Commit 17e63b7e authored by Leonard Guetta's avatar Leonard Guetta
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Break. I need to wrap up the proof od Theorem 1.6.18

parent 535b2c1a
......@@ -2,7 +2,7 @@
\usepackage{mystyle}
\title{Homology of $\omega$-categories}
\title{Homology of strict $\omega$-categories}
\author{Léonard Guetta}
\begin{document}
......
......@@ -175,11 +175,19 @@ year={2020}
journal={arXiv preprint arXiv:1202.3359},
year={2012}
}
@article{guetta2018polygraphs,
title={Polygraphs and discrete Conduch{\'e} functors},
@article{guetta2020polygraphs,
title={Polygraphs and discrete Conduch{\'e} $\omega$-functors},
author={Guetta, L{\'e}onard},
journal={arXiv preprint arXiv:1812.05332},
year={2018}
journal={Higher Structures},
volume={4},
number={2},
year={2020}
}
@book{hopcroft1979introduction,
title={Introduction to automata theory, languages, and computation},
author={Hopcroft, John E and Ullman},
publisher={Addison-Wesley},
year={1979}
}
@book{hovey2007model,
title={Model categories},
......@@ -187,6 +195,15 @@ year={2020}
year={2007},
publisher={American Mathematical Society}
}
@article{johnstone1999note,
title={A note on discrete {C}onduch{\'e} fibrations},
author={Johnstone, Peter},
journal={Theory and Applications of Categories},
volume={5},
number={1},
pages={1--11},
year={1999}
}
@article{lafont2009polygraphic,
title={Polygraphic resolutions and homology of monoids},
author={Lafont, Yves and M{\'e}tayer, Fran{\c{c}}ois},
......
......@@ -941,7 +941,7 @@ We wish now to see how to prove the existence of a congruence defined with a con
\]
which will obviously be the smallest congruence satisfying the desired condition. To see that $I$ is not empty, it suffices to notice that the binary relation ``being parallel $n$-cells'' is a congruence, which obviously is in $I$.
\end{proof}
\begin{proposition}
\begin{proposition}\label{prop:smallestcategoricalcongruence}
Let $X$ be an $n$-magma with $n\geq 1$. There exists a smallest categorical congruence on $X$.
\end{proposition}
\begin{proof}
......@@ -1220,30 +1220,37 @@ We can now prove the expected result.
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.
Definition \ref{def:conduche} admits an obvious truncated version as follows.
\begin{definition}\label{def:mconduche}
Let $m \in \mathbb{N}$. An $m$-functor $F : C \to D$ is a \emph{discrete Conduché $m$-functor} if 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{definition}
\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 Definition \ref{def: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}.
Let $n \in \mathbb{N}$. An $n$-functor $F : C \to D$ is a Conduché discrete $n$\nbd-functor in the sense of Definition \ref{def: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}
In practise, we will use the following criterion to detect discrete Conduché $n$-functor.
\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{proof}
It is an immediate consequence of Lemmae \ref{lemma:redundencies} and \ref{lemma:technicalorthogonality}.
\end{proof}
\begin{remark}
In the case $n=1$, we recover the usual definition of what's commonly referred to as discrete Conduché fibration (see for example \cite{johnstone1999note}).
\end{remark}
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}
\begin{lemma}\label{lemma:conducheindecomposable}
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 $n>0$.
When $n=0$, there is nothing to prove since every $0$-cell is indecomposable. We assume 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
Suppose that $x$ is indecomposable. Since $x$ is not a unit on a lower dimensional 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
\[
F(x)=y'\comp_k y'',
\]
......@@ -1251,5 +1258,455 @@ We can now prove the expected result.
\[
x=x'\comp_k x''
\]
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 $C$.
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 $C$, and thus $y'$ or $y''$ is of the form $\1^n_{F(z)}$. This proves that $F(x)$ is indecomposable.
