we deduce that $x=\1^n_{\src_k(x)}=\1^n_{\trgt_k(x)}$. Thus, if we set $z=\src_k(x)=\trgt_k(x)$, we have $x=\1^n_z$ and $F(z)=y$, which is what we needed to prove.
\end{proof}
\begin{remark}
In light of the previous lemma, the reader might wonder why the right orthogonality to $\kappa^n_k$ was included in Definition \ref{def:conduche}. The motivation for such a choice is that is \emph{should} be possible to apply this definition \emph{mutatis mutandis} to weak $\oo$-categories where Lemma \ref{lemma:redundencies} might not hold anymore.
\end{remark}
Somewhat related to Lemma \ref{lemma:redundencies} is the following.
\begin{lemma}\label{lemma:technicalorthogonality}
Let $F : C \to D$ an $\oo$-functor and $k<m<n \in \mathbb{N}$. If $F$ is right orthogonal to $\nabla^n_k$ and $\kappa^n_m$, then it is right orthogonal to $\nabla^m_k$.
\end{lemma}
\begin{proof}
Let $x$ be an $m$-cell of $C$ and suppose that
\[
F(x)=y'\comp_k y''
\]
with $(y',y'')$ a pair of $k$-composable $m$-cells of $D$. Then, we have
\[
F(\1^n_x)=\1^n_{y'} \comp_k \1^n_{y''}.
\]
From the right orthogonality to $\nabla^n_k$, we know that there exist $z$ and $z''$ $k$\nbd-composable $n$-cells of $C$ such that $F(z')=\1^n_{y'}$, $F(z'')=\1^n_{y''}$ and
\[
\1^n_x=z' \comp_k z''.
\]
From the right orthogonality to $\kappa^n_m$, we know that there exist $x'$ and $x''$ $m$\nbd-cells of $C$ such that $z'=\1^n_{x'}$, $z''=\1^n_{x''}$, $F(x')=y'$ and $F(x'')=y''$. It follows that $\src_k(x')=\src_k(x'')$ and
\[
\1^n_x=\1^n_{x'} \comp_k \1^n_{x''}=\1^n_{x'\comp_k x''},
\]
hence $x=x'\comp_k x''$. This proves the existence part of the right orthogonality to $\nabla^m_k$.
Now suppose that there are two pairs $(x_1',x_1'')$ and $(x_2',x_2'')$ that lift the pair $(y',y'')$ in the required way. It follows that the pairs $(\1^n_{x_1'},\1^n_{x_1''})$ and $(\1^n_{x_2'},\1^n_{x_2''})$ lift the pair $(\1^n_{y'},\1^n_{y''})$ in the required way.
From the uniqueness part of the right orthogonality to $\nabla^n_k$, we deduce that $\1^n_{x_1'}=\1^n_{x_2'}$ and $\1^n_{x_1''}=\1^n_{x_2''}$, hence $x'_1=x'_2$ and $x''_1=x''_2$.
\end{proof}
\begin{paragr}\label{paragr:mconduche}
Let $m \in \mathbb{N}$. Definition \ref{def:conduche} admits an obvious truncated version as follows. An $m$-functor $F : C \to D$ is a \emph{discrete Conduché $m$-functor} it is right orthogonal to $\kappa^n_k$ and $\nabla^n_k$ for all $k,n \in \mathbb{N}$ such that $0 \leq k < n \leq m$.
\end{paragr}
As an immediate consequence of Lemmae \ref{lemma:redundencies} and \ref{lemma:technicalorthogonality}, we have the following proposition, which gives the criterion we will use in practise to detect discrete Conduché $m$-functors.
\begin{proposition}
Let $n \in \mathbb{N}$. An $n$-functor $F : C \to D$ is a discrete Conduché $n$-functor if and only if it is right orthogonal $\nabla^n_k$ for every $k \in \mathbb{N}$ such that $k<n$.
\end{proposition}
\begin{paragr}
Let us dwell on a subtlety here. Since we have considered $n\Cat$ as a full subcategory of $\oo\Cat$ for any $n \in \mathbb{N}$ (see \ref{paragr:defncat}), an $n$\nbd-functor is a particular case of $\oo$\nbd-functor. Hence, it also makes sense to call an $n$-functor a \emph{discrete Conduché $n$-functor} when, seen as an $\oo$-functor, it is a discrete Conduché $\oo$-functor in the sense of Definition \ref{def:conduche}. This might have been conflicting with the definition we gave in \ref{paragr:mconduche} but the following lemma tells us that, in fact, the two notions are equivalent. In consequence, there is no distinction between the notions to make.
\end{paragr}
\begin{lemma}
Let $n \in \mathbb{N}$. An $n$-functor $F : C \to D$ is a Conduché discrete $n$\nbd-functor in the sense of \ref{paragr:mconduche} if and only if, seen as an $\oo$-functor, it is a Conduché discrete $\oo$\nbd-functor in the sense of Definition \ref{def:conduche}.
\end{lemma}
\begin{proof}
We only have to prove that if $F$ is an Conduché discrete $n$-functor then, when seen as an $\oo$-functor, it is a Conduché discrete $\oo$-functor; the other implication being trivial. From Lemma \ref{lemma:redundencies}, this amounts to show that if $F$ is right orthogonal to $\nabla^m_k$ for all $k,m \in \mathbb{N}$ such that $0 \leq k < m \leq n$, then it is also right orthogonal to $\nabla^m_k$ for all $k,m \in \mathbb{N}$ such that $0 \leq k < m$ and $m > n$. 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 $0 \leq k <n$ implies that $F(x)$ is also not a unit on a lower dimensional cell. Now, if we have
\[
F(x)=y'\comp_k y'',
\]
with $(y',y'')$ a pair of $k$-composable $n$-cells of $D$, then the right orthogonality to $\nabla^n_k$ implies that
\[
x=x'\comp_k x''
\]
with $F(x')=y'$ and $F(x'')=y''$. Since $x$ is indecomposable, $x'$ or $x''$ has to be of the form $\1^n_z$ with $z$ and $k$-cell of $C$.
\end{proof}
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