### 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} ... ...
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