Conversely, suppose that $F(x)$ is indecomposable. The $n$-cell $x$ cannot be a unit on a lower dimensional cell as if it were, then $F(x)$ would also be a unit on a lower dimensional cell, which is impossible since it is indecomposable. Now, if
\[
x=x'\comp_k x''
\]
with $(x',x'')$ a pair of $k$-composable $n$-cells, then
\[F(x)=F(x')\comp_k F(x'').\]
Hence, either $F(x')$ or $F(x'')$ is a unit on a lower dimensional cell and from the right orthogonality to $\kappa^n_k$ we deduce that either $x'$ or $x''$ is a unit on a lower dimensional cell. This proves that $x$ is indecomposable.
\end{proof}
From the previous lemma and Proposition \ref{prop:uniquebasis}, we deduce the following proposition.
\begin{proposition}
Let $F : C \to D$ be an $\oo$-functor with $C$ and $D$ free $\oo$-categories. If $F$ is a discrete Conduché $\oo$-functor then it is rigid.
\end{proposition}
\begin{remark}
The converse of the above proposition does not hold. For details, see \cite[Appendix A]{guetta2020polygraphs}.
\end{remark}
We now turn to the main result concerning the relation between discrete Conduché $\oo$-functors and free $\oo$-categories that we were aiming for. As a matter of fact, a significant amount of results and definitions that we have seen so far were geared towards the statement and proof of the following theorem. \iffalse To avoid any ambiguity, let us precise that given an $\oo$\nbd-functor $F : C \to D$ and $\Sigma$ a subset of the $n$-cells of $D$ with $n \in \mathbb{N}$, we use the notation $F^{-1}(\Sigma)$ for the set
\[
F^{-1}(\Sigma):=\left\{ x \in C_n \vert F(x) \in \Sigma\right\}.
\]
\fi
\begin{theorem}\label{thm:conduche}
Let $F : C \to D$ be an discrete Conduché $\oo$-functor.
\begin{enumerate}
\item If $D$ is free then so is $C$.
\item If $C$ is free and if for every $n \in \mathbb{N}$, $F_n : C_n \to D_n$ is surjective, then $D$ is also free.
\end{enumerate}
\end{theorem}
The proof of the previous theorem is long and technical and the next two sections are devoted to it. Before that, let us make Theorem \ref{thm:conduche} a little more precise.
\begin{paragr}
Let $F : C \to D$ be a discrete Conduché $\oo$-functor.
\begin{enumerate}
\item If $D$ is free, we know from the previous theorem that $C$ is also free. Now, if we write $\Sigma^D_n$ for the $n$-basis of $D$, then it follows from Proposition \ref{prop:uniquebasis} and Lemma \ref{lemma:conducheindecomposable} that the $n$-basis of $C$ is given by
\[
\Sigma^C_n :=\left\{ x \in C_n \vert F(x) \in \Sigma^D_n\right\}.
\]
\item Similarly, if $C$ is free and if $F_k : C_k \to D_k$ is surjective for every $k \in \mathbb{N}$, we know that $D$ is also free. Now, if we write $\Sigma^C_n$ for the $n$-basis of $C$, then the $n$-basis of $D$ is given by
\[
\Sigma^D_n :=\left\{ y \in D_n \vert \exists x \in C_n \text{ with } F(x)=y \in \Sigma^D_n\right\} .
\]
\end{enumerate}
\end{paragr}
\section{Proof of Theorem \ref{thm:conduche}: part I}
This first part of the proof of Theorem \ref{thm:conduche} consists of several technical results on words. They lay a preliminary foundation on which the key arguments of the proof will later rely.
\emph{For the whole section, we fix a cellular $n$\nbd-extension $\E=(C,\Sigma,\sigma,\tau)$. A ``word'' always means an element of $\W[\E]$ and a ``well formed word'' always means an element of $\T[\E]$.}
\begin{paragr}
For a word $w$, the number of symbols that appear in $w$ is referred to as the \emph{length of $w$} and denoted by $\mathcal{L}(w)$. Recall also from Definition \ref{def:sizeword} that when $w$ is well formed, the number of occurences of symbols $\fcomp_k$ for any $0 \leq k \leq n$ that appear in $w$ is referred to as the \emph{size of $w$} and denoted by $\vert w \vert$.
