### edited a LOT of typos. I should do some proofreading of the section on the homology of 1-categories

parent 22df770c
 ... ... @@ -271,7 +271,7 @@ From the previous proposition, we deduce the following very useful corollary. \begin{paragr} By working a little more, we obtain the more general result stated in the proposition below. Let us say that a morphism of reflexive graphs, $\alpha : A \to B$, is \emph{quasi-injective on arrows} when graphs $\alpha : A \to B$ is \emph{quasi-injective on arrows} when for all $f$ and $g$ arrows of $A$, if $\alpha(f)=\alpha(g), ... ... @@ -312,11 +312,11 @@ From the previous proposition, we deduce the following very useful corollary. only have to treat the case when \alpha is injective on objects and \beta is quasi-injective on arrows; the remaining case being symmetric. Let use denote by E the set of objects of B that lies in the image of Let use denote by E the set of objects of B that are in the image of \beta. For each element x of E, we denote by F_x the fiber'' of x, that is the set of objects of A that \beta sends to x. We consider the set E and each F_x as discrete reflexive graphs, i.e. reflexive graphs with no non-unital arrow. Now, let G be the reflexive graph defined with the set E and each F_x as discrete reflexive graphs, i.e.\ reflexive graphs with no non-unit arrows. Now, let G be the reflexive graph defined with the following cocartesian square \[ \begin{tikzcd} ... ... @@ -369,17 +369,16 @@ From the previous proposition, we deduce the following very useful corollary. L of square \textcircled{\tiny \textbf{3}} is homotopy cocartesian. Hence, in virtue of Lemma \ref{lemma:pastinghmtpycocartesian}, all we have to show is that the image by L of square \textcircled{\tiny \textbf{2}} is homotopy cocartesian. On the other hand, we know that both morphisms cocartesian. On the other hand, the morphism \[ \coprod_{x \in E}F_x \to A \text{ and } A \to C \coprod_{x \in E}F_x \to A$ are injective on arrows, but since $\coprod_{x \in E}F_x$ does not have non-units arrows, both these morphisms are actually monomorphisms. Hence, using Corollary \ref{cor:hmtpysquaregraph}, we deduce that image by $L$ of square \textcircled{\tiny \textbf{1}} and of the pasting of squares \textcircled{\tiny \textbf{1}} and \textcircled{\tiny \textbf{2}} are homotopy cocartesian. This proves that the image by $L$ of square \textcircled{\tiny \textbf{2}} is homotopy cocartesian. is a monomorphism and thus, using Corollary \ref{cor:hmtpysquaregraph}, we deduce that the image by $L$ of square \textcircled{\tiny \textbf{1}} and of the pasting of squares \textcircled{\tiny \textbf{1}} and \textcircled{\tiny \textbf{2}} are homotopy cocartesian. This proves that the image by $L$ of square \textcircled{\tiny \textbf{2}} is homotopy cocartesian. \end{proof} We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \ref{prop:hmtpysquaregraphbetter} to a few examples. ... ... @@ -391,6 +390,7 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \begin{tikzcd} \sS_0 \ar[d] \ar[r,"{\langle A,B \rangle}"] & C \ar[d] \\ \sD_0 \ar[r] & C'. \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \] Then, this square is Thomason homotopy cocartesian. Indeed, it is obviously ... ... @@ -405,12 +405,13 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \begin{tikzcd} \sS_0 \ar[d,"i_1"] \ar[r,"{\langle A, B \rangle}"] & C \ar[d] \\ \sD_1 \ar[r] & C'. \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \] Then, this square is Thomason homotopy cocartesian in $\Cat$ (when equipped with the Thomason equivalences). 