Commit 0fe53bbc authored by Leonard Guetta's avatar Leonard Guetta
Browse files

end of working day

parent 5094e675
......@@ -164,7 +164,7 @@ From the previous proposition, we deduce the following very useful corollary.
\]
then either $f=g$ or $f$ and $g$ are both units. In other words, $\alpha$ never send a non-unit arrow to a unit arrow and $\alpha$ never identifies two non-unit arrows. It follows that if $\alpha$ is quasi-injective on arrows and injective on objects, then it is also injective on arrows and hence, a monomorphism of $\Rgrph$.
\end{paragr}
\begin{proposition}
\begin{proposition}\label{prop:hmtpysquaregraphbetter}
Let
\[
\begin{tikzcd}
......@@ -187,7 +187,7 @@ From the previous proposition, we deduce the following very useful corollary.
is a \emph{homotopy} cocartesian square of $\Cat$ when equipped with the Thomason weak equivalences.
\end{proposition}
\begin{proof}
The case where $\alpha$ or $\beta$ is injective both on objects and arrows is Corollary \ref{cor:hmtpysquaregraph}. Hence, we only have to treat the case when $\alpha$ is injective on objects and $\beta$ is injective on arrows. The other case being symmetric.
The case where $\alpha$ or $\beta$ is both injective on objects and quasi-injective on arrows is Corollary \ref{cor:hmtpysquaregraph}. Hence, we 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 $\beta$. 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 following cocartesian square
\[
......@@ -197,9 +197,38 @@ From the previous proposition, we deduce the following very useful corollary.
\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]
\end{tikzcd}
\]
where the morphism \[ \coprod_{x \in E}F_x \to A\] is induced by the inclusion of each $F_x$ in $A$, and the morphism \[\coprod_{x \in E}F_x \to E\] is the only one that sends an element $a \in F_x$ to $x$. In other words, $G$ is obtained from $A$ by collapsing the objects that are identified through $\beta$.
Notice importantly that the morphism \[ \coprod_{x \in E}F_x \to A\] is a monomorphism, i.e. injective on objects and arrows, and the morphism \[A \to G\] is injective on arrows.
where the morphism \[ \coprod_{x \in E}F_x \to A\] is induced by the inclusion of each $F_x$ in $A$, and the morphism \[\coprod_{x \in E}F_x \to E\] is the only one that sends an element $a \in F_x$ to $x$. In other words, $G$ is obtained from $A$ by collapsing the objects that are identified through $\beta$. It admits the following explicit description: $G_0$ is (isomorphic to) $E$ and the set of non-units arrows of $G$ is (isomorphic to) the set of non-units arrows of $A$; the source (resp. target) of a non-unit arrow $f$ of $G$ is the source (resp. target) of $\beta(f)$. This completely describe $G$.
% Notice also for later reference that the morphism \[ \coprod_{x \in E}F_x \to A\] is a monomorphism, i.e. injective on objects and arrows.
Now, we have the following solid arrow commutative diagram
\[
\begin{tikzcd}
\displaystyle\coprod_{x \in E}F_x \ar[r] \ar[d] & E \ar[ddr,bend left]\ar[d]&\\
A \ar[drr,bend right,"\beta"'] \ar[r] & G \ar[dr, dotted]&\\
&&B
\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]
\end{tikzcd},
\]
where the arrow $E \to B$ is the canonical inclusion. Hence, by universal property, the dotted arrow exists and makes the whole diagram commutes. A thorough verification easily shows that the morphism $G \to B$ is a monomorphism of $\Rgrph$.
By forming successive cocartesian square and combining with the square obtained earlier, we obtain a diagram of three cocartesian square:
\[
\begin{tikzcd}[row sep = large]
\displaystyle\coprod_{x \in E}F_x \ar[r] \ar[d] & E \ar[d]&\\
A \ar[d,"\alpha"] \ar[r] & G \ar[d] \ar[r] & B \ar[d,"\delta"]\\
C \ar[r] & H \ar[r] & D
\ar[from=1-1,to=2-2,phantom,"\ulcorner" very near end,"\textcircled{\tiny \textbf{1}}" near start, description]
\ar[from=2-1,to=3-2,phantom,"\ulcorner" very near end,"\textcircled{\tiny \textbf{2}}", description]
\ar[from=2-2,to=3-3,phantom,"\ulcorner" very near end,"\textcircled{\tiny \textbf{3}}", description]
\end{tikzcd}.
\]
What we want to prove is that the image by the functor $L$ of the pasting of squares \textcircled{\tiny \textbf{2}} and \textcircled{\tiny \textbf{3}} is homotopy cocartesian. Since the morphism $G \to B$ is a monomorphism, we deduce from Corollary \ref{cor:hmtpysquaregraph} that the image by the functor $L$ of square \textcircled{\tiny \textbf{3}} is homotopy cocartesian. Hence, all we have to show is that the image by $L$ of square \textcircled{\tiny \textbf{2}} is homotopy cocartesian. \todo{Parler du pasting lemma ?} On the other hand, we know that both morphisms
\[
\coprod_{x \in E}F_x \to A \text{ and } A \to C
\]
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.
\end{proof}
We now apply Corollary \ref{cor:hmtpysquaregraph} and Proposition \ref{prop:hmtpysquaregraphbetter} to a few examples.
\begin{example}[Adding a generator]
Let $C$ be a free category, $A$ and $B$ (possibly equal) two objects of $C$ and let $C'$ be the category obtained from $C$ by adding a generator $A \to B$, i.e. defined with the following cocartesian square:
\[
......@@ -208,10 +237,10 @@ From the previous proposition, we deduce the following very useful corollary.
\sD_1 \ar[r] & C'.
\end{tikzcd}
\]
Then, this square is 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 $i_1 : \partial\sD_1 \to \sD_1$ comes from a monomorphism of $\Rgrph$.
Then, this square is 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 $i_1 : \partial\sD_1 \to \sD_1$ comes from a monomorphism of $\Rgrph$. Hence, we can apply Corollary \ref{cor:hmptysquaregraph}.
\end{example}
\begin{remark}
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 free (1-)categories are \good{} (which we already knew since we have seen that all (1-)categories are \good{}).
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 already knew since we have seen that \emph{all} (1-)categories are \good{}).
\end{remark}
\begin{example}[Identyfing two generators]
Let $C$ be a free category and $f,g : A \to B$ parallel generating arrows of $C$ such that $f\neq g$. Now consider the category $C'$ obtained from $C$ by ``identifying'' $f$ and $g$, i.e. defined with the following cocartesian square
......@@ -221,16 +250,7 @@ From the previous proposition, we deduce the following very useful corollary.
\sD_1 \ar[r] & C',
\end{tikzcd}
\]
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 $\sD_1$. Let us prove that the previous square is homotopy cocartesian in $\Cat$ (when equipped with Thomason weak equivalences). Notice that this square is the image of a cocartesian square of $\Rgrph$ by the functor $L$ as in the previous example. We now distinguish two cases. First, if $A \neq B$, then the map $\sS_1 \to C$ comes from a monomorphism of reflexive graphs and we conclude as in the previous example. Now if $A=B$, notice that the map $\sS_1 \to C$ factorizes as
\[
\sS_1 \to F_2 \to C
\]
where $F_2$ is the free monoid with two generators, seen as a category. In particular, it is free and notice that the map on the left comes from a monomorphism of reflexive graphs. Now, this factorization yields a factorization of our cocartesian square into two cocartesian squares
\[
\begin{tikzcd}
a
\end{tikzcd}
\]
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 homotopy cocartesian in $\Cat$ (when equipped with Thomason weak equivalences). 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 Proposition \ref{prop:hmtpysquaregraphbetter}. Note that since we did \emph{not} suppose that $A\neq B$, the top morphism is not necessarily a monomorphism and we cannot always apply Corollary \ref{cor:hmtpysquaregraph}.
\end{example}
\begin{example}[Killing a generator]
Let $C$ be a free category and let $f : A \to B$ one of its generating arrow such that $A \neq B$. Now consider the category $C'$ obtained from $C$ by ``killing'' $f$, i.e. defined with the following cocartesian square:
......@@ -246,3 +266,4 @@ From the previous proposition, we deduce the following very useful corollary.
Note that in the previous example, we see that it was useful to consider the category of reflexive graphs and not only the category of graphs because the map $\sD_1 \to \sD_0$ does not come from a morphism in the category of graphs. \todo{À mieux dire ?}
\end{remark}
\todo{Préciser que si A=B dans l'exemple précédent ça ne marche pas ?}
......@@ -15,6 +15,10 @@
\usepackage{mathtools}
\usepackage{tikz-cd}
% Typography
\usepackage{pifont} % I use it for circled numbers
% List
\usepackage{enumitem}
......
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