\end{paragr}
\begin{definition}\label{def:subword}A word $v$ is a \emph{subword} of a word $w$ if there exist words $a$ and $b$ such that $w$ can be written as
\[w=avb.\]
\end{definition}
\begin{remark}
Beware that in the previous definition, none of the words were supposed to be well formed. In particular, a subword of a well formed word is not necessarily well formed.
\end{remark}
\begin{paragr} Since a word $w$ is a finite sequence of symbols, it makes sense to write $w(i)$ for the symbol at position $i$ of $w$, with $0 \leq i \leq \mathcal{L}(w)-1$.
For any $0 \leq i \leq \mathcal{L}(w)-1$, define $P_{w}(i)$ to be the number of opening parenthesis in $w$ with position $ \leq i$ minus the number of closing parenthesis in $w$ with position $\leq i$. This defines a function
\[
P_{w} : \{0,\dots,\mathcal{L}(w)-1\} \to \mathbb{Z}.
\]
\end{paragr}
\begin{remark}
Such a counting function is standard in the literature about formal languages. For example see \cite[chapter 1, exercice 1.4]{hopcroft1979introduction}.
\end{remark}
\begin{definition}
A word $w$ is \emph{well parenthesized} if:
\begin{enumerate}
\item it is not empty,
\item $P_w(i) \geq 0$ for any $0 \leq i \leq \mathcal{L}(w)-1$,
\item $P_w(i) = 0$ if and only if $i = \mathcal{L}(w)-1$.
\end{enumerate}
\end{definition}
\begin{paragr}
It follows from the previous definition that the first letter of a well parenthesized word is necessarily an opening parenthesis and that the last letter is necessarily a closing parenthesis. Thus, the length of a well parenthesized word is not less than 2.
Moreover, it is immediate that if $w_1$ and $w_2$ are well parenthesized words then, for any $0 \leq k \leq n$,
\[
(w_1 \fcomp_k w_2)
\]
is well parenthesized.
\end{paragr}
\begin{lemma}\label{lemmawellformedispseudo}
A well formed word is well parenthesized.
\end{lemma}
\begin{proof}
Let $w$ be a well formed word. We proceed by induction on $|w|$. If $|w|=0$, then $w$ is either of the form
\[
(\cc_{\alpha})
\]
or of the form
\[
(\ii_{x}).
\]
In either case, the assertion is trivial.
Now suppose that $|w|>0$, we know by definition that
\[
w=(w_1 \fcomp_k w_2)
\]
with $w_1,w_2$ well formed words such that $|w_1|, |w_2| < |w|$. The desired properties follow easily from the induction hypothesis. Details are left to the reader.
\end{proof}
The converse of the previous lemma is obviously not true. However, Lemma \ref{lemmapartialconverse} below is a partial converse.
\begin{lemma}\label{lemmasubwordpseudo}
Let $w$ be a well parenthesized word of the form
\[
w=(w_1 \fcomp_k w_2)
\]
with $w_1$ and $w_2$ well parenthesized words, and $0 \leq k \leq n$ and let $v$ be a subword of $w$. If $v$ is well parenthesized then one the following holds:
\begin{enumerate}
\item $v=w$,
\item $v$ is a subword of $w_1$,
\item $v$ is a subword of $w_2$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $a$ and $b$ be words such that
\[
avb=w=(w_1\fcomp_k w_2).
\]
Let $l_1, l_2, l, l_a, l_b, l_v$ respectively be the length of $w_1,w_2,w,a,b,v$. Notice that
\[
l_a+l_v+l_b=l=l_1+l_2+3.
\]
Notice that since $v$ is well parenthesized, the following cases are forbidden:
\begin{enumerate}
\item $l_1 \leq l_a \leq l_1 +1$,
\item $l_2 \leq l_b \leq l_2 +1$,
\item $l_a\geq l-1$,
\item $l_b\geq l-1$.