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 Then, this square is Thomason homotopy cocartesian in $\Cat$. 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} ... ... @@ -428,6 +429,7 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \begin{tikzcd} \sS_1\ar[d] \ar[r,"{\langle f, g \rangle}"] &C \ar[d] \\ \sD_1 \ar[r] & C', \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \] where the morphism $\sS_1 \to \sD_1$ is the one that sends the two generating ... ... @@ -449,11 +451,14 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \begin{tikzcd} \sD_1 \ar[d] \ar[r,"\langle f \rangle"] & C \ar[d] \\ \sD_0 \ar[r] & C'. \ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end] \end{tikzcd} \] Then, this above square is Thomason homotopy cocartesian. Indeed, it obviously is the image of a square in $\Rgrph$ by the functor $L$ and since the source and target of $f$ are different, the top map comes from a monomorphism of $\Rgrph$. Then, this above square is Thomason homotopy cocartesian. Indeed, it obviously is the image of a cocartesian square in $\Rgrph$ by the functor $L$ and since the source and target of $f$ are different, the top map comes from a monomorphism of $\Rgrph$. Hence, we can apply Corollary \ref{cor:hmtpysquaregraph}. \end{example} \begin{remark} Note that in the previous example, we see that it was useful to consider the ... ... @@ -472,7 +477,7 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition X : \Delta^{\op} \times \Delta^{\op} \to \Set. \] In a similar fashion as for simplicial sets (\ref{paragr:simpset}), for $n,m \geq 0$, we use the notation \geq 0, we use the notations \begin{align*} X_{n,m} &:= X([n],[m]) \\ \partial_i^h &:=X(\delta^i,\mathrm{id}) : X_{n+1,m} \to X_{n,m}\\ ... ... @@ -535,9 +540,9 @@ We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition $\delta_! \dashv \delta^* \dashv \delta_*.$ We say that a morphism a bisimplicial sets,f : X \to Y$, is a \emph{diagonal We say that a morphism$f : X \to Y$of bisimplicial sets is a \emph{diagonal weak equivalence} (resp.\ \emph{diagonal fibration}) when$\delta^*(f)$is a weak equivalence of simplicial sets (resp.\ fibration of simplicial sets). By weak equivalence (resp.\ fibration) of simplicial sets. By definition,$\delta^*$induces a morphism of op-prederivators $\overline{\delta^*} : \Ho(\Psh{\Delta\times \Delta}^{\mathrm{diag}}) \to ... ... @@ -687,7 +692,7 @@ In practice, we will use the following corollary. C \ar[r,"v"] & D \end{tikzcd}$ be a commutative square in the category of bisimplicial sets satisfying either of the be a commutative square in the category of bisimplicial sets satisfying at least one of the two following conditions: \begin{enumerate}[label=(\alph*)] \item For every$n\geq 0$, the square of simplicial sets ... ... @@ -866,7 +871,7 @@ an equivalent definition of the bisimplicial nerve which uses the other directio For$k=0$, we define$V_0(C)$to be the category obtained from$C$by simply forgetting the$2$\nbd{}cells (which is nothing but$\tau^{s}_{\leq 1}(C)$with the notations of \ref{paragr:defncat}). The correspondence$n \mapsto V_n(C)$defines to a simplicial object in$\Cat$V_n(C)$ defines a simplicial object in $\Cat$ $V(C) : \Delta^{\op} \to \Cat,$ ... ... @@ -926,7 +931,7 @@ of $2$-categories. $(\binerve(C))_{\bullet,m} = NS(C).