Commit 5e7b0879 authored by Leonard Guetta's avatar Leonard Guetta
Browse files

idem

parent 39423129
......@@ -929,7 +929,7 @@ of $2$-categories.
\[
(\binerve(C))_{\bullet,m} = NS(C).
\]
The result follows then from Lemma \ref{bisimpliciallemma} and the fact that the
The result then follows from Lemma \ref{bisimpliciallemma} and the fact that the
weak equivalences of simplicial sets are stable by coproducts and finite
products.
\end{proof}
......@@ -943,7 +943,7 @@ of $2$-categories.
\[
\binerve(C)_{\bullet,m}=N(V_m(C)).
\]
The result follows them from Lemma \ref{bisimpliciallemma}.
The result then follows from Lemma \ref{bisimpliciallemma}.
\end{proof}
\begin{paragr}
It also follows from Lemma \ref{lemma:binervthom} that the bisimplicial nerve
......@@ -990,7 +990,7 @@ of $2$-categories.
From Proposition \ref{prop:streetvsbisimplicial}, we deduce the proposition
below which contains two useful criteria to detect Thomason homotopy
cocartesian square of $2\Cat$.
cocartesian squares of $2\Cat$.
\end{paragr}
\begin{proposition}\label{prop:critverthorThomhmtpysquare}
Let
......@@ -1036,13 +1036,14 @@ of $2$-categories.
\ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end]
\end{tikzcd}
\]
of 2\nbd{}categories. If $A$,$B$ and $C$ are free and \good{}, if at least $u$
of 2\nbd{}categories. If $A$, $B$ and $C$ are free and \good{}, if at least
one of $u$
or $f$ is a folk cofibration and if the square is Thomason homotopy
cocartesian, then $D$ is \good{}.
\end{paragr}
\begin{paragr}
Let $n,m \geq 0$. We denote by $A_{(m,n)}$ the free $2$-category with only one
generating $2$-cell whose source is a chain of length $m$ and its target a
generating $2$-cell whose source is a chain of length $m$ and whose target is a
chain of length $n$:
\[
\underbrace{\overbrace{\begin{tikzcd}[column sep=small, ampersand
......@@ -1128,7 +1129,7 @@ of $2$-categories.
non-trivial $1$\nbd{}cell of $\Delta_1$ to the target of the generating
$2$-cell of $A_{(1,1)}$. It is not hard to check that $\tau$ is strong
deformation retract and thus, a co-universal Thomason equivalence (Lemma
\ref{lemma:pushoutstrngdefrtract}). Hence, the morphism $A_{(1,1)} \to
\ref{lemma:pushoutstrngdefrtract}). Hence, the morphism $\Delta_n \to
A_{(1,n)}$ is also a (co-universal) Thomason equivalence and the square is
Thomason homotopy cocartesian (Lemma \ref{lemma:hmtpycocartsquarewe}). Now,
the morphism $\tau : \Delta_1 \to A_{(1,1)}$ is also a folk cofibration and
......@@ -1147,7 +1148,7 @@ of $2$-categories.
\end{tikzcd}
\]
where $\sigma : \Delta_1 \to A_{(1,1)}$ is the $2$-functor that sends the
unique non trivial $1$\nbd{}cell of $\Delta_1$ the source of the generating
unique non trivial $1$\nbd{}cell of $\Delta_1$ to the source of the generating
$2$\nbd{}cell of $A_{(1,1)}$, we can prove that $A_{(m,1)}$ is \good{} and has
the homotopy type of a point.
......@@ -1163,7 +1164,7 @@ of $2$-categories.
where $\tau$ is the $2$-functor that sends the unique non-trivial $1$-cell of
$\Delta_1$ to the target of the generating $2$-cell of $A_{(m,1)}$. This
$2$-functor is once again a folk cofibration, but it is \emph{not} in general
a co-universal Thomason equivalence (it would if we had made the hypothesis that
a co-universal Thomason equivalence (it would be if we had made the hypothesis that
$m\neq 0$, but we did not). However, since we made the hypothesis that $n\neq
0$, it follows from Lemma \ref{lemma:istrngdefrtract} that $i : \Delta_1 \to
\Delta_n$ is a co-universal Thomason equivalence. Hence, the previous square
......@@ -1193,7 +1194,7 @@ results at the end of the previous section.
