Commit 25e266fb authored by Leonard Guetta's avatar Leonard Guetta
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parent dcee846c
......@@ -688,7 +688,7 @@ 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}
......
......@@ -777,7 +777,7 @@ We can now prove the following proposition, which is the key result of this sect
\]
is not free on a $2$-graph. The reason is that the source (resp. the target) of $\alpha$ is $g \comp_0 f$ (resp. $i \comp_0 h$) which are not generating $1$-cells.
\end{example}
\begin{example}
\begin{example}\label{example:freemonoid}
Let $n \geq 1$ and $M$ be a monoid (commutative if $n>1$). The $n$-category $B^nM$ is free if and only if $M$ is a free monoid (free commutative monoid if $n>1$). If so, it has exactly one generating cell of dimension $0$, no generating cells of dimension $0 <k< n$, and the set of generators of the monoid (which is unique) as generating $n$-cells.
\end{example}
\section{Cells of free $\oo$-categories as words}
......@@ -1021,20 +1021,20 @@ We can now prove the expected result.
Let us start with the surjectivity. Let $H : \E^+/{\equiv} \to \D$ be a $(n+1)$-functor. We define a map $\varphi : \Sigma \to D_{n+1}$ as
\begin{align*}
\varphi : \Sigma &\to D_{n+1}\\
\alpha &\mapsto H_{n+1}([\cc_{\alpha}]),
\alpha &\mapsto H_{n+1}([(\cc_{\alpha})]),
\end{align*}
where $[w]$ is the equivalence class under $\equiv$ of an element $w \in \T[\E]$. All we need to show is that
\[
\widehat{\varphi}=H_{n+1}.
\]
Let $z$ be an element of $\T[\E]/{\equiv}$ and let us choose $w \in \T[\E]$ such that $z=[w]$. We proceed to show that $\widehat{\varphi}(z)=H_{n+1}(z)$ by induction on the size of $w$ (Definition \ref{def:sizeword}).
If $|w|=0$, then either $w=\cc_{\alpha}$ for some $\alpha \in \Sigma$ or $w = \ii_x$ for some $x \in C_n$. In the first case, we have
If $|w|=0$, then either $w=(\cc_{\alpha})$ for some $\alpha \in \Sigma$ or $w = (\ii_x)$ for some $x \in C_n$. In the first case, we have
\[
\widehat{\varphi}([\cc_{\alpha}])=\overline{\varphi}(\cc_{\alpha})=\varphi(\alpha)=H_{n+1}([\cc_{\alpha}]),
\widehat{\varphi}([(\cc_{\alpha})])=\overline{\varphi}((\cc_{\alpha}))=\varphi(\alpha)=H_{n+1}([(\cc_{\alpha})]),
\]
and in the second case we have,
\[
\widehat{\varphi}([\ii_x])=\overline{\varphi}(\ii_x)=1_{H_n(x)}=H_{n+1}([\ii_x])
\widehat{\varphi}([(\ii_x)])=\overline{\varphi}((\ii_x))=1_{H_n(x)}=H_{n+1}([(\ii_x)])
\]
where for the last equality, we used the fact that $[\ii_x]$ is the unit on $x$ in $\E^+/{\equiv}$.
......@@ -1048,9 +1048,48 @@ We can now prove the expected result.
\]
But, by definition, for every $\alpha \in \Sigma$ we have
\[
\varphi(\alpha)=\widehat{\varphi}([\cc_{\alpha}])=\widehat{\varphi'}([\cc_{\alpha}])=\varphi'(\alpha).
\varphi(\alpha)=\widehat{\varphi}([(\cc_{\alpha})])=\widehat{\varphi'}([(\cc_{\alpha})])=\varphi'(\alpha).\qedhere
\]
\end{proof}
\begin{paragr}
In particular, the previous proposition tells us that $n$-cells in a free $\oo$-category can be represented as a (well formed) words made up of the generating $n$-cells and units on lower dimensionan cells.
In other words, the previous proposition tells us that for an $n$-cellular extension $\E$, the $(n+1)$-cells of $\E^*$ can be represented as equivalence classes of (well formed) words made up of the indeterminates of $\E$ and units on $n$-cells. Note, however, that the equivalences classes are enormous. For example, for an indeterminate $\alpha$, all three following words are in the equivalence class of $(\cc_{\alpha})$:
\[
((\cc_{\alpha})\fcomp_n(\ii_{\sigma(\alpha)})) \quad (((\cc_{\alpha})\fcomp_n(\ii_{\sigma(\alpha)}))\fcomp_n (\ii_{\sigma(\alpha)})) \quad ((\cc_{\alpha})\fcomp_n((\ii_{\sigma(\alpha)})\fcomp_n (\ii_{\sigma(\alpha)}))).
\]
% Besides, the way we have proven the existence of the smallest congruence on an $n$-magma was abstract and will turn out to be unsufficient
Note that when $n=0$, we have already said that a $0$-cellular extension $\E$ is nothing but a graph and that $\E^*$ is the free category on the graph. In particular, the $1$-cells of $\E^*$ can simply be encoded in strings of composable arrows of the graph, which is a much simpler description that the one obtained from the previous proposition since no equivalence relation is involved. A natural question is whether it would be possible to obtain a similar simple description for $n>0$. While it is certainly possible to have a simpler description of the $(n+1)$-cells of $\E^*$ than the one obtained in the present section (for example by getting rid of some ``pathological'' well formed words and reduce the size of the equivalence classes), it seems not possible to completely avoid the use of an equivalence relation, at least not in a canonical way. Indeed, we have already seen in Example \ref{example:freemonoid} that a free \emph{commutative} monoid $M$ gives rise to a free $n$-category $B^nM$ whose generating $n$-cells are in bijection with the generators of the monoid. And, as soon as there are at least two generators of the monoid, let say $a$ and $b$, there is no canonical way of representing elements of the monoid by a unique word on the generators, for we have
\[
ab = ba.
\]
\todo{À finir ? Il faudrait peut-être dire qu'en rajoutant des hypothèses on doit pouvoir quand même avoir une forme normale. Dans le cas n=2, quand on fait l'hypothèse que le 2-polygraphe est positif, je sais que c'est vrai.}
\end{paragr}
\section{Discrete Conduché $\oo$-functors}
\begin{paragr}
Recall that given a category $\C$ and a class $M$ of arrows of $\C$, we say that an arrow $f : X \to Y$ of $\C$ is \emph{left orthogonal} to $M$ if for every $m : A \to B$ in $M$ and every solid arrow square
\[
\begin{tikzcd}
X \ar[d,"f"'] \ar[r] & A \ar[d,"m"] \\
Y \ar[r] & B,
\ar[from=2-1,to=1-2,dashed,"u"]
\end{tikzcd}
\]
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.}
\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:
\[
\begin{tikzcd}[column sep = large, row sep=large]
\sD_k \ar[r,"\langle \src_k(e_n) \rangle"] \ar[d,"\langle \trgt_k(e_n) \rangle"'] & \sD_n \ar[d] \\
\sD_n \ar[r] & \displaystyle\sD_n \coprod_{\sD_k} \sD_n. \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end]
\end{tikzcd}
\]
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
\begin{align*}
\oo\Cat &\to \Set \\
C &\mapsto C_n \times_{C_k} C_n .
\end{align*}
The $k$-compostion operation $(\shortminus) \comp_k (\shortminus)$
\end{paragr}
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