$ The result follows then from Lemma \ref{bisimpliciallemma} and the fact that The result follows then from Lemma \ref{bisimpliciallemma} and the fact that the weak equivalences of simplicial sets are stable by coproducts and finite products. \end{proof} ... ... @@ -995,7 +1000,7 @@ of $2$-categories. A \ar[r,"u"]\ar[d,"f"] & B \ar[d,"g"] \\ C \ar[r,"v"] & D \end{tikzcd}\end{equation} be a square in $2\Cat$ satisfying either of the following conditions: be a square in $2\Cat$ satisfying at least one of the two following conditions: \begin{enumerate}[label=(\alph*)] \item For every $n\geq 0$, the square $... ... @@ -1106,7 +1111,7 @@ of 2-categories. By definition we have r \circ i = 1_{\Delta_1}. Now, the natural order on \Delta_n induces a natural transformation \[ \alpha : i\circ r \Rightarrow \mathrm{id}_{\Delta_n}, \alpha : \mathrm{id}_{\Delta_n} \Rightarrow i\circ r,$ and it is straightforward to check that $\alpha \ast i = \mathrm{id}_i$. \end{proof} ... ... @@ -1153,6 +1158,8 @@ of $2$-categories. \begin{tikzcd} \Delta_1 \ar[r,"i"] \ar[d,"\tau"] & \Delta_n \ar[d] \\ A_{(m,1)} \ar[r] & A_{(m,n)}, \ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"] \end{tikzcd} \] where $\tau$ is the $2$-functor that sends the unique non-trivial $1$-cell of ... ... @@ -1160,7 +1167,7 @@ of $2$-categories. $2$-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 $m\neq 0$, but we did not). However, since we made the hypothesis that $n\neq 0$, it follows from Lemma \ref{lemma:istrngdefrtract} that $i : \Delta \to 0$, it follows from Lemma \ref{lemma:istrngdefrtract} that $i : \Delta_1 \to \Delta_n$ is a co-universal Thomason equivalence. Hence, the previous square is Thomason homotopy cocartesian and $A_{(m,n)}$ has the homotopy type of a point. Since $A_{(m,1)}$, $\Delta_1$ and $\Delta_n$ are \good{}, this shows ... ... @@ -1185,8 +1192,8 @@ results at the end of the previous section. \begin{itemize}[label=-] \item generating $0$\nbd{}cell: $A$, \item generating $1$\nbd{}cells: $f,g : A \to A$, \item generating $2$\nbd{}cells: $\alpha : f \Rightarrow 1_A$, $\beta : 1_A \Rightarrow g$. \item generating $2$\nbd{}cells: $\alpha : f \Rightarrow 1_A$, $\beta : g \Rightarrow 1_A$. \end{itemize} In picture, this gives $... ... @@ -1229,9 +1236,9 @@ results at the end of the previous section.$ where $\tau : \Delta_1 \to A_{(1,1)}$ has already been defined in \ref{paragr:Amn}. We have seen that $\tau$ is a co\nbd{}universal Thomason equivalence. In particular, so is $\sD_0 \to A_{(0,1)}$ (as $\sD_0$ and $\Delta_0$ are two different names for the same category) and thus, square \eqref{square:lemniscate} is Thomason homotopy cocartesian. This proves that equivalence and thus, so is $\sD_0 \to A_{(0,1)}$ (as $\sD_0$ and $\Delta_0$ are two different names for the same category). Hence, square \eqref{square:lemniscate} is Thomason homotopy cocartesian and this proves that $P$ is \good{} and has the homotopy type of a point. All the variations by reversing the direction of $\alpha$ or $\beta$ work ... ... @@ -1363,7 +1370,7 @@ Let us now get into more sophisticated examples. \item $H(\delta)=\lambda$ and $H(\epsilon)=\lambda$. \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 $V(H)_k : \ref{cor:criterionThomeqII}. In order to do so, we have to compute$V_k(H) : V_k(\sS_2) \to V_k(P'')$for every$k\geq 0$. For$k=0$, the category$V_0(\sS_2)$is the free category on the graph $... ... @@ -1381,10 +1388,10 @@ Let us now get into more sophisticated examples. the generator j''. Hence, it is a Thomason equivalence of categories. For k>0, the category V_k(\sS_2) has two objects \overline{A} and \overline{B} and an arrow \overline{A} \to \overline{B} is a k\nbd{}tuple of either one of the following form k\nbd{}tuple of one of the following forms: \begin{itemize}[label=-] \item (1_i,\cdots,1_i,\delta,1_j,\cdots,1_j) \item (1_i,\cdots,1_i,\epsilon,1_j,\cdots,1_j) \item (1_i,\cdots,1_i,\delta,1_j,\cdots,1_j), \item (1_i,\cdots,1_i,\epsilon,1_j,\cdots,1_j), \item (1_i,\cdots,1_i), \item (1_j,\cdots,1_j), \end{itemize} ... ... @@ -1394,14 +1401,14 @@ Let us now get into more sophisticated examples. every 2\nbd{}cell of P'' (except for \1^2_{A''}) is uniquely encoded as a finite word on the alphabet that has three symbols : 1_l, \lambda and \mu. Concatenation corresponding to the 0\nbd{}composition of these cells. This means exactly that V_1(P'') is free on the graph that has one objects This means exactly that V_1(P'') is free on the graph that has one object 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 objects A'' and 2k+1 arrows which are of either one of the following form free category on the graph that has one object A'' and 2k+1 arrows which are of either 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,\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 ... ... @@ -1561,11 +1568,11 @@ Now let \sS_2 be labelled as \[ \begin{tikzcd} C \ar[r,bend left=75,"f",""{name=A,below,pos=9/20},""{name=C,below,pos=11/20}] left=75,"h",""{name=A,below,pos=9/20},""{name=C,below,pos=11/20}] \ar[r,bend right=75,"g"',""{name=B,above,pos=9/20},""{name=D,above,pos=11/20}] & D. \ar[from=C,to=D,bend left,"\alpha",Rightarrow] \ar[from=A,to=B,bend right,"\beta"',Rightarrow] right=75,"i"',""{name=B,above,pos=9/20},""{name=D,above,pos=11/20}] & D. \ar[from=C,to=D,bend left,"\gamma",Rightarrow] \ar[from=A,to=B,bend right,"\delta"',Rightarrow] \end{tikzcd}$ Notice that$P$is obtained as the following amalgamated sum: ... ... @@ -1583,9 +1590,8 @@ Now let$\sS_2$be labelled as have to show that the induced square of$\Cat$\begin{equation} \begin{tikzcd} V_k(\sS_0) \ar[d] \ar[r,"i_1"] & V_k(\sD_1) \ar[d]\\ V_k(\sS_0) \ar[r,"{V_k(\langle C,D \rangle)}"] \ar[d,"V_k(i_1)"] & V_k(\sS_2) \ar[d]\\ V_k(\sD_1) \ar[r] & V_k(P) \ar[from=1-1,to=2-2,"\ulcorner",phantom,very near end] \end{tikzcd}\label{eq:squarebouquethybridvertical} \end{equation} is Thomason homotopy cocartesian for every$k \geq 0$. Notice first that we ... ... @@ -1603,21 +1609,22 @@ Now let$\sS_2$be labelled as A. \ar[loop above,"f"] \ar[loop below,"g"] \end{tikzcd} \] In particular, square \eqref{eq:squarebouquethybridvertical} is cocartesian for$k=0$and we are in the situation of identification of two objects of a free category (see Example In particular, square \eqref{eq:squarebouquethybridvertical} is cocartesian for$k=0$and we are in the situation of identification of two objects of a free category (see Example \ref{example:identifyingobjects}). Hence, square \eqref{eq:squarebouquethybridvertical} is Thomason cocartesian for$k=0$. Similarly, for$k>0$, we have already seen the$V_k(\sS_2)$is the free category on the graph that has$2$objects and$2k+2$parallel arrows between these two objects and the category$V_k(P)$is the free category on the graph that has one object and whose arrows are$k$\nbd{}tuples of one of the following form: \eqref{eq:squarebouquethybridvertical} is Thomason cocartesian for$k=0$. Similarly, for$k>0$, we have already seen that$V_k(\sS_2)$is the free category on the graph that has$2$objects and$2k+2$parallel arrows between these two objects and we leave as an easy exercice to the reader to check that the category$V_k(P)$is the free category on the graph that has one object and$2k+2$arrows, which are the$k$\nbd{}tuples of one of the following forms: \begin{itemize}[label=-] \item$(1_f,\cdots,\alpha,\cdots,1_g)$, \item$(1_f,\cdots,\beta,\cdots,1_g)$, \item$(1_f,\cdots,1_f)$, \item$(1_f,\cdots,1_f)$. \item$(1_g,\cdots,1_g)$. \end{itemize} In particular, square \eqref{eq:squarebouquethybridvertical} is again a cocartesian square of identification of two objects of a free category, and ... ... @@ -1788,8 +1795,8 @@ Now let$\sS_2$be labelled as \] Let us prove that$P$is \good{}. Let$P'$be the sub-$2$\nbd{}category of$P$spanned by$A$,$B$,$f$,$g$,$\alpha$and$\beta$and let$P''$be the sub-$2$\nbd{}category of$P$spanned by$B$,$C$,$h$,$i$,$\beta$and$\gamma$. These two$2$\nbd{}categories are copies of$\sS_2$and we have a sub-$2$\nbd{}category of$P$spanned by$B$,$C$,$h$,$i$,$\gamma$and$\delta$. These two$2$\nbd{}categories are copies of$\sS_2$and we have a cocartesian square \begin{equation}\label{squarebouquetbis} \begin{tikzcd} ... ... @@ -1807,8 +1814,8 @@ Now let$\sS_2$be labelled as V_k(P'') \ar[r] & V_k(P) \end{tikzcd} \end{equation} is Thomason homotopy cocartesian. For$k=0$, we have$V_0(\sD_0) \simeq \sD_0$and the categories$V_0(P')$,$V_0(P'')$and$V_0(P)$are respectively free on is Thomason homotopy cocartesian. For every$k\geq 0$, we have$V_k(\sD_0) \simeq \sD_0$and for$k=0$the categories$V_0(P')$,$V_0(P'')$and$V_0(P)$are respectively free on the graphs $\begin{tikzcd} ... ... @@ -1826,9 +1833,9 @@ Now let \sS_2 be labelled as A \ar[r,"f",shift left] \ar[r,"g"',shift right] & B \ar[r,"h",shift left] \ar[r,"i"',shift right] & C. \end{tikzcd}$ For$k=0$, this implies that square \ref{squarebouquetvertical} is cocartesian and in This implies that square \ref{squarebouquetvertical} is cocartesian for$k=0$and in virtue of Corollary \ref{cor:hmtpysquaregraph} it is also Thomason homotopy cocartesian. For$k>0$, the category$V_k(P')$has two objects$A$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 \begin{itemize}[label=-] \item$(1_f,\cdots,1_f,\alpha,1_g,\cdots,1_g)$, ... ... @@ -1841,7 +1848,7 @@ Now let$\sS_2$be labelled as arrows between these two objects. The same goes for$V_k(P'')$since$P'$and$P''$are isomorphic. Similarly, the category$V_k(P)$is free on the graph that has three objects$A$,$B$,$C$, whose arrows from$A$to$B$are$k$\nbd{}tuple of one of the following form are$k$\nbd{}tuples of one of the following form \begin{itemize}[label=-] \item$(1_f,\cdots,1_f,\alpha,1_g,\cdots,1_g)$, \item$(1_f,\cdots,1_f,\beta,1_g,\cdots,1_g)$, ... ... @@ -1857,8 +1864,8 @@ Now let$\sS_2$be labelled as \item$(1_i,\cdots,1_i)$, \end{itemize} and with no other arrows. This implies that square \ref{squarebouquetvertical} is cocartesian for$k>0$and in virtue of Corollary \ref{cor:hmtpysquaregraph} it is also Thomason homotopy cocartesian for$k>0$. 