\end{enumerate}
Indeed, the first case would imply that the first letter of $v$ is a closing parenthesis or the symbol $\fcomp_k$. Similarly, the second case would imply that the last letter of $v$ is an opening parenthesis or the symbol $\fcomp_k$. The third and fourth cased would imply that $l_v<2$ which is also impossible.
That leaves us with the following cases:
\begin{enumerate}
\item $l_a=0$,
\item $l_b=0$,
\item $0<l_a<l_1$ and $0<l_b < l_2$,
\item $0<l_a<l_1$ and $l_b > l_2+1$,
\item $l_1 +1 < l_a$ and $0 < l_b < l_2$.
\end{enumerate}
If we are in the first case, then
\[
P_w(j)=P_v(j)
\]
for $0 \leq j \leq l_v-1$. That implies that $P_w(l_v-1)=0$ which means that $l=l_v$, hence $w=v$.
By a similar argument that we leave to the reader, we can show that the second case implies that $w=v$.
If we are in the fourth (resp.\ fifth) case, then it is clear that $v$ is a subword of $w_1$ (resp.\ $w_2$).
Suppose now that we are in the third case. Intuitively, it means that the first letter of $v$ is inside $w_1$ and the last letter of $v$ is inside $w_2$. Notice first that
\begin{equation}\label{usefulinequality}\tag{$\star$}
l_a<l_1 <l_a+l_v-3,
\end{equation}
where the inequality on the right comes from the fact that $l_v \geq 2$ because $v$ is well formed.
Besides, by definition of $P_w$,
\[
P_w(j)=P_v(j-l_a)+P_w(l_a)
\]
for $l_a \leq j < l_v+l_a$. In particular, we have
\[
1=P_{w_1}(l_1-1)+1=P_w(l_1-1)=P_v(l_1-1)+P_w(l_a).
\]
From \eqref{usefulinequality} and since $v$ is well parenthesized, we deduce that
\[
P_v(l_1-1)>0.
\]
Hence, $P_w(l_a)\leq 0$ which is impossible because $w$ is well formed and ${l_a < l-1}$.\qedhere
\end{proof}
\begin{lemma}\label{lemmapartialconverse}
Let $w$ be a well parenthesized word. If $w$ is a subword of a well formed word, then it is also well formed.
\end{lemma}
\begin{proof}
Let $u$ be a well formed word such that $w$ is a subword of $u$. We proceed by induction on $|u|$. If $|u|=0$, then $u$ is either of the form
\[
(\cc_{\alpha})
\]
or of the form
\[
(\ii_x).
\]
In both cases, $w=u$ since the only well parenthesized subword of $u$ is $u$ itself.
Suppose now that $|u|>0$. By definition,
\[
u=(u_1 \fcomp_k u_2)
\]
with $|u_1|,|u_2| < |u|$. By Lemmas \ref{lemmawellformedispseudo} and \ref{lemmasubwordpseudo}, we have that either:
\begin{itemize}
\item[-] $w=u$ in which case $w$ is well formed by hypothesis,
\item[-] $w$ is a subword of $u_1$ and from the induction hypothesis we deduce that $w$ is well formed,
\item[-] $w$ is a subword of $u_2$ which is similar to previous case.\qedhere
\end{itemize}
\end{proof}
\begin{proposition}\label{prop:subwords} Let $w$ be a well formed word of the form
\[
w = (w_1 \fcomp_k w_2)
\]
with $w_1$ and $w_2$ well formed words, and $0 \leq k \leq n$, and let $v$ be a subword of $w$. If $v$ is well formed, then we are in one of the following cases:
\begin{enumerate}
\item $v=w$,
\item $v$ is a subword of $w_1$,
\item $v$ is a subword of $w_2$.
\end{enumerate}
\end{proposition}
\begin{proof}
This follows immediately from Lemma \ref{lemmawellformedispseudo} and Lemma \ref{lemmasubwordpseudo}.
\end{proof}
\begin{proposition}\label{prop:subwordsubstitution}
Let $u$ be a well formed word of the form
\[
vew
\]
with $v$, $w$ and $e$ words and such that $e$ is well formed. If $e'$ is a well formed word that is parallel to $e$, then the word
\[
ve'w
\]
is also well formed.
