Commit 2ac942db authored by Leonard Guetta's avatar Leonard Guetta
Browse files

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 $k<n$:] Since every generating cell of $\gamma$ of $\Or_k$ is of dimension non-greater than $k$, the cell $\varphi(\gamma)$ is of dimension strictly lower than $n$. Since all cells of dimension strictly lower than $n$ are both in the image of $j^+_n$ and in the image of $j^-_n$, $\varphi$ obviously factors through $j^+_n$ (and $j^+_n$).
\item[Case $k=n$:] The image of $\alpha_n$ is either a non-trivial $n$\nbd{}cell of $\sS_n$ or a unit on a strictly lower dimensional cell. In the second situation, everything works like the case $k<n$. Now suppose for example that $\varphi(\alpha_n)$ is $h^+_n$, which is in the image of $j^+_n$. Since all of the other generating cells of $\Or_n$ are of dimension strictly lower than $n$, their images by $\varphi$ are also of dimension strictly lower than $n$ and hence, are all contained in the image of $j^+_n$. Altogether this proves that $\varphi$ factors through $j^+_n$. The case where $\varphi(\alpha_n)=h^-_n$ is symmetric.
\item[Case $k>n$:] 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 $k<n$. If not, this means that $x$ is a non-trivial $n$\nbd{}cell of $\sS_n$. Suppose for example that $x=h^+_n$. Now let $\gamma$ be a generator of $\Or_k$ of dimension $k-1$. Either we have $\varphi(\gamma)=1^n_y$ for $y$ a cell of dimension strictly lower than $n$, or we have that $\varphi(\gamma)$ is a non-degenerate $n$\nbd{}cell of $\sS_n$. In the first situation, $y$ is in the image of $j^+_n$ as in the case $k<n$, and thus, so is $1^n_y$. In the second situation, this means \emph{a priori} that either $y=h_n^+$ or $y=h_n^-$. But we know that $\gamma$ is part of a composition that is equal to either the source or the target of $\alpha_k$ (see \ref{paragr:orientals}) and thus, $f(\gamma)$ is part of a composition that is equal to either the source or the target of $x=h^+_n$. Since no composition involving $h^-_n$ can be equal to $h^+_n$ (one could invoke the function introduced in \ref{prop:countingfunction}), this implies that $y=h_n^+$ and hence, $f(\gamma)$ is in the image of $j^n_+$. This goes for all generating cells of dimension $k-1$ of $\Or_k$ and we can recursively apply the same reasoning for generating cells of dimension $k-2$, then $k-3$ and so forth. Altogether, this proves that $\varphi$ factorizes through $j^+_n$. The case where $x=h^-_n$ and $\varphi$ factorizes through $j^-_n$ is symmetric.
\item[Case $k>n$:] 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 $k<n$. If not, this means that $x$ is a non-trivial $n$\nbd{}cell of $\sS_n$. Suppose for example that $x=h^+_n$. Now let $\gamma$ be a generator of $\Or_k$ of dimension $k-1$. We have $\varphi(\gamma)=\1^{k-1}_y$ with $y$ which is either a unit on a cell of dimension strictly lower than $n$, or a non-degenerate $n$\nbd{}cell of $\sS_n$ (if $k-1=n$, recall the convention that $\1^{k-1}_y=y$). In the first situation, $y$ is in the image of $j^+_n$ as in the case $k<n$, and thus, so is $\1^{k-1}_y$. In the second situation, this means \emph{a priori} that either $y=h_n^+$ or $y=h_n^-$. But we know that $\gamma$ is part of a composition that is equal to either the source or the target of $\alpha_k$ (see \ref{paragr:orientals}) and thus, $f(\gamma)$ is part of a composition that is equal to either the source or the target of $x=h^+_n$. Since no composition involving $h^-_n$ can be equal to $h^+_n$ (one could invoke the function introduced in \ref{prop:countingfunction}), this implies that $y=h_n^+$ and hence, $f(\gamma)$ is in the image of $j^+_n$. This goes for all generating cells of dimension $k-1$ of $\Or_k$ and we can recursively apply the same reasoning for generating cells of dimension $k-2$, then $k-3$ and so forth. Altogether, this proves that $\varphi$ factorizes through $j^+_n$. The case where $x=h^-_n$ and $\varphi$ factorizes through $j^-_n$ is symmetric.
