the abstract proof
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draft/draft.tex
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draft/draft.tex
@ -949,6 +949,59 @@ We can define $(-)^\dagger$ as a functor from $\gra(\BC)\to\rel(\BC)$, then we d
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\arrow["{{\brks{{(Fp_1)^\dagger},{(Fp_2)^\dagger}}}}"{description}, dashed, tail, from=1-2, to=3-2]
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\end{tikzcd}
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\end{equation*}
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\begin{lemma}\label{lem:norm-simp}
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Assuming that we have the following commutative diagram:
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\begin{equation*}
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\begin{tikzcd}[ampersand replacement=\&]
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X \& R \& X \\
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FX \& FR \& FX
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\arrow["\alpha"', from=1-1, to=2-1]
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\arrow["{{p_1}}"', from=1-2, to=1-1]
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\arrow["{{p_2}}", from=1-2, to=1-3]
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\arrow["{{\sigma}}", from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["{Fp_1}", from=2-2, to=2-1]
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\arrow["{Fp_2}"', from=2-2, to=2-3]
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\end{tikzcd}
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\end{equation*}
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Then there exists $\sigma^\dagger\c R\to(FR)^\dagger$ in the following diagram that is also commutative:
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\begin{equation*}
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\begin{tikzcd}[ampersand replacement=\&]
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X \& R \& X \\
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FX \& {(FR)^\dagger} \& FX
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\arrow["\alpha"', from=1-1, to=2-1]
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\arrow["{{p_1}}"', from=1-2, to=1-1]
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\arrow["{{p_2}}", from=1-2, to=1-3]
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\arrow["{{\sigma^\dagger}}", from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["{Fp_1^\dagger}", from=2-2, to=2-1]
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\arrow["{Fp_2^\dagger}"', from=2-2, to=2-3]
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\end{tikzcd}
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\end{equation*}
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\end{lemma}
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\begin{proof}
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The proof is trivial considering that $\sigma^\dagger=e_{FR}\comp\sigma$, where $e_{FR}$ is the epimorphism in the epi-mono factorization of $\brks{Fp_1,Fp_2}$, as depicted in the following diagram:
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\begin{equation*}
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\begin{tikzcd}[ampersand replacement=\&]
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X \& R \& X \\
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FX \& FR \& FX \\
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FX \& {(FR)^\dagger} \& FX
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\arrow["\alpha"', from=1-1, to=2-1]
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\arrow["{{p_1}}"', from=1-2, to=1-1]
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\arrow["{{p_2}}", from=1-2, to=1-3]
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\arrow["{{\sigma}}", from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["\id"', from=2-1, to=3-1]
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\arrow["{Fp_1}", from=2-2, to=2-1]
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\arrow["{Fp_2}"', from=2-2, to=2-3]
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\arrow["{e_{FR}}", from=2-2, to=3-2]
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\arrow["\id", from=2-3, to=3-3]
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\arrow["{Fp_1^\dagger}", from=3-2, to=3-1]
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\arrow["{Fp_2^\dagger}"', from=3-2, to=3-3]
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\end{tikzcd}
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\end{equation*}
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\qed
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\end{proof}
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We show this relation with $F_\rel(R,X)$.
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\begin{definition}[Simulation]\label{def:sim}
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A coalgebra $\sigma\c R\to (FR)^\dagger$ is a simulation over the $F$-coalgebra $\alpha\c X\to FX$ iff the following diagram is lax-commutative:
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@ -1757,8 +1810,8 @@ At the moment we have limited the discussion to the category of sets and we are
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\arrow["{p_2}", from=1-2, to=1-3]
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\arrow["\sigma", dashed, from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["{{\mathcal{P}p_1^\dagger}_\subseteq}", from=2-2, to=2-1]
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\arrow["{{\mathcal{P}p_2^\dagger}_\subseteq}"', from=2-2, to=2-3]
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\arrow["{{\mathcal{P}p_1}_\subseteq}", from=2-2, to=2-1]
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\arrow["{{\mathcal{P}p_2}_\subseteq}"', from=2-2, to=2-3]
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\end{tikzcd}
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\end{equation*}
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It is defined as $\sigma(x_1,x_2)=(\alpha(x_1),\alpha(x_2))$.
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@ -1792,8 +1845,8 @@ Because assuming we have $\sigma$, for every given $\sigma'$ we can define a $\d
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\arrow["id"', from=2-1, to=3-1]
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\arrow["\delta", from=2-2, to=3-2]
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\arrow["id"', from=2-3, to=3-3]
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\arrow["{{\mathcal{P}p_1^\dagger}_\subseteq}", from=3-2, to=3-1]
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\arrow["{{\mathcal{P}p_2^\dagger}_\subseteq}"', from=3-2, to=3-3]
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\arrow["{{\mathcal{P}p_1}_\subseteq}", from=3-2, to=3-1]
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\arrow["{{\mathcal{P}p_2}_\subseteq}"', from=3-2, to=3-3]
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\end{tikzcd}
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\end{equation*}
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$\delta$ is defined as the following:
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@ -1804,6 +1857,69 @@ $\delta$ is defined as the following:
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w & \mathsf{o.w}
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\end{cases}
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\end{gather*}
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Now, we want to prove an abstraction of this statement. First, we need to spell out what $\appr;(FR)^\dagger;\appr$ is. We define relation compositions with pullbacks, so we have the following diagram:
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\begin{equation*}
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\begin{tikzcd}[ampersand replacement=\&]
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{\appr;(FR)^\dagger;\appr} \& {\appr;(FR)^\dagger} \& {\appr} \& {FX} \\
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\& (FR)^\dagger \& {FX} \\
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{\appr} \& {FX} \\
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{FX}
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\arrow["{{{{{{{\pi_1}}}}}}}", from=1-1, to=1-2]
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\arrow["{{{{{{{\pi_2}}}}}}}"', from=1-1, to=3-1]
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\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-2]
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\arrow["{{{{{{{\varphi_1}}}}}}}", from=1-2, to=1-3]
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\arrow["{{{{{{{\varphi_2}}}}}}}", from=1-2, to=2-2]
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\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=2-3]
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\arrow["{{{{{{{i_1}}}}}}}", from=1-3, to=1-4]
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\arrow["{{{{{{{i_2}}}}}}}", from=1-3, to=2-3]
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\arrow["{{{{{{{Fp_1^\dagger}}}}}}}", from=2-2, to=2-3]
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\arrow["{{{{{{{Fp_2^\dagger}}}}}}}"', from=2-2, to=3-2]
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\arrow["{{{{{{{i_1}}}}}}}"', from=3-1, to=3-2]
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\arrow["{{{{{{{i_2}}}}}}}"', from=3-1, to=4-1]
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\end{tikzcd}
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\end{equation*}
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Additionally, we make the abbreviations that ${Fp_1}_\appr=i_1\comp\varphi_1\comp\pi_1$ and ${Fp_2}_\appr=i_2\comp\pi_2$.
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\begin{prop}
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Assuming we have a morphism $\sigma'\c R\to(FR)^\dagger$ then exists $\delta\c(FR)^\dagger\to(\appr;(FR)^\dagger;\appr)^\dagger$ such that $\sigma=\delta\comp\sigma'$ that commutes in the following diagram:
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\begin{equation*}
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\begin{tikzcd}[ampersand replacement=\&]
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X \& R \& X \\
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{FX} \& {(FR)^\dagger} \& {FX} \\
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{FX} \& {(\appr;(FR)^\dagger;\appr)^\dagger} \& {FX}
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\arrow["\alpha"', from=1-1, to=2-1]
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\arrow["{p_1}"', from=1-2, to=1-1]
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\arrow["{p_2}", from=1-2, to=1-3]
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\arrow["{\sigma'}", from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["id"', from=2-1, to=3-1]
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\arrow["\delta", from=2-2, to=3-2]
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\arrow["id"', from=2-3, to=3-3]
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\arrow["{({Fp_1}_\appr)^\dagger}", from=3-2, to=3-1]
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\arrow["{({Fp_2}_\appr)^\dagger}"', from=3-2, to=3-3]
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\end{tikzcd}
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\end{equation*}
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\end{prop}
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\begin{proof}
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Ultimately, we need to define $\delta'\c(FR)\to\appr;(FR)^\dagger;\appr$, where $\delta=e_{\appr;(FR)^\dagger;\appr}\comp\delta'$, and $e_{\appr;(FR)^\dagger;\appr}\c\appr;(FR)^\dagger;\appr\to(\appr;(FR)^\dagger;\appr)^\dagger$ is the epimorphism in the epi-mono factorization of $\brks{{Fp_1}_\appr,{Fp_2}_\appr}$ because by~\autoref{lem:norm-simp} it suffices to show that the following diagram commutes:
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\begin{equation*}
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\begin{tikzcd}[ampersand replacement=\&]
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X \& R \& X \\
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FX \& {(FR)^\dagger} \& FX \\
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FX \& {\appr;(FR)^\dagger;\appr} \& FX
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\arrow["\alpha"', from=1-1, to=2-1]
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\arrow["{{p_1}}"', from=1-2, to=1-1]
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\arrow["{{p_2}}", from=1-2, to=1-3]
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\arrow["{{\sigma'}}", from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["id"', from=2-1, to=3-1]
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\arrow["{\delta'}", from=2-2, to=3-2]
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\arrow["id"', from=2-3, to=3-3]
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\arrow["{{{Fp_1}_\appr}}", from=3-2, to=3-1]
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\arrow["{{{Fp_2}_\appr}}"', from=3-2, to=3-3]
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\end{tikzcd}
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\end{equation*}
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So, we need to define $\delta'$.
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\end{proof}
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\end{document}
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