\end{proposition}
\begin{proof}
We proceed by induction on $|u|$.
\begin{description}
\item[Base case] If $|u|=0$, then necessarily $v$ and $w$ are both the empty word and the assertion is trivial.
\item[Inductive step] If $|u| \geq 1$, then
\[
u=(u_1 \fcomp_k u_2)
\]
with $u_1$ and $u_2$ well formed words such that $|u_1|,|u_2| < |u|$. By hypothesis, $e$ is a subword of $u$ and from Proposition \ref{prop:subwords}, we are in one of the following cases.
\begin{itemize}
\item[-] $u=e$ in which case the assertion is trivial.
\item[-] $e$ is a subword of $u_1$, which means that there exist words $\tilde{v}, \tilde{w}$ such that
\[
u_1=\tilde{v}e\tilde{w}.
\]
Moreover, we have \[v=(\tilde{v}\] and \[w=\tilde{w}\ast_k u_2).\] By induction hypothesis, the word
\[
\tilde{v}e'\tilde{w}
\]
is well formed and thus
\[
(\tilde{v}e'\tilde{w}\ast_k u_2)=vew
\]
is well formed.
\item[-] $e$ is a subword of $u_2$, which is symmetric to the previous case.\qedhere
\end{itemize}
\end{description}
\end{proof}
\begin{lemma}\label{lemmaunicitydecompositionpseudo}
Let $w_1,w_2,w_1',w_2'$ be well parenthesized words, $0 \leq k \leq n$ and $0 \leq k' \leq n$ be such that
\[
(w_1\fcomp_k w_2) = (w_1' \fcomp_{k'} w_2').
\]
Then, $w_1=w_1'$, $w_2=w_2'$ and $k=k'$.
\end{lemma}
\begin{proof}
Let us define $l:=\mathrm{min}(\mathcal{L}(w_1),\mathcal{L}(w_1'))$. Notice that
\[
P_w(j)=P_{w_1}(j-1)+1=P_{w_1'}(j-1)+1
\]
for $0 < j \leq l$ hence
\[
P_{w_1}(l-1)=P_{w_1'}(l-1).
\]
Since $w_1$ and $w_1'$ are well parenthesized, one of the members of the last equality (and thus both) is equal to $0$. That implies that $\mathcal{L}(w_1)=\mathcal{L}(w_1')$ and the desired properties follow immediately from that.
\end{proof}
\begin{proposition}\label{prop:unicitydecomposition}
Let $w_1,w_1',w_2,w_2'$ be well formed words, and $0 \leq k \leq n$ and $0 \leq k' \leq n$ be such that $(w_1\ast_k w_2)$ and $(w_1'\ast_{k'} w_2')$ are well formed.
If
\[
(w_1 \fcomp_k w_2) = (w_1' \fcomp_{k'} w_2'),
\]
then
\[
w_1=w_1', w_2=w_2' \text{ and }k=k'.
\]
\end{proposition}
\begin{proof}
This follows from Lemma \ref{lemmawellformedispseudo} and Lemma \ref{lemmaunicitydecompositionpseudo}.
\end{proof}
\begin{corollary}\label{cor:compatiblesubwords}
Let $w$ be a well formed word and suppose that it can be written as
\[w=(w_1 \fcomp_k w_2)\]
with $w_1$ and $w_2$ well formed words and $0\leq k \leq n$. Then $\src_k(w_1)=\trgt_k(w_2)$.
\end{corollary}
\begin{proof}
By hypothesis, $|w| \geq 1$. From the definition of well formed words, we know that $w$ is of the form
\[
(w_1'\fcomp_{k'}w_2')
\]
with $w_1'$ and $w_2'$ well formed words and $0 \leq k' \leq n$ such that
\[
\src_{k'}(w_1')=\trgt_{k'}(w_2').
\]
From Proposition \ref{prop:unicitydecomposition}, we have that $w_1'=w_1$, $w_2'=w_2$ and $k=k'$.
