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