@@ -2161,7 +2161,7 @@ Recall from Proposition \ref{prop:conduchepractical} that for an $\omega$-functo
\[
\wt{F}_a : \G[\Sigma^C]_a \to\G[\Sigma^D]_{F(a)}
\]
for any $a \in C_{n}$.
for every $a \in C_{n}$.
\end{paragr}
\begin{lemma}\label{lemmafaithful} With the notations of the above paragraph, the map
\[
...
...
@@ -2213,7 +2213,7 @@ Recall from Proposition \ref{prop:conduchepractical} that for an $\omega$-functo
\mu=(v_1,v_2,e,e') : v \to v'
\]
be an elementary move. Since, by definition,
\[\wt{F}(u)= v = v_1ev_2\]
\[\wt{F}(u)= v = v_1ev_2,\]
$u$ is necessarily of the form
\[
u=u_1\overline{e}u_2
...
...
@@ -2248,12 +2248,12 @@ Recall from Proposition \ref{prop:conduchepractical} that for an $\omega$-functo
\wt{F}(\lambda)=\mu.
\]
All that is left to prove now is ythe existence of $\overline{e'}$ with the desired properties.
All that is left to prove now is the existence of $\overline{e'}$ with the desired properties.
\begin{description}
\item[First case:] The word $e$ is of the form
\[((x\fcomp_k y )\fcomp_k z)\]
and $e'$ is of the form
and the word $e'$ is of the form
\[(x\fcomp_k (y \fcomp_k z))\]
with $x,y,z \in\T[\Sigma^D]$.
The word $\overline{e}$ is then necessarily of the form
...
...
@@ -2278,7 +2278,7 @@ Recall from Proposition \ref{prop:conduchepractical} that for an $\omega$-functo
\[
(x\fcomp_k(\ii_{\1^{n-1}_z}))
\]
and $e'$ is of the form
and the word $e'$ is of the form
\[
x
\]
...
...
@@ -2296,7 +2296,7 @@ Recall from Proposition \ref{prop:conduchepractical} that for an $\omega$-functo
\[
\wt{F}(y)=\1^{n-1}_z.
\]
Then we set
Then, we set
\[
\overline{e'}:=x.
\]
...
...
@@ -2312,7 +2312,7 @@ If $k <n-1$, we need first to use the fact that $f$ is right orthogonal to $\kap
\[
((\ii_x)\fcomp_k (\ii_y))
\]
and $e'$ is of the form
and the word $e'$ is of the form
\[
(\ii_{x\comp_k y})
\]
...
...
@@ -2334,7 +2334,7 @@ If $k <n-1$, we need first to use the fact that $f$ is right orthogonal to $\kap
\[
((x\fcomp_ky)\fcomp_l(z \fcomp_k t))
\]
and $e'$ is of the form
and the word $e'$ is of the form
\[
((x\fcomp_lz)\fcomp_k(y \fcomp_l t))
\]
...
...
@@ -2383,7 +2383,7 @@ If $k <n-1$, we need first to use the fact that $f$ is right orthogonal to $\kap
In the proof of the previous theorem, we only used the hypothesis that $F$ is right orthogonal to $\kappa_k^n$ for any $k$ such that $0\leq k < n-1$.
\end{remark}
\begin{lemma}\label{lemma:isomorphismgraphs}
Let $F : C \to D$ be an $\omega$-functor, $n>0$, $\Sigma^D \subseteq D_{n}$ and $\Sigma^C = F^{-1}(\Sigma^D)$. If $\tau^s_{\leq n}(F)$ is a discrete Conduché $n$\nbd{}functor, then for every $a \in C_{n}$
Let $F : C \to D$ be an $\omega$-functor, $n>0$, $\Sigma^D \subseteq D_{n}$ and $\Sigma^C := F^{-1}(\Sigma^D)$. If $\tau^s_{\leq n}(F)$ is a discrete Conduché $n$\nbd{}functor, then for every $a \in C_{n}$
\[
\wt{F}_a : \G[\Sigma^C]_a\to\G[\Sigma^D]_{F(a)}
\]
...
...
@@ -2417,13 +2417,13 @@ If $k <n-1$, we need first to use the fact that $f$ is right orthogonal to $\kap
we conclude that $v=u'$.
\end{proof}
\begin{proposition}\label{prop:conduchenbasis}
Let $F : C \to D$ be an $\omega$-functor, $n \in\mathbb{N}$, $\Sigma^D \subseteq D_{n}$ and $\Sigma^C = F^{-1}(\Sigma^D)$. If $\tau_{\leq n}^s(f)$ is a discrete Conduché $n$\nbd{}functor, then:
Let $F : C \to D$ be an $\omega$-functor, $n \in\mathbb{N}$, $\Sigma^D \subseteq D_{n}$ and $\Sigma^C := F^{-1}(\Sigma^D)$. If $\tau_{\leq n}^s(f)$ is a discrete Conduché $n$\nbd{}functor, then:
\begin{enumerate}
\item if $\Sigma^D$ is an $n$\nbd{}basis then so is $\Sigma^C$,
\item if $F_{n} : C_{n}\to D_{n}$ is surjective and $\Sigma^C$ is an $n$\nbd{}basis then so is $\Sigma^D$.
\end{enumerate}
\end{proposition}
\begin{proof}The case $n=0$ is trivial. We know suppose that $n>0$.
\begin{proof}The case $n=0$ is trivial. We now suppose that $n>0$.
From Lemma \ref{lemma:isomorphismgraphs} we have that for every $a \in C_n$, the map
\[
\wt{F}_a : \G[\Sigma^C]_a \to\G[\Sigma^D]_{F(a)}
...
...
@@ -2437,7 +2437,7 @@ Putting all the pieces together, we finally have the awaited proof.
\item In the case that $D$ is free, it follows immediately from the first part of the Proposition \ref{prop:conduchenbasis} that $C$ is free.
\item In the case that $C$ is free and $F_n : C_n \to D_n$ is surjective for every $n \in\mathbb{N}$, let us write $\Sigma_n^C$ for the $n$\nbd{}basis of $C$. It follows from Proposition \ref{prop:uniquebasis} and Lemma \ref{lemma:conducheindecomposable} that
\[
F^{-1}(F(\Sigma^C_n)=\Sigma_n^C.
F^{-1}(F(\Sigma^C_n))=\Sigma_n^C.
\]
Hence, we can apply the second part of Proposition \ref{prop:conduchenbasis} and $C$ is free.