\end{proof}
\section{Proof of Theorem \ref{thm:conduche}: part II}
Recall that we have seen in Proposition \ref{prop:smallestcategoricalcongruence} that there was a smallest categorical congruence on any $n$-magma $X$ with $n \geq 1$. However, the description of this congruence we obtained when proving its existence was rather abstract. The main goal of this section is to give a more concrete description of the smallest categorical congruence in the case that $X= \E^+$ for an $n$-cellular extension $\E$.
\begin{definition}\label{def:elementarymove}
Let $n \geq 0$, $\E=(C,\Sigma,\sigma,\tau)$ be an $n$\nbd-cellular extension and $u,u' \in \T[\E]$. An \emph{elementary move} from $u$ to $u'$ is a quadruple $\mu=(v,w,e,e')$ with $v,w \in \W[\E]$ and $e,e' \in \T[\E]$ such that
\[
u=vew,
\]
\[
u'=ve'w,
\]
and one of the following holds:
\begin{enumerate}[label=(\arabic*)]
\item
$e$ is of the form
\[((x\fcomp_ky)\fcomp_kz)\]
and $e'$ is of the form
\[
(x\fcomp_k(y\fcomp_kz))\]
with $x,y,z \in \T[E]$ and $0\leq k \leq n$,
\item $e$ is of the form \[((\ii_c)\fcomp_kx)\]
and $e'$ is of the form
\[x\]
with $x \in \T[E]$, $0 \leq k \leq n$ and $c=\1^n_{\trgt_k(x)}$,
\item $e$ is of the form
\[(x\fcomp_k(\ii_{c}))\]
and $e'$
is of the form
\[x\]
with $x \in \T[E]$, $0 \leq k \leq n$ and $c=\1^n_{\src_k(x)}$,
\item $e$ is of the form \[((\ii_{c})\fcomp_k(\ii_{d}))\] and $e'$ is of the form \[(\ii_{c\comp_kd})\]with $(c,d)$ a pair of $k$-compasable $n$-cells of $C$ with $0 \leq k < n$,
\item $e$ is of the form \[((x\fcomp_ky)\fcomp_l(z\fcomp_kt))\]
and $e'$ is of the form
\[((x\fcomp_lz)\fcomp_k(y\fcomp_lt))\]
with $x,y,z,t \in \T[E]$ and $0\leq l < k \leq n$.
\end{enumerate}
\end{definition}
\begin{paragr}
We will use the notation
\[
\mu : u \rightarrow u'
\]
to say that $\mu$ is an elementary move from $u$ to $u'$.
We write $\G[\E]$ for the graph (or $1$-graph in the terminology of \ref{paragr:defncat}) defined as:
\begin{itemize}
\item[-] the set of objects of $\G[\E]$ is $\T[\E]$,
\item[-] for all $u,u'$ in $\T[E]$, an arrow of $\G[\E]$ from $u$ to $u'$ is an elementary move from $u$ to $u'$.
\end{itemize}
We will use the categorical notation
\[
\G[\E](u,u')
\]
for the set of arrows from $u$ to $u'$.
Finally, we will also sometimes write
\[
u \leftrightarrow u'
\]
to say that there exists an elementary move from $u$ to $u'$ or from $u'$ to $u$.
\end{paragr}
\begin{definition}\label{definitionequivalence}
Let $\E=(C,\Sigma,\sigma,\tau)$ be an $n$-cellular extension and $u,u' \in \T[E]$. We say that the well formed words $u$ and $u'$ are \emph{equivalent} and write
\[
u \sim u'
\]
if they are in the same connected component of $\G[E]$. More precisely, this means that there exists a finite sequence $(u_j)_{0\leq j \leq N}$ of well formed words with $u_0=u$, $u_N=u'$ and $u_j \leftrightarrow u_{j+1}$ for $0 \leq j < N$. The equivalence class of a well formed word $u$ will be denoted by $[u]$.