\end{description}
\end{proof}
From these two lemmas, follows the important proposition below.
......@@ -186,7 +188,7 @@ is a homotopy cocartesian square of simplicial sets. Since $N_{\oo}$ induces an
\end{tikzcd}
\]
is Thomason homotopy cocartesian for every $n\geq 0$. Finally, since $i_n : \sS_{n-1} \to \sD_{n}$ is
a folk cofibration and $\sS_{n-1}$ and $\sD_{n}$ are folk cofibrant for every $n\geq0$, we deduce the result from Corollary \ref{cor:usefulcriterion} and an immediate induction. The base case being simply that $\sS_{-1}=\emptyset$ is obviously \good{}.
a folk cofibration and $\sS_{n-1}$ and $\sD_{n}$ are folk cofibrant for every $n\geq0$, we deduce the desired result from Corollary \ref{cor:usefulcriterion} and an immediate induction. The base case being simply that $\sS_{-1}=\emptyset$ is obviously \good{}.
\end{proof}
\begin{paragr}
The previous proposition implies what we claimed in Paragraph \ref{paragr:prelimcriteriongoodcat}, which is that the morphism of op-prederivators
......@@ -204,16 +206,17 @@ is a homotopy cocartesian square of simplicial sets. Since $N_{\oo}$ induces an
\ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end]
\end{tikzcd}
\]
where the map $\sD_2 \to B^2\mathbb{N}$ points the unique generating $2$-cell of $B^2\mathbb{N}$ and
$\sD_0 \to B^2\mathbb{N}$ points to the only object of $B^2\mathbb{N}$. It is easily checked that this
square is cocartesian and since $\sS_1$, $\sD_0$ and $\sD_2$ are free and
$i_2$ is a cofibration for the canonical model structure, the square is also
homotopy cocartesian with respect to folk weak equivalences. If $\J$ was
homotopy cocontinuous, then this square would also be homotopy cocartesian
with respect to Thomason equivalences. Since we know that $\sS_1$, $\sD_0$ and $\sD_2$ are \good{}, this would imply that $B^2\mathbb{N}$ is \good{}.
where the map $\sD_2 \to B^2\mathbb{N}$ points the unique generating
$2$-cell of $B^2\mathbb{N}$ and $\sD_0 \to B^2\mathbb{N}$ points to
the only object of $B^2\mathbb{N}$. Since $\sS_1$, $\sD_0$ and $\sD_2$ are
free and $i_2$ is a folk cofibration,
the square is also folk homotopy cocartesian. If $\J$ was homotopy
cocontinuous, then this square would also be Thomason homotopy
cocartesian. Since we know that $\sS_1$, $\sD_0$ and $\sD_2$ are
\good{}, this would imply that $B^2\mathbb{N}$ is \good{}.
From Proposition \ref{prop:spheresaregood}, we also deduce the proposition
below which gives a criterion to detect \good{} $\oo$\nbd{}category when we
below which gives a criterion to detect \good{} $\oo$\nbd{}categories when we
already know that they are free.
\end{paragr}
\begin{proposition}
......@@ -231,20 +234,22 @@ is a homotopy cocartesian square of simplicial sets. Since $N_{\oo}$ induces an
is homotopy cocartesian with respect to Thomason equivalences, then $C$ is \good{}.
\end{proposition}
\begin{proof}
Since the morphisms $i_k$ are folk cofibration and the $\oo$\nbd{}categories
$\sS_{k-1}$ and $\sD_{k}$ are folk cofibrant and \good{}, it follows
from Corollary \ref{cor:usefulcriterion} that all $\sk_k(C)$ are \good{}. The
result follows then from Lemma \ref{lemma:filtration} and Proposition \ref{prop:sequentialhmtpycolimit}.
