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Changement de la propriété universelle, changement des isos, coupé le gros diagramme en deux

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Samy Avrillon 2024-08-18 12:39:58 +02:00
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Report/M2Report.tex
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@ -1,5 +1,5 @@
% !TeX spellcheck = en_US
\documentclass[10pt,a4paper]{article}
\documentclass[11pt,a4paper]{article}
\input{./header.tex}
@ -19,6 +19,7 @@
\hsep
\setcounter{tocdepth}{2}
\tableofcontents
\newpage
@ -509,7 +510,7 @@
\[
R_0^i := R^1_0 \circ \dots \circ R_{i-1}^i
\]
\item An operator $\tl^i : \BB_i \times \BB_0 \to \BB_i$ along with a morphism $\inj_\tl^i : X \to X \tl^i Y$ for every $X,Y$ in $\BB_i \times \BB_0$, such that the canonical morphism $\en_{i-1}^i : (R_{i-1}^i X) \tl^{i-1} Y \to R_{i-1}^i (X \tl^i Y)$ is an isomorphism. This operation follows a specific universal property: For every morphisms $g : X \to Z$ and $h : Y \to R_0^iZ$, there is an unique morphism $\{g;h\}$ such that the two following diagrams commute:
\item An operator $\tl^i : \BB_i \times \BB_0 \to \BB_i$ along with two morphism $\inj_1^i : X \to X \tl^i Y$ and $\inj_2^i : Y \to R_0^i(X \tl^i Y)$ for every $X,Y$ in $\BB_i \times \BB_0$, such that the canonical morphism $\en_{i-1}^i : (R_{i-1}^i X) \tl^{i-1} Y \to R_{i-1}^i (X \tl^i Y)$ is an isomorphism. This operation follows a specific universal property: For every morphisms $g : X \to Z$ and $h : Y \to R_0^iZ$, there is an unique morphism $\{g;h\}$ such that the following diagrams commute:
\label{sec:tlUniversalProperty}
\begin{center}
@ -519,19 +520,16 @@
% END OF GENERATED LATEX
\end{center}
where $\en_0^i$ is the following composition:
We will also denote a composition morphism as such:
\[\en_0^i := R_0^{i-1} \en_{i-1}^i \circ \dots \circ R_0^1\en_1^2 \circ \en_0^1 : (R_0^i X) \oplus Y = (R_0^i X) \tl^0 Y \to R_0^i (X \tl^i Y)\]
And where $\inj_2$ is the second injector of $\oplus$, the coproduct of the category $\BB_0 = \TSet$. We will define $\tl^0$ to be $\oplus$, so the equality above holds.
The operator is also defined on morphisms, such that for any morphism $g : X \to X'$ in $\BB_i$ and $h : Y \to Y'$ in $\BB_0$, there is a morphism $g \tl^i h : X \tl^i Y \to X' \tl^i Y'$ in $\BB_i$.
\item A functor $F_i : \BB_i \to \CC_i$
\item A functor $G_i : \CC_i \to \BB_i$
\item An adjunction between them $F_i \vdash G_i$
\item A proof that $F_iG_i \cong \Id_{\CC_i}$ (i.e. $F_i \vdash G_i$ make up a coreflection)
\item A proof that $F_i\inj_\tl^i$ is an isomorphism. The inverse isomorphism will be denoted as $(F_i\inj_\tl^i)^{-1}$
\item A proof that $F_i\inj_1^i$ is an isomorphism. The inverse isomorphism will be denoted as $(F_i\inj_1^i)^{-1}$
\end{itemize}
Here is a figure that describes the recursive construction of some of the above objects
@ -625,7 +623,7 @@
\setlength\itemsep{-1ex}
\item $\CC_0$ is $\one$, the category with only one object $\star$ and one trivial morphism (i.e. the terminal category of $\Cat$)
\item $\BB_0$ is $\TSet$, the category of models of the $(\mathcal{O},\El)$ sort specification.
\item The universal property of $\tl^0$ is that of the coproduct. Therefore, we will define $\tl^0 : \BB_0 \times \BB_0 \to \BB_0$ to be the coproduct $\oplus$ of $\TSet$, with $\inj_\tl^0 : X \to X \oplus Y$ being the first injector of the coproduct. The second injector of the coproduct $\oplus$ will be refered as $\inj_2$.