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 homotopy cocartesian. Hence,$P$is \good{} and has the homotopy type of a bouquet of two$2$\nbd{}spheres. ... ... @@ -1870,7 +1877,7 @@ homotopy type of the torus. \begin{itemize}[label=-] \item generating$0$\nbd{}cell:$A$, \item generating$1$\nbd{}cells:$f , g : A \to A$, \item generating$2$\nbd{}cells:$\alpha : g\comp_0 f \Rightarrow f \comp_0 \item generating $2$\nbd{}cell: $\alpha : g\comp_0 f \Rightarrow f \comp_0 g$. \end{itemize} In picture, this gives: ... ... @@ -1892,8 +1899,7 @@ homotopy type of the torus. $f \cdots fg\cdots g$ with $n$ occurrences of $f$ and $m$ occurrences of $g$. where $f$ is repeated $n$ times and $g$ is repeated $m$ times. Recall that the category $B^1(\mathbb{N}\times\mathbb{N})$ is nothing but the monoid $\mathbb{N}\times\mathbb{N}$ considered as a category with only one object, and let $F : P \to B^1(\mathbb{N}\times\mathbb{N})$ be the unique ... ... @@ -1963,13 +1969,13 @@ homotopy type of the torus. \sH^{\pol}(B^1(\mathbb{N}\times\mathbb{N})) \end{tikzcd} \] Since $F$ is a Thomason equivalence, the bottom horizontal arrow is an Since $F$ is a Thomason equivalence, the top horizontal arrow is an isomorphism and since $B^1(\mathbb{N}\times\mathbb{N})$ is a $1$\nbd{}category, it is \good{} (Theorem \ref{thm:categoriesaregood}), which means that the right vertical arrow is an isomorphism. The $1$\nbd{}category $B^1(\mathbb{N}\times \mathbb{N})$ is not free but since it has the homotopy type of the torus, we have $H^{\sing}_k(B^1(\mathbb{N}\times \mathbb{N}))=0$ for $k\geq 0$ and it follows then from Corollary \ref{cor:polhmlgycofibrant} type of the torus, we have $H^{\sing}_k(B^1(\mathbb{N}\times \mathbb{N}))=0=H_k^{\pol}(\mathbb{N}\times \mathbb{N}))$ for $k\geq 2$ and it follows then from Corollary \ref{cor:polhmlgycofibrant} and Paragraph \ref{paragr:polhmlgylowdimension} that the map $\sH^{\pol}(F)$ may be identified with the image in $\ho(\Ch)$ of the map $... ...  \chapter{Homology of contractible \texorpdfstring{\oo}{ω}-categories its consequences} \chapter{Homology of contractible \texorpdfstring{\oo}{ω}-categories and its consequences} \section{Contractible \texorpdfstring{\oo}{ω}-categories} Recall that for every \oo\nbd{}category C, we write p_C : C \to \sD_0 the canonical morphism to the terminal object of \sD_0. Recall that for every \oo\nbd{}category C, we write p_C : C \to \sD_0 for the canonical morphism to the terminal object of \sD_0. \begin{definition}\label{def:contractible} An \oo\nbd{}category C is \emph{oplax contractible} when the canonical morphism p_C : C \to \sD_0 is an oplax homotopy equivalence (Definition \ref{def:oplaxhmtpyequiv}). ... ... @@ -61,7 +61,7 @@ Consider the commutative square \sH^{\pol}(\sD_0) \ar[r,"\pi_{\sS_0}"] & \sH^{\sing}(\sD_0). \end{tikzcd}$ It follows respectively from Proposition \ref{prop:oplaxhmtpyisthom} and Proposition \ref{prop:oplaxhmtpypolhmlgy} that the right and left morphisms of the above square are isomorphisms. Then, an immediate computation left to the reader shows that $\sD_0$ is \good{} and that $\sH^{\pol}(\sD_0)\simeq \sH^{\sing}(\sD_0)\simeq \mathbb{Z}$. By a 2-out-of-3 property, we deduce that $\pi_C : \sH^{\sing}(C)\to \sH^{\pol}(C)$ is an isomorphism. It follows respectively from Proposition \ref{prop:oplaxhmtpyisthom} and Proposition \ref{prop:oplaxhmtpypolhmlgy} that the right and left vertical morphisms of the above square are isomorphisms. Then, an immediate computation left to the reader shows that $\sD_0$ is \good{} and that $\sH^{\pol}(\sD_0)\simeq \sH^{\sing}(\sD_0)\simeq \mathbb{Z}$. By a 2-out-of-3 property, we deduce that $\pi_C : \sH^{\sing}(C)\to \sH^{\pol}(C)$ is an