Since the morphisms $i_k$ are folk cofibration and the
$\oo$\nbd{}categories $\sS_{k-1}$ and $\sD_{k}$ are folk cofibrant
and \good{}, it follows from Corollary \ref{cor:usefulcriterion} and
an immediate induction that all $\sk_k(C)$ are \good{}. The result
follows then from Lemma \ref{lemma:filtration} and Proposition
\ref{prop:sequentialhmtpycolimit}.
\end{proof}
\section{The miraculous case of 1-categories}
Recall that the terms \emph{1-category} and \emph{(small) category} are
synonymous. While we have used the latter one more often so far, in this section
we will mostly use the former one. As usual, the canonical functor $\iota_1 :
\Cat \to \oo\Cat$ is treated as an inclusion functor and hence we always consider
$(1-)$categories as particular cases of $\oo$\nbd{}categories.
$1$\nbd{}categories as particular cases of $\oo$\nbd{}categories.
The goal of what follows is to show that every $1$-category is \good{}. In order to do that, we will prove that every 1-category
is a canonical colimit of contractible $1$-categories and that this colimit is
The goal of what follows is to show that every $1$\nbd{}category is \good{}. In order to do that, we will prove that every 1-category
is a canonical colimit of contractible $1$\nbd{}categories and that this colimit is
homotopic both
with respect to the folk weak equivalences and with respect to the Thomason equivalences.
We call the reader's attention to an important subtlety here: even though the
......@@ -258,67 +263,63 @@ then $P$ is not necessarily a $1$\nbd{}category. In particular, polygraphic
homology groups of a $1$\nbd{}category need \emph{not} be trivial in dimension
higher than $1$.
\begin{paragr}
Recall that for an object $a_0$ of an $\oo$\nbd{}category $A$, we denote by $A/a_0$ the slice $\oo$\nbd{}category over $a_0$ (Paragraph \ref{paragr:slices}). When $A$ is a $1$-category, $A/a_0$ is also a $1$-category whose description is as follows:
\begin{itemize}[label=-
]
\item an object of $A/a_0$ is a pair $(a, p : a \to a_0)$ where $a$ is an object of $A$ and $p$ is an arrow of $A$,
\item an arrow $(a,p) \to (a',p')$ of $A/a_0$ is an arrow $ q : a \to a'$ of $A$ such that $p'\circ q = p$.
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 $p'\circ q = p$,
\end{itemize}
We denote by
\[
and we write $\pi_a$ for the canonical forgetful functor
\[
\begin{aligned}
\pi_{a_0} : A/a_0 &\to A \\
(a,p) &\mapsto a
\pi_{a} : A/a &\to A \\
(a',p) &\mapsto a'.
\end{aligned}
\]
the canonical forgetful functor.
Recall also that given an $\oo$\nbd{}functor $f : X \to A$ and an object $a_0$
of $A$, we have defined the $\oo$\nbd{}category $A/a_0$ and the $\oo$\nbd{}functor
This is special case of the more general notion of slice $\oo$\nbd{}categories introduced in \ref{paragr:slices}. In particular, given an $\oo$\nbd{}category $X$ and an $\oo$\nbd{}functor $f : X \to A$, we have defined the $\oo$\nbd{}category $X/a$ and the $\oo$\nbd{}functor
\[
f/a_0 : X/a_0 \to A//a_0
f/a : X/a \to A/a
\]
as the following pullback
\[
\begin{tikzcd}
X/a_0 \ar[r]\ar[dr, phantom, "\lrcorner", very near start] \ar[d,"f/a_0"'] & X \ar[d,"f"] \\
A/a_0 \ar[r,"\pi_{a_0}"] &A.
X/a \ar[r]\ar[dr, phantom, "\lrcorner", very near start] \ar[d,"f/a"'] & X \ar[d,"f"] \\
A/a \ar[r,"\pi_{a}"] &A.
\end{tikzcd}
\]
When $A$ is a $1$-category, the $n$-cells of $X/a_0$ 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
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*}
<