\item generating $2$\nbd{}cells: $\alpha : f \Rightarrow 1_A$, $\beta : g
\Rightarrow 1_A$.
\end{itemize}
In picture, this gives
In pictures, this gives
\[
\begin{tikzcd}[column sep=huge]
A \ar[r,bend left=75,"f",""{name=A,below}]\ar[r,bend
......@@ -1204,8 +1205,8 @@ results at the end of the previous section.
\end{tikzcd}
\text{ or }
\begin{tikzcd}
A. \ar[loop,in=50,out=130,distance=1.5cm,"f"',""{name=A,below}]
\ar[loop,in=-50,out=-130,distance=1.5cm,"g",""{name=B,above}]
A. \ar[loop,in=50,out=130,distance=1.5cm,"f",""{name=A,below}]
\ar[loop,in=-50,out=-130,distance=1.5cm,"g"',""{name=B,above}]
\ar[from=A,to=1-1,Rightarrow,"\alpha"]
\ar[from=B,to=1-1,Rightarrow,"\beta"]
\end{tikzcd}
......@@ -1281,7 +1282,7 @@ Let us now get into more sophisticated examples.
\]
and let $F : P \to P'$ be the unique $2$\nbd{}functor such that
\begin{itemize}[label=-]
\item $F(A)=A'$ and $F(B')=B$,
\item $F(A)=A'$ and $F(B)=B'$,
\item $F(f)=F(g)=h$,
\item $F(\alpha)=\gamma$ and $F(\beta)=1_h$.
\end{itemize}
......@@ -1363,9 +1364,9 @@ Let us now get into more sophisticated examples.
\]
Let $H : \sS_2 \to P''$ be the unique $2$\nbd{}functor such that:
\begin{itemize}[label=-]
\item $H(\overline{A})=(\overline{B})=A''$,
\item $H(\overline{A})=H(\overline{B})=A''$,
\item $H(i)=l$ and $H(j)=1_{A''}$,
\item $H(\delta)=\lambda$ and $H(\epsilon)=\lambda$.
\item $H(\delta)=\lambda$ and $H(\epsilon)=\mu$.
\end{itemize}
Let us prove that $H$ is a Thomason equivalence using Corollary
\ref{cor:criterionThomeqII}. In order to do so, we have to compute $V_k(H) :
......@@ -1403,13 +1404,13 @@ Let us now get into more sophisticated examples.
and three arrows. More generally, it is a tedious but harmless exercise to
prove that for every $k>0$, the category $V_k(P'')$ is the
free category on the graph that has one object $A''$ and $2k+1$
arrows which are of either one of the following forms:
arrows which are of one of the following forms:
\begin{itemize}[label=-]
\item $(1_l,\cdots,1_l,\lambda,1^2_{A''},\cdots,1^2_{A''})$,
\item $(1_l,\cdots,1_l,\mu,1^2_{A''},\cdots,1^2_{A''})$,
\item $(1_l,\cdots,1_l)$.
\end{itemize}
Once again, the functor $V_k(H)$ comes from a morphism a reflexive graphs and
Once again, the functor $V_k(H)$ comes from a morphism of reflexive graphs and
is obtained by ``killing the generator $(1_j,\cdots,1_j)$''. Hence, it is a
Thomason equivalence and thus, so is $H$. This proves that $P''$ has the
homotopy type of $\sS_2$.
......@@ -1490,7 +1491,7 @@ Let us now get into more sophisticated examples.
Notice that we have $F\circ G = \mathrm{id}_{P'}$, which means that $P'$ is a
retract of $P$. In particular, $\sH^{\sing}(P)$ is a retract of
$\sH^{\sing}(P')$ and since $P'$ has the homotopy type of a
$K(\mathbb{Z},2)$ (see \ref{paragr:bubble}), this proves that $P$ have
$K(\mathbb{Z},2)$ (see \ref{paragr:bubble}), this proves that $P$ has
non-trivial singular homology groups in all even dimension. But since it is a
free $2$\nbd{}category, all its polygraphic homology groups are trivial strictly above
dimension $2$, which means that $P$ is \emph{not} \good{}.