\item The universal property of $\tl^0$ is that of the coproduct. Therefore, we will define $\tl^0 : \BB_0 \times \BB_0 \to \BB_0$ to be the coproduct $\oplus$ of $\TSet$, with $\inj_1^0 : X \to X \tl^0 Y$ being the first injector of the coproduct and $\inj_2^0 : Y \to X \tl^0 Y$ being the second injector.
\item $F_0 : \TSet \to \one$ is the terminal functor of $\Cat$
\item $G_0 : \one \to \TSet$ is the functor that sends the only object of $\one$ to the initial object of $\TSet$: $0_\TSet$
\item For $X$ an object of $\TSet$, we have
@ -808,7 +806,7 @@
\paragraph{Constructing the objects}
We will define the $\tl^i$ operator of two objects $X$ from $\BB_i$ and $Y$ from $\BB_0$ as follows:
\[
X \tl_i Y := \left((R_{i-1}^i X) \tl^{i-1} Y, (R_0^{i-1} \inj_\tl^{i-1})_\UU \circ \Cstr_i^X \circ (H_iF_{i-1}\inj_\tl^{i-1})^{-1}\right)
X \tl_i Y := \left((R_{i-1}^i X) \tl^{i-1} Y, (R_0^{i-1} \inj_1^{i-1})_\UU \circ \Cstr_i^X \circ (H_iF_{i-1}\inj_1^{i-1})^{-1}\right)
\]
The constructor goes as follows:
@ -820,9 +818,9 @@
% END OF GENERATED LATEX
\end{center}
The injector $\inj_\tl^i : X \to X \tl^i Y$ is defined as follows:
The first injector $\inj_1^i : X \to X \tl^i Y$ is defined as follows:
\[
\inj_\tl^i := \inj_\tl^{i-1} : R_{i-1}^i X \to R_{i-1}^i (X \tl^i Y) = R_{i-1}^i X \tl^{i-1} Y
\inj_1^i := \inj_1^{i-1} : R_{i-1}^i X \to R_{i-1}^i (X \tl^i Y) = R_{i-1}^i X \tl^{i-1} Y
\]
It is a morphism of $\BB_i$ as it makes the following diagram commute:
@ -833,9 +831,14 @@
% END OF GENERATED LATEX
\end{center}
The second injector is defined as
\[
\inj_2^i := \inj_2^{i-1} : Y \to R_0^{i-1} (R_{i-1}^iX \tl^{i-1} Y) = R_0^i (X \tl^i Y)
\]
\paragraph{Universal Property}
We will now show that that the universal property stated in \autoref{sec:tlUniversalProperty} holds.
To that end, we take two objects $X$ and $Z$ in $\BB_i$, $Y$ in $\BB_0$, a morphism $g : X \to Z$ in $\BB_i$ and a morphism $h : Y \to R_0^iZ$ in $\BB_0$. We want to build a morphism $\{g,h\}$ of $\BB_i$ such that the following two diagrams commute.
To that end, we take two objects $X$ and $Z$ in $\BB_i$, $Y$ in $\BB_0$, a morphism $g : X \to Z$ in $\BB_i$ and a morphism $h : Y \to R_0^iZ$ in $\BB_0$. We want to build a morphism $\{g,h\}$ of $\BB_i$ such that the following diagram commute.
\begin{center}
% YADE DIAGRAM TlUniversal.json
@ -853,6 +856,8 @@
% END OF GENERATED LATEX
\end{center}
And this morphism $\{g;h\}_i$ is such that the universal property for it is exactly the same as the one of $\{R_{i-1}^i g, h\}$.
With this definition, the isomorphism $\en_{i-1}^i : (R_{i-1}^i X) \tl^{i-1} Y \to R_{i-1}^i(X \tl^i Y)$ is simply the identity morphism.
\paragraph{Building morphisms}
@ -866,25 +871,27 @@
This is indeed a morphism of $\BB_i$ as it makes the following diagram commute:
\begin{center}
\begin{scaletikzpicturetowidth}{.9\textwidth}
% YADE DIAGRAM TlDefOnMorphisms.json
% GENERATED LATEX
\input{graphs/TlDefOnMorphisms.tex}
% END OF GENERATED LATEX
\end{scaletikzpicturetowidth}
\end{center}
\paragraph{Composition with $F_i$}
We finally need to prove, for any objects $X$ in $\BB_i$ and $Y$ in $\TSet$, that the morphism
$F_i(\inj_\tl^i) : F_iX \to F_i(X \tl^i Y)$ is an isomorphism.
$F_i(\inj_1^i) : F_iX \to F_i(X \tl^i Y)$ is an isomorphism.
Morphisms $\BB_{i}$ are morphisms of $\BB_{i-1}$ that follows some equations, and composition is the same. Moreover, we know from the definition above that $R_{i-1^i} \inj_\tl^i := \inj_\tl^{i-1}$.
Morphisms $\BB_{i}$ are morphisms of $\BB_{i-1}$ that follows some equations, and composition is the same. Moreover, we know from the definition above that $R_{i-1^i} \inj_1^i := \inj_1^{i-1}$.
So, we just need to know that $R_{i-1}^i(F_{i-1}\inj_\tl^i)$ is an isomorphism.
So, we just need to know that $R_{i-1}^i(F_{i-1}\inj_1^i)$ is an isomorphism.
\[
R_{i-1}^iF_i\inj_\tl^i = \left[(F_{i-1} \times \id) (E(\inj_\tl^i))\right]_1 = \left[(F_{i-1} \times \id) (R_{i-1}^i \inj_\tl^i,!)\right]_1 = F_{i-1} \inj_\tl^{-1}
R_{i-1}^iF_i\inj_1^i = \left[(F_{i-1} \times \id) (E(\inj_1^i))\right]_1 = \left[(F_{i-1} \times \id) (R_{i-1}^i \inj_1^i,!)\right]_1 = F_{i-1} \inj_1^{i-1}
\]
And we know from the previous induction step that $F_{i-1}\inj_\tl^{i-1}$ is an isomorphism. Therefore, $F_i\inj_\tl^i$ is an isomorphism.
And we know from the previous induction step that $F_{i-1}\inj_1^{i-1}$ is an isomorphism. Therefore, $F_i\inj_1^i$ is an isomorphism.
\section{Other Properties}
@ -1093,7 +1100,7 @@
\section{Summary}
\lipsum[2-3]
\lipsum[1-3]
@ -1112,7 +1119,7 @@
\section{$W \dashv E$ adjunction}
\label{apx:adjunction}
\subsubsection{Constructing $\phi_{XYZ}$}
\subsection{Constructing $\phi_{XYZ}$}
Let $f$ be in $\Hom(W(X,Y),Z)$.
We want to construct $\phi_{XYZ}(f) : (X,Y) \to E(Z)$.
@ -1133,10 +1140,9 @@
% GENERATED LATEX
\input{graphs/PhiXYZSndComponentPullback.tex}
% END OF GENERATED LATEX
\end{center}
\subsubsection{Constructing $\phi^{-1}_{XYZ}$}
\subsection{Constructing $\phi^{-1}_{XYZ}$}
Now, we take $(g,h)$ a morphism from $(X,Y)$ to $E(Z)$.
@ -1165,14 +1171,13 @@
% GENERATED LATEX
\input{graphs/PhiXYZ-1MorphismOfBi.tex}
% END OF GENERATED LATEX
\end{center}
\subsection{Composition $\phi_{XYZ} \circ \phi_{XYZ}^{-1}$}
The first component of $\phi_{XYZ} (\phi_{XYZ}^{-1}(g,h))$ is
\[
R_{i-1}^i(\phi_{XYZ}^{-1}(g,h)) \circ \inj_\tl^{i-1} = \left\{g ; \square \right\} \circ \inj_\tl^{i-1} = g
R_{i-1}^i\left(\phi_{XYZ}^{-1}(g,h)\right) \circ \inj_1^{i-1} = \left\{g ; \square \right\} \circ \inj_1^{i-1} = g
\]
The second component of $\phi_{XYZ} (\phi_{XYZ}^{-1}(g,h))$ follows the following diagram
@ -1187,7 +1192,7 @@
\todo{Justifier que la partie du haut commute, i.e. que \[
(R_0^{i-1}\{g,\square\})_\El \circ (\en_0^i)_\El \circ (\inj_2)_\El = \square_\El
(R_0^{i-1}\{g,\square\})_\El \circ (\inj^{i-1}_2)_\El = \square_\El
\]}
The diagram commutes, and so we can deduce that the second component of $\phi_{XYZ}(f)$ is $h$, by proprty of the pullback $E(Z)$
@ -1197,7 +1202,7 @@
Now, the converse composition is
\[
\phi_{XYZ}^{-1} (\phi_{XYZ}(f)) = \left\{R_{i-1}^i f \circ \inj_\tl^{i-1} ; \square \right\}
\phi_{XYZ}^{-1} (\phi_{XYZ}(f)) = \left\{R_{i-1}^i f \circ \inj_1^{i-1} ; \square \right\}
\]
where $\square$ follows the following diagram
@ -1208,13 +1213,13 @@
% END OF GENERATED LATEX
\end{center}
We want to show that $\phi_{XYZ}^{-1} (\phi_{XYZ}(f)) = f$. By definition of $\{;\}$ in $\BB_i$, it suffices to show that $\square = R_0^i f \circ \en_0^{i-1} \circ \inj_2$.
We want to show that $\phi_{XYZ}^{-1} (\phi_{XYZ}(f)) = f$. By definition of $\{;\}$ in $\BB_i$, it suffices to show that $\square = R_0^i f \circ \inj^{i-1}_2$.
The two required equalities are proved by the diagram above:
\begin{align*}
\square_\El = \left(R_0^i f \circ \en_0^{i-1} \circ \inj_2\right)_\El \\
\square_\UU = \left(R_0^i f \circ \en_0^{i-1} \circ \inj_2\right)_\UU
\square_\El = \left(R_0^i f \circ \inj^{i-1}_2\right)_\El \\
\square_\UU = \left(R_0^i f \circ \inj^{i-1}_2\right)_\UU
\end{align*}
\subsection{Naturality}
@ -1234,23 +1239,30 @@
\[\begin{array}{rcl}
\phi_{X'Y'Z'}(f \circ \ii \circ W(g,h))_1
&=& R_{i-1}^i\left(f \circ \ii \circ (g \tl^{i-1} K_\bullet(g,h))^+\right) \circ \inj_\tl^{i-1} \\
&=& R_{i-1}^i f \circ R_{i-1}^i \ii \circ (g \tl^{i-1} K_\bullet(g,h)) \circ \inj_\tl^{i-1} \\
&=& R_{i-1}^i f \circ R_{i-1}^i \ii \circ \inj_\tl^{i-1} \circ g \\
&=& R_{i-1}^i\left(f \circ \ii \circ (g \tl^{i-1} K_\bullet(g,h))^+\right) \circ \inj_1^{i-1} \\
&=& R_{i-1}^i f \circ R_{i-1}^i \ii \circ (g \tl^{i-1} K_\bullet(g,h)) \circ \inj_1^{i-1} \\
&=& R_{i-1}^i f \circ R_{i-1}^i \ii \circ \inj_1^{i-1} \circ g \\
&=& \left[E(f) \circ \phi_{XYZ}(\ii) \circ (g,h)\right]_1
\end{array}\]
The second components are defined as described by the following diagram
The second components are defined by the pullback properties of $E(Z)$ and $E(Z')$. The two sides that define each morphism are give separately in the two following diagrams.
\begin{center}
% YADE DIAGRAM NaturalityDoublePullbackDefinition.json
% YADE DIAGRAM NaturalityDoublePullbackDefinition1.json
% GENERATED LATEX
\input{graphs/NaturalityDoublePullbackDefinition.tex}
\input{graphs/NaturalityDoublePullbackDefinition1.tex}
% END OF GENERATED LATEX
\end{center}
The second projection of $\phi_{XYZ}(f \circ \ii \circ W(g,h))$ is defined by the pullback of $E(Z')$ commuting with the two path highlighted by the inner blue line. And that of $\phi_{X'Y'Z'}(\ii)$ is defined by the red outer line.
The outer squares commute, and therefore, by the pullback property, we get that
\begin{center}
% YADE DIAGRAM NaturalityDoublePullbackDefinition2.json
% GENERATED LATEX
\input{graphs/NaturalityDoublePullbackDefinition2.tex}
% END OF GENERATED LATEX
\end{center}
The second projection of $\phi_{XYZ}(f \circ \ii \circ W(g,h))$ is defined by the pullback of $E(Z')$ commuting with the blue bottom path. And that of $\phi_{X'Y'Z'}(\ii)$ is defined by the red top path.
both diagrams commute, and therefore, by the pullback property, we get that
\[
\phi_{X'Y'Z'}(f\circ\ii\circ W(g,h))_2 = \left[E(f) \circ \phi_{XYZ}(\ii) \circ (g,h)\right]_2
\]
@ -1279,7 +1291,7 @@
% END OF GENERATED LATEX
\end{center}
We can extend this pullback with the two isomorphisms $\en_0^i$ and $H_iF_{i-1}(\inj_\tl^{i-1})$ in another pullback. This new pullback is over the injection morphism $\inj_2$ of the coproduct of $\TSet$, so the new pullback object is the second component of the $\bullet \oplus B$ i.e. $B$.
We can extend this pullback with the isomorphisms $H_iF_{i-1}(\inj_\tl^{i-1})$ into another pullback. This new pullback is over the injection morphism $\inj_2$ of the coproduct of $\TSet$, so the new pullback object is the second component of the $\bullet \oplus B$ i.e. $B$, by property of the coproduct.
\begin{center}
% YADE DIAGRAM ReflectionFGExtendedPullback.json
@ -1288,6 +1300,8 @@
% END OF GENERATED LATEX
\end{center}
\todo{Il manque l'info $\inj_2^{i-1} \circ \en_0^{i-1} = \inj_2^0$ OU ALORS que $\inj_2^{i-1}$ se transpose dans les pullbacks aussi.}
The first component of the isomorphism is the following isomorphism, where $\eta_{i-1}^{FG}$ if the counit of the adjunction $F_{i-1} \vdash G_{i-1}$, that we know to be an isomorphism from the induction hypothesis.
\begin{center}
@ -1307,153 +1321,3 @@
\end{document}

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Report/header.tex
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@ -42,6 +42,7 @@
\usepackage{newunicodechar}
\usepackage{txfonts}
\usepackage{yade}
\usepackage{environ}
\usepackage[backend=biber,style=numeric]{biblatex}
\usepackage{hyperref}
@ -126,13 +127,13 @@
\node (El) at (0,1) {\El};
\draw[->] (El) -- node[anchor=east] {\ensuremath{\scriptstyle p}} (U);
\node (XU) at (1,0) {\ensuremath{#3}};
\node (XEl) at (1,1) {\ensuremath{#1}};
\node (XU) at (1.5,0) {\ensuremath{#3}};
\node (XEl) at (1.5,1) {\ensuremath{#1}};
\draw[->] (XEl) -- node[anchor=west] {\ensuremath{\scriptstyle #2}} (XU);
\draw[|->] (.3,0) -- (.7,0);
\draw[|->] (.3,0.5) -- (.7,0.5);
\draw[|->] (.3,1) -- (.7,1);
\draw[|->] (.3,0) -- (1,0);
\draw[|->] (.3,0.5) -- (1,0.5);
\draw[|->] (.3,1) -- (1,1);
\coordinate (base) at (.5,.4);
\end{tikzpicture}
@ -169,5 +170,18 @@
% Fixing Yade green
\definecolor{green}{RGB}{11,102,35}
% Environ for scaling tikz pictures
\makeatletter
\newsavebox{\measure@tikzpicture}
\NewEnviron{scaletikzpicturetowidth}[1]{%
\def\tikz@width{#1}%
\def\tikzscale{1}\begin{lrbox}{\measure@tikzpicture}%
\BODY
\end{lrbox}%
\pgfmathparse{#1/\wd\measure@tikzpicture}%
\edef\tikzscale{\pgfmathresult}%
\BODY
}
\makeatother
\addbibresource{Bilibibio.bib}