M2Internship/Report/M2Report.tex

353 lines
12 KiB
TeX
Raw Blame History

This file contains invisible Unicode characters

This file contains invisible Unicode characters that are indistinguishable to humans but may be processed differently by a computer. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

% !TeX spellcheck = en_US
\documentclass[10pt,a4paper]{article}
\input{./header.tex}
\title{Categorical semantics of the reduction of GATs to two-sorted GATs.
\\[1ex] \large Notes on my 4.5-month internship at the Laboratoire d'Informatique de l'École Polytechnique (Palaiseau, France)}
\hypersetup{pdftitle={Categorical semantics of the reduction of GATs to two-sorted GATs}}
\author{Samy Avrillon, supervised by
\\[1ex] Ambroise Lafont (LIX, Palaiseau, France)}
\begin{document}
\doparttoc
\maketitle
\hsep
\tableofcontents
\newpage
\section{Introduction}
\lipsum[7-9]
\section{Content}
\subsection{Infinite construction of $\BB_i$}
\[
\BB_i := \left(X : \TSet, \Cstr : (a : S_{i-1}) \to \Hom_{\BB_{a-1}}(G_{a-1}\Gamma_a,R_{a-1}^i(\this)) \to X(\UU)\right)
\]
A morphism from $(X,\Cstr)$ to $(X', \Cstr')$ is a morphism from $X$ to $X'$ in $\TSet$ (i.e. a natural transformation $X \implies X'$) which makes the following diagram commute, for all $a$ in $S_{i-1}$.
\diagram{D1}
Identities and compositions are that of the base category $\TSet$, and categorical equalities are trivially derived from the diagram above.
The diagram expresses as
\[
\forall \gamma \in \Hom_{G_{a-1}\Gamma_a,X}, \quad \Cstr'_a(f \circ \gamma) = f_\UU(\Cstr_a \gamma)
\]
\todo{Define properly the use of \this}
\subsection{$H$ functors}
For every object $X$ of a locally small category $C$, we define the functor $H_X : (X/\Set) \to \TSet$
\[
H_X(Y,f) = \TSetObject{X}{f}{Y}
\]
Dually, we make another functor $\Hbar_X : (\Set/X) \to \TSet$
\[
\Hbar_X(Y,f) = \TSetObject{Y}{f}{X}
\]
The morphisms translate as-is, and composition and identity relies on that of $(X/\Set)$ or $(\Set/X)$.
\todo{(small) Show that it is actually a functor (should be trivial), potentially add a diagram}
\subsection{$G$ and $F$ infinite constructions}
\[
G_i(Y) = \left(\sum_{a\in S_i}H_{Y(a)}\left(\lim_{(a/S_i)*}\overline{Y_a}\right), λa.λ\eta.(a,u (\inj_2 \star))\right)
\]
where $u : \left(Y(a) \oplus 1, \inj_1\right) \to (\lim_{(a/S_i)*} \overline{Y_a})$
so that $\varphi_{b,f} u (\star) = \eta_\El^b(f)$
\[
F_i(X,\Cstr)(a) = X(p)^{-1}\left(\Cstr_a\left(\Hom_{\BB_{a-1}}(G_{a-1}\Gamma_a,X)\right)\right)
\]
\[
F_i(X,\Cstr)(f : a \to b)(X(p)(x); x \in \Cstr_a(\eta)) = \eta_\El^b(f)
\]
\subsection{Hypohesis}
\begin{property}[H3]
\[
F_i(X \oplus L_0^i Y) \cong F_i X
\]
\end{property}
\todo{Do we need to specify that the adjunction is (rtl) $F(inj_1^i)$ ? And prove it ?}
\begin{property}[H3']
\[
\Hom(G_i\Gamma, X) \cong \Hom(G_i\Gamma,X \oplus L_0^i Y)
\]
With left to right isomorphism being $(inj_1^i \circ \dash)$
\end{property}
\begin{property}[H1]
\[
R_{i-1}^i(X \oplus_i L_0^i Y) \cong R_{i-1}^i X \oplus L_0^{i-1} Y
\]
\end{property}
\begin{property}[H1r]
\[
R_{0}^i(X \oplus_i L_0^i Y) \cong R_{0}^i X \oplus_{i-1} L_0^{i-1} Y
\]
We use the sum that makes this equivalence to be an equality.
\end{property}
\subsection{Notations}
\[
R_i^j = R_{i}^{i+1} \circ R_{i+1}^j = R_{i}^{j-1} \circ R_{j-1}^{j} = R_{i}^{i+1} \circ ... \circ R_{j-1}^{j}
\]
\[
L_i^j = L_{j-1}^{j} \circ L_{i}^{j-1} = L_{i+1}^{j} \circ L_{i}^{i+1} = L_{j-1}^{j} \circ ... \circ L_{i}^{i+1}
\]
In any category with a coproduct, for every morphisms $f : X \to Z$ and $g : Y \to Z$, we denote with $\{f;g\}$ the unique morphism from $X \oplus Y$ to $Z$ such that $\{f;g\} \circ \inj_1 = f$ and $\{f;g\} \circ \inj_2 = g$.
\subsection{Recursive definition of $\BB_i$}
\[
\BB_i := \left(X : \BB_{i-1}, \Cstr_i : \Hom_{\BB_{i-1}} (G_{i-1}\Gamma_i,X) \to (R_0^{i-1}X)_\UU \right)
\]
A morphism $(X,\Cstr_i) \to (X',\Cstr'_i)$ is a morphism $f : X \to X'$ in $\BB_{i-1}$ such that the following diagram commutes.
\diagram{D1 $[a \mapsfrom i]$}
Identities and compositions are that of the category $\BB_{i-1}$, and categorical equalities are trivially derived from the diagram above.
\subsection{W definition}
We define a functor $W : \displaystyle\prod_{X : \BB_{i-1}} (\Set/\Hom_{\BB_{i-1}}(G_{i-1}\Gamma_i,X)) \to \BB_{i}$
\[
W(X,Y) := \left(X \oplus L_0^{i-1} \Hbar_{\Hom(G_{i-1}\Gamma_i,\dash)}(X,Y), \widetilde{\inj_2} \right)
\]
With $\widetilde{\inj_2}$ being defined by
\[
\begin{array}{lcl}
\Hom(G_{i-1}\Gamma_i,X \oplus L_0^{i-1} \Hbar_\bullet(X,Y)) & \to^{\text{H3'}} & \Hom(G_{i-1}\Gamma_i,X)\\
& = & \Hbar_\bullet(X,Y)_\UU \\
& \to^{\inj_2^0} & \left(R_0^{i-1}X \oplus \Hbar_\bullet(X,Y)\right)_\UU \\
& \to^{\text{H1r}} & \left(R_0^{i-1}(X \oplus L_0^{i-1}\Hbar_\bullet(X,Y))\right)_\UU
\end{array}
\]
The action on a morphism $(g,h)$ from $(X,Y)$ to $(X',Y')$ is the following:
\[
W(g,h) := \left(g \oplus L_0^{i-1} \Hbar_{\Hom(G_{i-1}\Gamma_i,\dash)}(g,h)\right)
\]
It is indeed a morphism from $\BB_{i}$ as it makes the following diagram commute.
\diagram{D2}
\subsection{E definition}
We define $E : \BB_{i} \to \displaystyle\prod_{X : \BB_{i-1}} (\Set/\Hom_{\BB_{i-1}}(G_{i-1}\Gamma_i,X))$
\[
E(X) = (R_{i-1}^i X, (A,h))
\]
Where $(A,h)$ is defined as the following pullback:
\diagram{D3a}
The action on a morphism $f$ from $X$ to $X'$ is defined as $E(f) = (R_{i-1}^i f, !)$, with $!$ being the only morphism making the following diagram commute:
\diagram{D3b}
\subsection{W E adjunction}
We have an adjunction $W \dashv E$.
Let $(X,Y)$ be in $\displaystyle\prod_{X : \BB_{i-1}} (\Set/\Hom_{\BB_{i-1}}(G_{i-1}\Gamma_i,X))$ and $Z$ be in $\BB_i$.
We will construct a natural isomorphism.
\[
\phi_{XYZ} : \Hom(W(X,Y),Z) \to \Hom((X,Y),E(Z))
\]
\subsubsection{Constructing $\phi_{XYZ}$}
Let $f$ be in $\Hom(W(X,Y),Z)$.
We want to construct $\phi_{XYZ}(f) : (X,Y) \to E(Z)$.
The first component of $\phi_{XYZ}(f)$ is $R_{i-1}^i f \circ inj_1^{i-1} : X \to R_{i-1}^i Z$, with $R_{i-1}^i f$ being a morphism of $\BB_{i-1}$ from $R_{i-1}^i(W(X,Y)) = X \oplus_{i-1} L_0^{i-1}$ to $R_{i-1}^i Z$.
The second component is defined through the universal property of the pullback defined by $E(Z)$ according to the following diagram:
\diagram{D6}
\subsubsection{Constructing $\phi^{-1}_{XYZ}$}
Now, we take $(g,h)$ a morphism from $(X,Y)$ to $E(Z)$.
We define $\phi^{-1}_{XYZ}(g,h)$ as a morphism of $\BB_i$ from $W(X,Y)$ to $Z$, i.e. a morphism of $\BB_{i-1}$ from $X \oplus_{i-1} L_0^{i-1} \Hbar (X,Y)$ to $R_0^{i-1}(Z)$ that makes a certain diagram commute:
\[
\phi^{-1}_{XYZ}(g,h) = \left\{g ; \eta_0^i \circ L_0^{i-1} \square \right\}
\]
Where $\square$ is a morphism $\Hbar (X,Y) \to R_0^i Z$ defined by the following diagram:
\diagram{D7}
This is indeed a morphism of $\BB_i$ from $W(X,Y)$ to $Z$ as it makes the following diagram commute
\diagram{D8}
We now have to show that $\phi_{XYZ}$ and $\phi_{XYZ}^{-1}$ do make a natural isomorphism.
\subsubsection{Composition $\phi_{XYZ} \circ \phi_{XYZ}^{-1}$}
The first component of $\phi_{XYZ} (\phi_{XYZ}^{-1}(g,h))$ is
\[
R_{i-1}^i(\left\{g ; \eta_0^i \circ L_0^{i-1} \square \right\}) \circ \inj_1^{i-1} = \left\{g ; \eta_0^i \circ L_0^{i-1} \square \right\} \circ \inj_1^{i-1} = g
\]
The second component of $\phi_{XYZ} (\phi_{XYZ}^{-1}(g,h))$ follows the following diagram
\diagram{D9}
The diagram commutes as the following equality holds:
\[
\left(\left\{g ; \eta_0^i \circ L_0^{i-1} \square \right\} \circ \inj^{i-1}_2\right)_\El = \left(\eta_0^i \circ L_0^{i-1} \square\right)_\El = (\eta_0^i)_\El \circ (L_0^{i-1} \square)_\El = \square_El = \pi_1 \circ h
\]
\todo{Justify $(\eta_0^i)_\El = id_{\BB_i}$ and $(L_0^i(f))_\El = f_\El$}
So, as the square is a pullback, we get the complete equality
\[
\phi_{XYZ} (\phi_{XYZ}^{-1}(g,h)) = (g,h)
\]
\subsubsection{Composition $\phi_{XYZ}^{-1} \circ \phi_{XYZ}$}
Now, the converse composition is
\[
\phi_{XYZ}^{-1} (\phi_{XYZ}(f)) = \left\{R_{i-1}^i f \circ \inj_1^{i-1} ; \eta_0^i \circ L_0^{i-1} \square \right\}
\]
where $\square$ follows the following diagram
\diagram{D10}
We want to show that $\phi_{XYZ}^{-1} (\phi_{XYZ}(f))$. By definition of $\{;\}$ in $\BB_1$, it suffices to show that $\eta_0^i \circ L_0^{i-1} \square = R_{i-1}^i f \circ \inj_1^{i-1}$.
So it suffices to show that
\[
\square = R_0^{i-1}(R_{i-1}^i f \circ \inj_2^{i-1}) \circ \varepsilon_0^{i-1} = R_0^i f \circ \left(R_0^{i-1} \inj_2^{i-1} \circ \varepsilon_0^{i-1}\right) = R_0^i f \circ \inj_2^0
\]
The two required equalities are proved by the diagram above:
\begin{align*}
\square_\El = \left(R_0^i f \circ \inj_2^0\right)_\El \\
\square_\UU = \left(R_0^i f \circ \inj_2^0\right)_\UU
\end{align*}
\todo{Expliciter à un endroit que $\Cstr^{W(X,Y)} = inj_2^\Set \circ \left(\inj_1^{i-1} \circ \dash\right)$ (déduit de la définition et de la forme de l'iso H3' et H1r=id)}
\todo{Show $R_0^{i-1} \inj_2^{i-1} \circ \varepsilon_0^{i-1} = \inj_2^0$}
\subsubsection{Naturality}
We want to show that the following diagram commutes, for any objects $X$,$Y$,$Z$,$X'$,$Y'$,$Z'$ and morphisms $f$,$g$,$h$.
\diagram{D10.0}
We take a morphism $\ii$ in $\Hom\left(W(X,Y),Z\right)$. We want to show that it is sent by the above diagram to the same morphism of $\Hom\left((X,Y),E(Z)\right)$.
We first look at the first component of the result morphism.
\[\begin{array}{rcl}
\phi_{XYZ}(f \circ \ii \circ W(g,h))_1
&=& R_{i-1}^i\left(f \circ \ii \circ (g \oplus L_0^{i-1} \Hbar(g,h))^+\right) \circ \inj_1^{i-1} \\
&=& R_{i-1}^i f \circ R_{i-1}^i \ii \circ \left\{\inj_1^{i-1} \circ g; \dots\right\} \circ \inj_1^{i-1} \\
&=& R_{i-1}^i f \circ R_{i-1}^i \ii \circ \inj_1^{i-1} \circ g \\
&=& \left[E(f) \circ \phi_{X'Y'Z'}(\ii) \circ (g,h)\right]_1
\end{array}\]
The second components are defined as described by the following diagram
\diagram{D10}
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 blue line. And that of $E(f) \circ \phi_{X'Y'Z'}(\ii) \circ (g,h)$ is defined by the circled path.
As the diagram commutes and by pullback property, we get the equality.
\subsection{Proof of H1 - Sum definition}
We will define the sums of the form $X \oplus_i L_0^i Y$ in $\BB_i$.
\[
X \oplus_i L_0^i Y := \left(R_{i-1}^i \oplus_{i-1} L_0^{i-1} Y, (R_0^{i-1} \inj_1^{i-1})_\UU \circ \Cstr_i^X \circ (\inj_1^{i-1} \circ \dash)^{-1}\right)
\]
Here, $(\inj_1^{i-1} \circ \dash)^{-1}$ is the inverse of the isomorphism of hypothesis H3', and
The constructor goes as follows:
\[
\Hom_{\BB_{i-1}}(G_{i-1}\Gamma_i,R_{i-1}^i X \oplus_{i-1} L_0^{i-1} Y) \equiv \Hom_{\BB_{i-1}}(G_{i-1}\Gamma_i,R_{i-1}^i X) \to (R_0^{i-1} X)_\UU \to (R_0^i (R_{i-1}^i X \oplus_{i-1} L_0^{i-1} Y))_\UU
\]
The first injector $X \to X \oplus_i L_0^i Y$ is defined as follows:
\[
\inj_1^i := \inj_1^{i-1} : R_{i-1}^i X \to R_{i-1}^i (X \oplus_i L_0^i Y) = R_{i-1}^i X \oplus_{i-1} L_0^{i-1} Y
\]
It is a morphism of $\BB_i$ as it makes the following diagram commute:
\diagram{D4}
The second injector is defined as follows:
\[
inj_2^i := (\eta_i \oplus_i \id_{L_0^i Y}) \circ L_{i-1}^i \inj_2^{i-1}
\]
Where $\eta_i$ is the counit of the adjunction $R_{i-1}^i \vdash L_{i-1}^i$, going from $L_{i-1}^i R_{i-1}^i X$ to $X$.
This goes from $L_0^i Y = L_{i-1}^i L_0^{i-1} Y$ to $L_{i-1}^i(R_{i-1}^i X \oplus_{i-1} L_0^{i-1} Y)$, which is equivalent to $L_{i-1}^i R_{i-1}^i X \oplus_i L_0^i Y)$ as $L_{i-1}^i$ is a left-adjunct functor and therefore it preserves colimits; then goes to $X \oplus_i L_0^i Y$.
We will now show that this definition is actually a definition of the coproduct in $\BB_i$.
To that extent, we take two objects $X$ and $Z$ in $\BB_i$, $Y$ in $\TSet$ and two morphisms of $\BB_i$ $\varphi_1 : X \to Z$ and $\varphi_2 : L_0^i Y \to Z$.
We define $\{\varphi_1 ; \varphi_2\}_{i } = \{R_{i-1}^i \varphi_1 ; R_{i-1}^i \varphi_2 \circ \varepsilon^i_{L_0^{i-1} Y}\}_{i-1}$ a morphism of $\BB_i$ as it makes the following diagram commute:
\diagram{D5}
\todo{Justifier $R_{i-1}^i(\eta_i \oplus_i \id_{L_0^i Y}) = R_{i-1}^i \eta_i \oplus_{i-1} \id_{L_0^{i-1} Y}$ (with H1 ?)}
\section{Summary}
\lipsum[2-3]
\section{Bibliography}
\begingroup
\renewcommand{\section}[2]{}%
\printbibliography
\endgroup
\newpage
\addappheadtotoc
\appendix
\addtocontents{toc}{\protect\setcounter{tocdepth}{-1}}
\appendixpage
\end{document}