\end{definition}
We wish now to establish that the equivalence relation $\sim$ defined above is the smallest congruence on $\E^+$. In order to do that, we need to prove several technical results on words.
\begin{lemma}\label{lemmaequivrelationsourcestargets}Let $u,u' \in \T[E]$. If $u \sim u'$ then $u$ and $u'$ are parallel.
\end{lemma}
\begin{proof}
Let
\[
\mu =(v,w,e,e') : u \to u'
\]
be an elementary move from $w$ to $w'$. We are going to prove that $s(u)=s(u')$ and $t(u)=t(u')$ with an induction on $\mathcal{L}(v)+\mathcal{L}(w)$(cf.\ \ref{paragrdefinitionwords}). Notice first that, by definition of elementary moves, $|u|\geq 1$ and thus
\[
u=(u_1 \ast_k u_2)
\]
with $u_1,u_2 \in \T[E]$.
\begin{description}
\item[Base case] If $\mathcal{L}(v)+\mathcal{L}(w)=0$, it means that both $v$ and $w$ are both the empty word. It is then straightforward to check the desired property using Definition \ref{definitionelementarymove}.
\item[Inductive step] Suppose now that $\mathcal{L}(v)+\mathcal{L}(w)\geq 0$. Since $e$ is a subword of $u$ that is well formed, from Lemma \ref{lemmasubwords} we are in one of the following cases:
\begin{itemize}
\item[-] $e=u$, which is exactly the base case.
\item[-] $e$ is a subword of $u_1$, which means that there exist $\tilde{v},\tilde{w} \in \T[E]$ such that
\[
u_1 = \tilde{v}e\tilde{w}.
\]
Moreover, we have
\[
v=(\tilde{v}
\]
and
\[
w=\tilde{w}\ast_ku_2).
\]
From Corollary \ref{corollarysubwordsubstitution}, the word
\[
u_1':=\tilde{v}e'\tilde{w}
\]
is well formed. Therefore we can use the induction hypothesis on
\[
\tilde{\mu} :=(\tilde{v},\tilde{w},e,e') : u_1 \to u_1'.
\]
This shows that $s(u_1)=s(u_1')$ and $t(u_1)=t(u_1')$ and since
\[
u=(u_1\ast_k u_2) \text{ and } u'=(u_1' \ast_k u_2)
\]
it follows easily that $s(u)=s(u')$ and $t(u)=t(u')$.
\item[-] $e$ is a subword of $u_2$, which is symmetric to the previous case.
\end{itemize}
\end{description}
By definition of $\sim$, this suffices to show the desired properties.
\end{proof}
\begin{lemma}\label{lemmaequivrelationiscongruence}
Let $v_1, v_2, v_1', v_2' \in \T[E]$ and $0 \leq k \leq n$ such that $v_1$ and $v_2$ are $k$-composable, and $v_1'$ and $v_2'$ are $k$-composable. If $v_1 \sim v_2$ and $v_1' \sim v_2'$ then
\[
(v_1 \ast_k v_2) \sim (v_1' \ast_k v_2').
\]
\end{lemma}
\begin{proof}
Let
\[
\mu = (v,w,e,e') : v_1 \to v_1'
\]
be an elementary move. Set
\[
\tilde{v} := (v
\]
and
\[
\tilde{w} := w\ast_k v_2).
\]
Then, by definition, $(\tilde{v},\tilde{w},e,e')$ is an elementary move from $(v_1 \ast_k v_2)$ to $(v_1' \ast_k v_2)$. Similarly, if we have an elementary move from $v_2$ to $v_2'$, we obtain an elementary move from $(v_1 \ast_k v_2)$ to $(v_1 \ast_k v_2')$.
By definition of $\sim$, this suffices to show the desired property.
\end{proof}
We can now prove the awaited result.
\begin{proposition}
Let $\E=(C,\Sigma,\sigma,\tau)$ be an $n$-cellular extension. The equivalence relation $\sim$ on $\T[\E]$ (Definition \ref{definitionequivalence}) is the smallest categorical congruence on $\T[\E]$.
\end{proposition}
\begin{proof}
The fact that $\simeq$ is a congruence follows from
\end{proof}
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