isomorphism. \end{proof} \begin{remark} Definition \ref{def:contractible} admits an obvious lax'' variation and Proposition \ref{prop:contractibleisgood} is also true for lax contractible $\oo$\nbd{}categories. ... ... @@ -137,11 +137,13 @@ In particular, for every $n \in \mathbb{N}$, $\sD_n$ is \good{}. Recall from \re \] is cartesian all of the four morphisms are monomorphisms. Since the functor $\Hom_{\oo\Cat}(\Or_k,-)$ preserves limits, the square \eqref{squarenervesphere} is a cartesian square of $\Set$ and all of whose four morphisms are monomorphisms. Hence, what we need to show is that for every $k \geq 0$ and every $\oo$\nbd{}functor $\varphi : \Or_k \to \sS_{n}$, there exists an $\oo$\nbd{}functor $\varphi' : \Or_k \to \sD_n$ such that either $j_n^+ \circ \varphi ' = \varphi$ or $j_n^- \circ \varphi' = \varphi$. \eqref{squarenervesphere} is a cartesian square of $\Set$ all of whose four morphisms are monomorphisms. Hence, in order to prove that square \eqref{squarenervesphere} is cocartesian, we only need to show that for every $k \geq 0$ and every $\oo$\nbd{}functor $\varphi : \Or_k \to \sS_{n}$, there exists an $\oo$\nbd{}functor $\varphi' : \Or_k \to \sD_n$ such that either $j_n^+ \circ \varphi ' = \varphi$ or $j_n^- \circ \varphi' = \varphi$. %% Notice now that the morphisms $j_n^+$ and $j_n^-$ trivially satisfy the following %% properties: ... ... @@ -158,7 +160,7 @@ In particular, for every $n \in \mathbb{N}$, $\sD_n$ is \good{}. Recall from \re \begin{description} \item[Case $kn$:] Since $\sS_n$ is an $n$\nbd{}category, the image of $\alpha_k$ is necessarily of the form $\varphi(\alpha_k)=\1^k_{x}$ with $x$ a cell of $\sS_n$ of dimension non-greater than $n$. If the dimension of $x$ is strictly lower than $n$, then everything works like in the case $kn$:] Since $\sS_n$ is an $n$\nbd{}category, the image of $\alpha_k$ is necessarily of the form $\varphi(\alpha_k)=\1^k_{x}$ with $x$ a cell of $\sS_n$ of dimension non-greater than $n$. If $x$ is a unit on a cell whose dimension is strictly lower than $n$, then everything works like in the case $k1$, the source and target of such an $n$\nbd{}cell are given by More explicitly, the $n$\nbd{}cells of $X/a$ can be described as pairs $(x,p)$ where $x$ is an $n$\nbd{}cell of $X$ and $p$ is an arrow of $A$ of the form $p : f(\trgt_0(x)) \to a$. When $n>1$, the source and target of such an $n$\nbd{}cell are given by $\src((x,p))=(\src(x),p) \text{ and } \trgt((x,p))=(\trgt(x),p).$ Moreover, the $\oo$\nbd{}functor $f/a_0$ is described as Moreover, the $\oo$\nbd{}functor $f/a$ is described as $(x,p) \mapsto (f(x),p),$ and the canonical $\oo$\nbd{}functor $X \to X/a_0$ as and the canonical $\oo$\nbd{}functor $X \to X/a$ as $(x,p) \mapsto x.$ \end{paragr} \begin{paragr}\label{paragr:unfolding} Let $f : X \to A$ be an $\oo$\nbd{}functor with $A$ a $1$-category. Every arrow $\beta : a_0 \to a_0'$ induces an $\oo$\nbd{}functor Let $f : X \to A$ be an $\oo$\nbd{}functor with $A$ a $1$\nbd{}category. Every arrow $\beta : a \to a'$ of $A$ induces an $\oo$\nbd{}functor \begin{align*} X/\beta : X/a_0 &\to X/{a_0'} \\ X/\beta : X/a &\to X/{a'} \\ (x,p) & \mapsto (x,\beta \circ p), \end{align*} which takes part of a commutative triangle $\begin{tikzcd}[column sep=tiny] X/{a_0} \ar[rr,"X/{\beta}"] \ar[dr] && X/{a'_0} \ar[dl] \\ X/{a} \ar[rr,"X/{\beta}"] \ar[dr] && X/{a'} \ar[dl] \\ &X&. \end{tikzcd}$ This defines a functor \begin{align*} <