Commit ba4e0f82 by Leonard Guetta

### Modifs suggérés par Garner presque terminées

parent 24f2551e
 ... ... @@ -296,14 +296,14 @@ From the previous proposition, we deduce the following very useful corollary. \item Either $\alpha$ or $\beta$ is injective on objects. \item Either $\alpha$ or $\beta$ is quasi-injective on arrows. \end{enumerate} Then, the square Then, the induced square of $\Cat$ $\begin{tikzcd} L(A) \ar[d,"L(\alpha)"] \ar[r,"L(\beta)"] &L(B) \ar[d,"L(\delta)"] \\ L(C) \ar[r,"L(\gamma)"] &L(D) \end{tikzcd}$ is a Thomason homotopy cocartesian square of $\Cat$. is Thomason homotopy cocartesian square. \end{proposition} \begin{proof} The case where $\alpha$ or $\beta$ is both injective on objects and ... ... @@ -408,16 +408,17 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \] Then, this square is Thomason homotopy cocartesian in $\Cat$. Indeed, it obviously is the image of a square of $\Rgrph$ by Then, this square is Thomason homotopy cocartesian. Indeed, it obviously is the image of a square of $\Rgrph$ by the functor $L$ and the morphism $i_1 : \sS_0 \to \sD_1$ comes from a monomorphism of $\Rgrph$. Hence, we can apply Corollary \ref{cor:hmtpysquaregraph}. \end{example} \begin{remark} Since every free category is obtained by recursively adding generators Since $i_1 : \sS_0 \to \sD_1$ is a folk cofibration% , since a Thomason homotopy % cocartesian square in $\Cat$ is also so in $\oo\Cat$ and since every free category is obtained by recursively adding generators starting from a set of objects (seen as a $0$-category), the previous example yields another proof that \emph{free} (1-)categories are \good{} (which we yields another proof that \emph{free} (1\nbd{})categories are \good{} (which we already knew since we have seen that \emph{all} (1-)categories are \good{}). \end{remark} \begin{example}[Identifying two generators] ... ... @@ -434,7 +435,7 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \] where the morphism $\sS_1 \to \sD_1$ is the one that sends the two generating arrows of $\sS_1$ to the unique generating arrow of $\sD_1$. Then this square is Thomason homotopy cocartesian in $\Cat$. is Thomason homotopy cocartesian. Indeed, it is the image by the functor $L$ of a cocartesian square in $\Rgrph$, the morphism $\sS_1 \to \sD_1$ is injective on objects and the morphism $\sS_1 \to C$ is quasi-injective on arrows. Hence, we can apply ... ... @@ -777,9 +778,8 @@ equivalent to the nerve defined in \ref{paragr:nerve}. S_n(C):= \coprod_{(x_0,\cdots,x_n)\in \Ob(C)^{\times (n+1)}}C(x_0,x_1) \times \cdots \times C(x_{n-1},x_n). \] Note that for $n=0$, the above formula reads $S_0(C)=C_0$. The face operators $\partial_i : S_{n}(C) \to S_{n-1}(C)$ are induced by horizontal composition and the degeneracy operators $s_i : S_{n}(C) \to S_{n+1}(C)$ are Note that for $n=0$, the above formula reads $S_0(C)=C_0$. For $n>0$, the face operators $\partial_i : S_{n}(C) \to S_{n-1}(C)$ are induced by horizontal composition for $0 < i  ... ... @@ -286,9 +286,11 @@ higher than$1$. Let$A$be a$1$\nbd{}category and$a$an object of$A$. Recall that we write$A/a$for the slice$1$\nbd{}category of$A$over$a$, that is the$1$\nbd{}category whose description is as follows: \begin{itemize}[label=-] \item an object of$A/a$is a pair$(a', p : a' \to a)$where$a'$is an object of$A$and$p$is an arrow of$A$, \item an arrow$(a',p) \to (a'',p')$of$A/a$is an arrow$ q : a' \to a''$of$A$such that$p'\circ q = p$, \item an arrow of$A/a$is a pair$(q,p : a' \to a)$where$p$is an arrow of$A$and$q$is an arrow of$A$of the form$q : a'' \to a'$. The target of$(q,p)$is given by$(a',p)$and the source by$(a'',p\circ q)$. %$ q : a' \to a''$of$A$such that$p'\circ q = p$. \end{itemize} and we write$\pi_a$for the canonical forgetful functor We write$\pi_afor the canonical forgetful functor \begin{aligned} \pi_{a} : A/a &\to A \\ ... ... @@ -433,7 +435,9 @@ higher than 1. we obtain that every 1\nbd{}category A is (canonically isomorphic to) the colimit \[ \colim_{a \in A} (A/a). \] % In other words, this simply say that the colimit of the Yoneda embeddingA \to % \Psh{A}$is the terminal presheaves We now proceed to prove that this colimit is homotopic with respect to the folk weak equivalences. \end{paragr} ... ... @@ -532,24 +536,32 @@ Up to Lemma \ref{lemma:basisofslice}, we fix once and for all an$\oo$\nbd{}func Beware that in the previous corollary, we did \emph{not} suppose that$X$was free. \begin{proof} Let$P$be a free$\omega$-category and$g : P \to X$a folk trivial fibration and consider the following commutative square of$\ho(\oo\Cat^{\folk})$and consider the following commutative diagram of$\ho(\oo\Cat^{\folk})$\label{comsquare} \begin{tikzcd} \displaystyle\hocolim^{\folk}_{a \in A}(P/a) \ar[d] \ar[r] & \displaystyle\colim_{a \in A}(P/a) \ar[d] \\ \displaystyle\hocolim^{\folk}_{a \in A}(X/a) \ar[r] & \displaystyle\colim_{a \in A}(X/a) \displaystyle\hocolim^{\folk}_{a \in A}(P/a) \ar[d] \ar[r] & \displaystyle\colim_{a \in A}(P/a) \ar[d] \ar[r] & P \ar[d]\\ \displaystyle\hocolim^{\folk}_{a \in A}(X/a) \ar[r] & \displaystyle\colim_{a \in A}(X/a) \ar[r] & X \end{tikzcd} where the vertical arrows are induced by the arrows where the middle and most left vertical arrows are induced by the arrows $g/a : P/a \to X/a. g/a : P/a \to X/a,$ Since trivial fibrations are stable by pullback,$g/a$is a trivial fibration. This proves that the left vertical arrow of square (\ref{comsquare}) is an isomorphism. and the most right vertical arrow is induced by$g$. Since trivial fibrations are stable by pullback,$g/a$is a trivial fibration. This proves that the most left vertical arrow of diagram \eqref{comsquare} is an isomorphism. Moreover, from Paragraph \ref{paragr:unfolding} and Lemma \ref{lemma:colimslice}, we deduce that the right vertical arrow of (\ref{comsquare}) can be identified with the image of$g : P \to X$in$\ho(\omega\Cat)$and hence is an isomorphism. Finally, from Proposition \ref{prop:sliceiscofibrant} and Corollary \ref{cor:cofprojms}, we deduce that the top horizontal arrow of (\ref{comsquare}) is an isomorphism. From Proposition \ref{prop:sliceiscofibrant} and Corollary \ref{cor:cofprojms}, we deduce that the arrow $\hocolim_{a \in A}^{\folk}(P/a)\to \colim_{a \in A}(P/a)$ is an isomorphism. Moreover, from Lemma \ref{lemma:colimslice}, we know that the arrows $\colim_{a \in A}(P/a)\to P$ and $\colim_{a \in A}(X/a)\to X$ are isomorphisms. By an immediate 2-out-of-3 property, this proves the result. Finally, since$g$is a folk weak equivalence, the most right vertical arrow of diagram \eqref{comsquare} is an isomorphism and by an immediate 2-out-of-3 property this proves that all arrows of \eqref{comsquare} are isomorphisms. In particular, so is the composition of the two bottom horizontal arrows, which is what we desired to show. \end{proof} We now move on to the next step needed to prove that every$1$\nbd{}category is \good{}. For that purpose, let us recall a construction commonly referred to as the Grothendieck construction''. \begin{paragr} ... ...  ... ... @@ -218,7 +218,7 @@ From now on, we will consider that the category$\Psh{\Delta}$is equipped with \end{corollary} We will speak of Thomason homotopy colimits'' and Thomason homotopy cocartesian squares'' for homotopy colimits and homotopy cocartesian squares in the localizer$(n\Cat^{\Th},\W_n^{\Th})$. the localizer$(n\Cat^{\Th},\W_n^{\Th})$. (See also \ref{paragr:thomhmtpycol} below.) Another consequence of Gagna's theorem is the following corollary. ... ... @@ -232,7 +232,7 @@ From now on, we will consider that the category$\Psh{\Delta}$is equipped with Corollaries \ref{cor:thomhmtpycocomplete} and \ref{cor:thomsaturated} would also follow from the existence of a model structure on$n\Cat$with$\W^{\Th}_n$as the weak equivalences. For$n=1$, this was established by Thomason \cite{thomason1980cat}, and for$n=2$, by Ara and Maltsiniotis \cite{ara2014vers}. For$n>3$, the existence of such a model structure is conjectured but not yet established. \end{remark} By definition, for all$1 \leq n \leq m \leq \omega$, the canonical inclusion $n\Cat \hookrightarrow m\Cat$ sends the Thomason equivalences of$n\Cat$to Thomason equivalences of$m\Cat$. Hence, it induces a morphism of localizers and then a morphism of op\nbd{}prederivators$\Ho(n\Cat^\Th) \to \Ho(m\Cat^{\Th})$. \begin{proposition} \begin{proposition}\label{prop:nthomeqder} For all$1 \leq n \leq m \leq \omega$, the canonical morphism $\Ho(n\Cat^\Th) \to \Ho(m\Cat^{\Th}) ... ... @@ -248,6 +248,21 @@ From now on, we will consider that the category \Psh{\Delta} is equipped with \end{tikzcd}$ \end{proof} \begin{paragr}\label{paragr:thomhmtpycol} It follows from the previous proposition that for all$1 \leq n \leq m \leq \omega$, the morphism$\Ho(n\Cat^\Th) \to \Ho(m\Cat^{\Th})$of op\nbd{}prederivators is homotopy cocontinuous and reflects homotopy colimits (in an obvious sense). Hence, given a diagram$d : I \to n\Cat$with$n>0$, we can harmlessly use the notation $\hocolim^{\Th}_{i \in I}(d)$ for both the Thomason homotopy colimits in$n\Cat$and in$\oo\Cat$(or any$m\Cat$with$n\leq m$). Similarly, a commutative square of$n\Cat$is Thomason homotopy cocartesian in$n\Cat$if and only if it is so in$\oo\Cat$. Hence, there is really no ambiguity when simply calling such a square \emph{Thomason homotopy cocartesian}. \end{paragr} \section{Tensor product and oplax transformations} Recall that$\oo\Cat$can be equipped with a monoidal product$\otimes$, introduced by Al-Agl and Steiner in \cite{al1993nerves} and by Crans in \cite{crans1995combinatorial}, commonly referred to as the \emph{Gray tensor product}. The implicit reference for this section is \cite[Appendices A and B]{ara2016joint}. \begin{paragr} ... ... @@ -291,7 +306,7 @@ From now on, we will consider that the category$\Psh{\Delta}$is equipped with \begin{tikzcd} & Y \\ X \ar[ru,"u"] \ar[r,"\alpha"] \ar[rd,"v"']& \homlax(\sD_1,Y) \ar[u,"\pi_0^Y"'] \ar[d,"\pi_1^Y"] \\ & Y & Y, \end{tikzcd} \] where$\pi^Y_0$and$\pi^Y_1$are induced by the two$\oo$\nbd{}functors$\sD_0 \to \sD_1$and where we implicitly used the isomorphism$\homlax(\sD_0,Y)\simeq Y\$, is commutative. ... ...
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!