......@@ -1703,7 +1704,7 @@ Now let $\sS_2$ be labelled as
\end{tikzcd}
\]
Let us prove that this $2$\nbd{}category is \good{}. Let $P_0$ be the
sub-$1$category of $P$ spanned by $A$, $B$ and $g$, let $P_1$ be the
sub-$1$\nbd{}category of $P$ spanned by $A$, $B$ and $g$, let $P_1$ be the
sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and
$\delta$ and let $P_2$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,
$B$, $f$, $g$, $\alpha$ and $\beta$. The $2$\nbd{}categories $P_1$ and $P_2$
......@@ -1831,7 +1832,7 @@ Now let $\sS_2$ be labelled as
A \ar[r,"f",shift left] \ar[r,"g"',shift right] & B \ar[r,"h",shift left] \ar[r,"i"',shift right] & C.
\end{tikzcd}
\]
This implies that square \ref{squarebouquetvertical} is cocartesian for $k=0$ and in
This implies that square \eqref{squarebouquetvertical} is cocartesian for $k=0$ and in
virtue of Corollary \ref{cor:hmtpysquaregraph} it is also Thomason homotopy
cocartesian for this value of $k$. For $k>0$, the category $V_k(P')$ has two objects $A$
and $B$ and an arrow $A \to B$ is a $k$\nbd{}tuple of one of the following form
......@@ -1861,7 +1862,7 @@ Now let $\sS_2$ be labelled as
\item $(1_h,\cdots,1_h)$,
\item $(1_i,\cdots,1_i)$,
\end{itemize}
and with no other arrows. This implies that square \ref{squarebouquetvertical}
and with no other arrows. This implies that square \eqref{squarebouquetvertical}
is cocartesian for every $k>0$ and in virtue of Corollary
\ref{cor:hmtpysquaregraph} it is also Thomason homotopy cocartesian for these values of $k$.
Altogether, this proves that square \eqref{squarebouquetbis} is Thomason
......@@ -1878,7 +1879,7 @@ homotopy type of the torus.
\item generating $2$\nbd{}cell: $\alpha : g\comp_0 f \Rightarrow f \comp_0
g$.
\end{itemize}
In picture, this gives:
In pictures, this gives:
\[
\begin{tikzcd}
A \ar[r,"f"] \ar[d,"g"'] & A \ar[d,"g"] \\
......@@ -1907,7 +1908,7 @@ homotopy type of the torus.
\item $F(\alpha)=1_{(1,1)}$.
\end{itemize}
This last equation makes sense since $(1,1)=(0,1)+(1,0)=(1,0)+(0,1)$. For every
$1$\nbd{}cell $w$ of $P$ (encoded as a finite words on the alphabet $\{f,g\}$)
$1$\nbd{}cell $w$ of $P$ (encoded as a finite words in the alphabet $\{f,g\}$)
such that $f$ appears $n$ times and $g$ appears $m$ times, we have
$F(w)=(n,m)$. Let us prove that $F$ is a Thomason equivalence using a dual of
\cite[Corollaire 5.26]{ara2020theoreme} (see Remark 5.20 of op.\ cit.). If we
......@@ -2044,7 +2045,7 @@ homotopy type of the torus.
\end{definition}
\begin{paragr}
The archetypal example of a $2$\nbd{}category that is \emph{not} bubble-free
is $B^2\mathbb{N}$. Another non-bubble $2$\nbd{}category if the one from Paragraph \ref{paragr:anothercounterexample}. It is
is $B^2\mathbb{N}$. Another non-bubble $2$\nbd{}category is the one from Paragraph \ref{paragr:anothercounterexample}. It is
remarkable that of all the free $2$\nbd{}categories we have seen so far, these
are the only examples that are non-\good{}. This motivates the following conjecture.
\end{paragr}
......
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment