Merge branch 'master' of git.wlog.site:pouya/coalgebraic-simulation
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@@ -292,7 +292,9 @@
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\newcommand{\spto}{\mathrel{\tikz{\draw[-{Stealth}] (0.0,0) -- (0.4,0); \draw (0.17,0.07) -- (0.17,-0.07);}}}
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\newcommand{\rto}{\mathrel{\tikz{\draw[-{Stealth}] (0.04,0) -- (0.4,0); \draw (0.17,0.07) -- (0.17,-0.07);\draw (0.04,0) -- (0,-0.07);\draw (0.04,0) -- (0,0.07);}}}
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\newcommand{\powf}{\mathcal{P}}
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\newcommand{\powfi}{\mathcal{P}_{\mathsf{inj}}}
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\newcommand{\sappr}{\sqsupseteq}
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\newcommand{\Dom}{\mathsf{Dom}}
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\newcommand{\simeet}{%
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\mathbin{%
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@@ -401,6 +403,117 @@ We want to say that a natural order structure entails an order over a functor de
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\begin{proof}
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We take $\tilde{F}X=\Hom(1,FX)$. Assuming that $U$ takes every poset to its carrier set, then we have $U\comp\tilde{F}X=FX$ for every object $X$.\qed
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\end{proof}
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\begin{definition}[Liftable Order Structure]\label{def:liftable-ord}
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A \emph{liftable order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ that is a natural order structure, and if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $h\appr Fg\comp k$ in $\Hom(X,FZ)$, then there is $k'\c X\to FY$ such that $k'\appr k$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
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\end{definition}
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\begin{definition}[Coliftable Order Structure]\label{def:coliftable-ord}
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A \emph{coliftable order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ that is a natural order structure, and if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $Fg\comp k\appr h $ in $\Hom(X,FZ)$, and $g$ is surjective, then there is $k'\c X\to FY$ such that $k\appr k'$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
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\end{definition}
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\begin{lemma}\label{lem:set-ord-str}
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In $\Set$, assuming that for every set $Y$ we have an order $\leq$ on $\Hom(1,FY)$ that satisfies \autoref{def:nat-ord}.\eqref{item:nat-ord:II}, and if for $A\in\Hom(1,FZ)$, $B\in\Hom(1,FY)$, and $g\c Y\to Z$, $h\leq Fg(B)$ in $\Hom(1,FZ)$ there exists $B'\in\Hom(1,FY)$ such that $B'\leq B$ and $B\in\Hom(1,FY)$ and $A=Fg(B')$, then there exists an order structure $\appr$ on $F$ that is a liftable order structure.
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\end{lemma}
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\begin{proof}
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For every $f,g\in\Hom(X,FY)$, we define $f\appr g$ whenever for all $x\in X$, $f(x)\leq g(x)$.
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Now, assuming $f\c X'\to X$, $\alpha,\beta\c X\to FY$, and $\alpha\appr\beta$, then for every $x'\in X'$, we have $\alpha(f(x'))\leq\beta(f(x'))$, so we have $\alpha\comp f\appr \beta\comp f$. Additionally, if we assume $g\c Y\to Y'$, then since $\leq$ satisfies \autoref{def:nat-ord}.\eqref{item:nat-ord:II}, for every $x\in X$ we have $Fg\comp\alpha(x)\leq Fg\comp\beta(x)$.
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Furthermore, if we have $g\c Y\to Z$, $k\c X\to FY$, $h\c X\to FZ$, and $Fg\comp k\appr h$, then for every $x\in X$ we have $Fg\comp k(x)\leq h(x)$, thus by the assumption of the lemma, for every $x$, there exists a set that we call $k'(x)$, for which we have $Fg(k'(x))\leq h(x)$. So, we have a function $k'\c X\to FY$, such that for every $x$ we have $Fg\comp k'(x)\leq h(x)$ that means $Fg\comp k'\appr h$.
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\qed
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\end{proof}
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\begin{remark}\label{rem:set-ord-str-co}
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The exact same argument in~\autoref{lem:set-ord-str} is true for coliftable order structures.
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\end{remark}
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\begin{prop}\label{prop:lift-gen-func}
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For a functor $F\c\Set\to\Set$, a functor of the form $\powf F$, where $\powf$ is the powerset functor, the set inclusion is a liftable order structure.
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\end{prop}
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\begin{proof}
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Using~\autoref{lem:set-ord-str} we only prove the case for every $h\in\Hom(1,\powf FZ)$, $k\in\Hom(1,\powf FY)$. Additionally, for every $g\c Y\to Z$, such that $h\subseteq\powf Fg(k)$, we define $k'\in\powf FY$ that that $k'=\{y\mid Fg(y)\in h\}$. We show that $k'\subseteq k$ and $\powf Fg(k')=h$.
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Assuming $y\in k'$ we have $Fg(y)\in h$. Since $h\subseteq \powf Fg(k)$ we have $Fg(y)\in\powf Fg(k)$, so we have $y\in k$ that means $k'\subseteq k$.
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Assuming $z\in\powf Fg(k')$, then there exists $y'\in k'$ that $z=Fg(y')$, so $z\in h$ and $\powf Fg(k')\subseteq h$.
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Assuming $z\in h$, since $h\subseteq \powf Fg(k)$, then $z\in\powf Fg(k)$. So, there exists $y'\in k$ such that $Fg(y')=z$. So, by the definition of $k'$ we have $y'\in k'$ that means $z\in\powf Fg(k')$.\qed
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\end{proof}
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\begin{example}
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In the category of sets, subset over the powerset functor is an example of a coliftable order structure if the $g$ in~\autoref{def:coliftable-ord} is a surjective. Using~\autoref{rem:set-ord-str-co} we only prove the case for every $h\in\Hom(1,\powf Z)$, $k\in\Hom(1,\powf Y)$. Additionally, $g\c Y\to Z$, such that $\powf g(k)\subseteq h$. We define $k'=k\cup\{y\mid g(y)\in h\}$, and we show that $\powf g(k')=h$.
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Obviously, $\powf g(k')\subseteq h$. Now, assuming $z\in h$ we prove that $z\in \powf g(k')$. Since $g$ is surjective, then exists $A\subseteq Y$ such that $\powf g(A)=h$. By the definition of $k'$, $A\subseteq k'$. So from $z\in h$ we have $z\in \powf g(A)$ that means that exists $a\in A$, such that $g(a)=z$. Now, since $A\subseteq k'$, then $a\in k'$. So, we have $z\in \powf g(k')$.\qed
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\end{example}
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\begin{remark}
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We give an example that justifies why the $g$ in~\autoref{def:coliftable-ord} should be surjective.
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If we take the surjectivity of $g$ out of~\autoref{def:coliftable-ord}, subset relation over the powerset functor is not an example of a coliftable order! For some set $X$ we take $h\c X\to\powf\mathbb{Z}$, for every $x\in X$, $h(x)=\mathbb{Z}$, $g\c\mathbb{Z}\to\mathbb{Z}$, and for every $z\in\mathbb{Z}$,
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\begin{gather*}
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g(z)=
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\begin{cases}
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0 & z\in\mathbb{Z}^+ \\
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1 & \mathsf{otherwise}
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\end{cases}
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\end{gather*}
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then
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\begin{gather*}
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\powf g(A)=
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\begin{cases}
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\emptyset & A=\emptyset \\
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\{0\} & A\subseteq \mathbb{Z}^+,A\neq\emptyset\\
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\{1\} & A\subseteq \mathbb{Z}^-\cup\{0\},A\neq\emptyset\\
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\{0,1\} & \mathsf{otherwise}
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\end{cases}
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\end{gather*}
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then no matter what $k\c X\to\powf\mathbb{Z}$ is there will be no $k'\c X\to\powf\mathbb{Z}$, for every $x\in X$, $\powf g(k'(x))=h(x)$, as $\powf g(k'(x))\subseteq\{0,1\}$, while $h(x)=\mathbb{Z}$, so $\powf g(k'(x))\subset h(x)$.
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\end{remark}
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\begin{remark}
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We could not prove a general statement like~\autoref{prop:lift-gen-func} for coliftability even if the surgectivity of $g$ remains in the definition. Perhaps more conditions on $F$ in~\autoref{prop:lift-gen-func} are needed. Conditions like preserving surjections and unions that make the statement extremely limited, and meaningless.
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\end{remark}
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\begin{lemma}\label{lem:liftable}
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Assuming that a a functor $F$ has an order structure $\appr$ that is liftable, then for every $f\in\Hom(X,FY)$ we have:
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\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
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\item $Ff\comp\sappr\quad=\quad\sappr\comp Ff$
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\item $(Ff)^\op\comp\appr\quad=\quad\appr\comp (Ff)^\op$
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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$(I)$ Assuming $t\mathrel{(Ff\comp\sappr)} x$, there exists $s$ such that $t\sappr s$ and $Ff(s)=x$. Since $\appr$ is liftable, and thus natural, by~\autoref{def:nat-ord}, from $s\appr t$ we get $x\appr Ff(t)$ that is $t\mathrel{(\sappr\comp Ff)} x$.
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Assuming $t\mathrel{(\sappr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\sappr x$. By~\autoref{def:liftable-ord} since $Ff(t)\sappr x$ there exists $s$ that $t\sappr s$ and $Ff(s)=x$ that is $t\mathrel{(Ff\comp\appr)}s$.
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$(II)$ Basically, by definition of $\op$ and relation composition we have
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\begin{gather*}
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(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
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(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
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\end{gather*}
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So it follows directly from applying $\op$ on both sides of $(I)$.\qed
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\end{proof}
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\begin{lemma}\label{lem:coliftable}
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Assuming that a a functor $F$ has an order structure $\appr$ that is coliftable, then for every surjective $f\in\Hom(X,FY)$ we have:
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\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
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\item $Ff\comp\appr\quad=\quad\appr\comp Ff$
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\item $(Ff)^\op\comp\sappr\quad=\quad\sappr\comp (Ff)^\op$
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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$(I)$ Assuming $t\mathrel{(Ff\comp\appr)} x$, there exists $s$ such that $t\appr s$ and $Ff(s)=x$. Since $\appr$ is coliftable, and thus natural, by~\autoref{def:nat-ord}, from $t\appr s$ we get $Ff(t)\appr x$ that is $t\mathrel{(\appr\comp Ff)} x$.
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Assuming $t\mathrel{(\appr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\appr x$. By~\autoref{def:coliftable-ord} since $Ff(t)\appr x$ there exists $s$ that $t\appr s$ and $Ff(s)=x$ that is $t\mathrel{(Ff\comp\appr)}s$.
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$(II)$ Basically, by definition of $\op$ and relation composition we have
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\begin{gather*}
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(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
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(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
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\end{gather*}
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So it follows directly from applying $\op$ on both sides of $(I)$.\qed
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\end{proof}
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\section{Coalgebraic Bisimulation}%\label{sec:}
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In this section, by $\spa(\BC)$ we refer to spans in a category $\BC$ that has
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products, and by $\rel(\BC)$ we refer to the category of relations in $\BC$, i.e.\
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@@ -1091,7 +1204,7 @@ A big concern with this approach is that Comma Objects are defined in a 2-catego
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\subsection{Choosing a suitable order for our setting}
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Maybe we can first choose a suitable order on $T(\Sigma_\val\mS\times D(\mS,\mS))$ and then prove that if a relation and its inverse is a simulation then it is a bisimulation as well. Maybe $T$ being $\omega$-continuous can give the ordering. It can be something easier that relates to termination as well! That if a term has a big-step evaluation, then it is bigger than or equal to any other term, and if it does not, then it is less than or equal to any other term.
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\section{Symmetric Simulation is Bisimulation}
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\section{Symmetric Simulation is a Bisimulation}
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\begin{definition}[Graph]
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In a category $\BC$ a graph is a tuple $(R,X)$ of the following form:
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\begin{equation*}
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@@ -2049,6 +2162,35 @@ We recall that in the above diagram $\sigma_3$ is a bisimulation, and the rest a
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% \end{gather*}
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% We have $(Fp_1)^\dagger\comp\sigma$.
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%\end{proof}
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\begin{example}
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And another counter-example!!!
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Assuming that the functor is the powerset endofunctor over the category of sets and injective maps. We show the functor with $\powfi$. For every $X$, we define the order on $\powfi X$ as $A\appr B$ whenever $|A|\leq |B|$, where $|A|$ and $|B|$ are just cardinalities of $|A|$ and $|B|$ respectively. This is a preorder. Then we define the order over every $\Hom(X,\powfi Y)$ pointwise. We have chosen the category of sets with injective maps because we could not define a functor from $\Set$ to $\preord$ with the mentioned ordering.\\
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We take $R=\{(1,1),(1,2),(2,1),(2,2)\}$, and $X=\{1,2,3\}$. $\alpha$ is defined as below:
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\begin{gather*}
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\alpha(x)=
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\begin{cases}
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\{1,2\} & x=1 \\
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\{2,3\} & x=2\\
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\{3\} & x=3
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\end{cases}
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\end{gather*}
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$\sigma$ is defined as below:
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\begin{gather*}
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\sigma(w)=
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\begin{cases}
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\{(1,2),(2,2)\} & w=(1,2) \\
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\{(2,1),(2,2)\} & w=(2,1) \\
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\{(1,2),(2,1)\} & w=(1,1) \\
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\{(2,2),(1,1)\} & w=(2,2)
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\end{cases}
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\end{gather*}
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$R$ is symmetric, and $\sigma$ is a witness for $R$ to be an AM-simulation, but $R$ is not a bisimulation in the traditional sense because $(2,1)\in R$, and $2\to 3$, but $(3,1)$ or $(3,2)$ are not in $R$. It is easy to see that it is not an AM-bisimulation as well because we can not define a function that can serve as an evidence for it as $3$ does not appear in any pair in $R$, while it exists in $\alpha(2)$.
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This counter-example also works as a counter-example for Hughes-Jacobs definition of simulation. Actually, $\appr\comp(FR)^\dagger\comp\appr=\powf X\times \powf X$, so $R\subseteq( \appr\comp(FR)^\dagger\comp\appr)$ that means that $R$ is a simulation. Worth noting that they claim that their setting works for an arbitrary category. So, unlike their definition, in the context of the relator-based definitions that only work in $\Set$, this counter-example does not live.
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The mentioned ordering is not liftable. Assuming $h\in\Hom(\nats,\powfi \nats)$, $g\c \nats\to \nats$, and $k\in\Hom(\nats,\powfi \nats)$, and they are defined for every $n$ in $\nats$ as $h(n)=\{2\times n\}$, $g(n)=3\times n$, and $k(n)=\{n\}$, then $|h(n)|=|\powfi g(k(n))|=1$ that means $h\appr\powfi g\comp k$ is satisfied, but there is no $k'$ that $h=\powfi g\comp k$ because we can never have $h(1)=\powfi g\comp k'(1)$, as assuming $k'(1)=\{n\}$, and $n$ must be a natural number, then we should have $2=3\times n$ that is impossible.
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\end{example}
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\subsection{The concrete proof}
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%\begin{lemma}\label{lem:sim-opsim-inc1}\ppnote{Actually, this lemma holds for every functor in an arbitrary category.}
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% Assuming that $\sigma\c R\to\powf R$ is witness for a symmetric relation $R$ to be an AM simulation on $\powf$-coalgebra $(X,\alpha)$, then for all $(x_1,x_2)\in R$ we have:
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@@ -2192,18 +2334,26 @@ Now, we prove our main statement.
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\end{cor}
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Now, we make the proof more abstract. We prove the statement for set-functors of the form $\powf F$, where $F$ is an arbitrary set-functor, and $\powf$ is the powerset functor.
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\subsection{Maybe Functor}
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We prove that symmetric simulation is a bisimulation for the case that $FX=X+1$. First, we prove it for $\Set$. The order structure that we can define for this functor is that for a set $X$, the order is $\id_{X+1}\cup\{(\bot,x)\mid x\in X\}$.\\
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Now, we prove that the given order on the maybe functor is a liftable one.
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We prove that symmetric simulation is a bisimulation for the case that $FX=X+1$. First, we prove it for $\Set$. The order structure that we can define for this functor is that for sets $X$ and $Y$, and functions $f,g\c X\to Y+1$ we have $f\appr g$ whenever $\Dom(f)\subseteq\Dom(g)$, and for every $x\in\Dom(f)$ we have $f(x)=g(x)$ ($\Dom(f)$ is the domain of a function $f$).
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Now, we prove that the given order on the maybe functor is liftable and coliftable.
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\begin{lemma}\label{lem:maybe-lif}
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The order structure on the set-functor $FX=X+1$ is a liftable order.
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\end{lemma}
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\begin{proof}
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By~\autoref{lem:set-ord-str}, assuming $h\in\Hom(1,Y+1)$, $k\in\Hom(1,X+1)$, $g\c X\to Y$, and $h\appr Fg(k)$, we need to prove that exists $k'\in\Hom(1,X+1)$ such that $k'\appr k$ and $Fg(k')=h$. Since $h\in\Hom(1,Y+1)$ we have two cases:
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\begin{itemize}
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\item $h=\bot$: In this case we take $k'=\bot$, so we have $k'\appr k$, and then we have $Fg(k')=\bot=h$.
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\item $h\in Y$: Since $h\appr Fg(k)$ and $h\neq\bot$, we have $h=Fg(k)$. In this case we take $k'=k$, so $k'\appr k$ and $Fg(k')=h$.\qed
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\end{itemize}
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% By~\autoref{lem:set-ord-str}, assuming $h\in\Hom(1,Y+1)$, $k\in\Hom(1,X+1)$, $g\c X\to Y$, and $h\appr Fg(k)$, we need to prove that exists $k'\in\Hom(1,X+1)$ such that $k'\appr k$ and $Fg(k')=h$. Since $h\in\Hom(1,Y+1)$ we have two cases:
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% \begin{itemize}
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% \item $h=\bot$: In this case we take $k'=\bot$, so we have $k'\appr k$, and then we have $Fg(k')=\bot=h$.
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% \item $h\in Y$: Since $h\appr Fg(k)$ and $h\neq\bot$, we have $h=Fg(k)$. In this case we take $k'=k$, so $k'\appr k$ and $Fg(k')=h$.\qed
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% \end{itemize}
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Assuming that $h\in\Hom(X,Z+1)$, $g\c Y\to Z$, $k\in\Hom(X,Y+1)$, such that $h\appr (g+1)\comp k$. We define $k'\in\Hom(X,Y+1)$ as follows:
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\begin{gather*}
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k'(x)=
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\begin{cases}
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\bot&x\notin\Dom(h)\\
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k(x)&x\in\Dom(h)
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\end{cases}
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\end{gather*}
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We have $\Dom((g+1)\comp k')=\Dom(k')$ and $\Dom(k')=\Dom(h)$, so we have $\Dom((g+1)\comp k')=\Dom(h)$. If $x\in\Dom(h)$, then $h(x)=(g+1)\comp k(x)$ and $(g+1)\comp k(x)=(g+1)\comp k'(x)$, so we have $h(x)=(g+1)\comp k'(x)$. So, we have $h=(g+1)\comp k'$. Additionally, $\Dom(k')\subseteq\Dom(k)$, and for every $x\in \Dom(k')$, we have $k'(x)=k(x)$, so we have $k'\appr k$. \qed
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\end{proof}
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\begin{lemma}\label{lem:maybe-colif}
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The order structure on the set-functor $FX=X+1$ is a coliftable order.
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@@ -2286,41 +2436,79 @@ So, proven by Dubut, for every AM-simulation relation over a coalgebra $(X,\alph
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% Now, we are left to prove that if $\alpha(x_1)\in X$ and $\alpha(x_2)\in X$ then $\beta(x_1,x_2)\in R$ that means that the codomain of $\beta$ is indeed $R+1$. We assume $\alpha(x_i)\in X$. Since $Fp_i\comp\beta(x_1,x_2)=\alpha(x_i)\in X$ we have $\beta(x_1,x_2)\in R$.
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% \qed
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%\end{proof}
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%%%%%%Second proof:
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\begin{prop}\label{prop:sym-sim-bis-may}
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For a $F$-coalgebra $(X,\alpha)$ in $\Set$, such that $FX=X+1$, a symmetric AM-simulation relation $R$ on $X$ is an AM-bisimulation.
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\end{prop}
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\begin{proof}
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||||
We prove that $\sigma$ also is a witness for $R$ to be an AM-bisimulation. For every $(x_1,x_2)\in R$ we have
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\begin{enumerate}
|
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\item $\alpha(x_1)=p_1+1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1},
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||||
\item $\alpha(x_2)\sappr p_2+1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4},\\
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% \end{enumerate}
|
||||
and since $R$ is symmetric, we have
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% \begin{enumerate}
|
||||
\item $\alpha(x_2)=p_1+1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2},
|
||||
\item $\alpha(x_1)\sappr p_2+1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3},
|
||||
\end{enumerate}
|
||||
Followed by $\alpha(x_2)\sappr p_2+1\comp\sigma(x_1,x_2)$ we have the following cases:
|
||||
\begin{itemize}
|
||||
\item $\alpha(x_2)=p_2+1\comp\sigma(x_1,x_2)$: In this case, we already have $\sigma$ as a witness for $R$ to be a bisimulation.
|
||||
\item $Fp_2\comp\sigma(x_1,x_2)=\bot$: In this case, we have
|
||||
\begin{align*}
|
||||
p_2+1\comp\sigma(x_1,x_2)=\bot,&\\
|
||||
\Rightarrow&\sigma(x_1,x_2)=\bot,\\
|
||||
\Rightarrow&p_1+1\comp\sigma(x_1,x_2)=\bot,\\
|
||||
\Rightarrow&\alpha(x_1)=\bot,&\eqref{eq:maybe-func-set-1}\\
|
||||
\Rightarrow&p_2+1\comp\sigma(x_2,x_1)=\bot,&\eqref{eq:maybe-func-set-3}\\
|
||||
\Rightarrow&\sigma(x_2,x_1)=\bot,\\
|
||||
\Rightarrow&p_1+1\comp\sigma(x_2,x_1)=\bot,\\
|
||||
\Rightarrow&\alpha(x_2)=\bot,&\eqref{eq:maybe-func-set-2}\\
|
||||
\Rightarrow&\alpha(x_2)=p_1+1\comp\sigma(x_1,x_2).
|
||||
\end{align*}
|
||||
\end{itemize}\qed
|
||||
% \begin{enumerate}
|
||||
% \item $\alpha(x_1)=(p_1+1)\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1},
|
||||
% \item $\alpha(x_2)\sappr (p_2+1)\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4},\\
|
||||
% % \end{enumerate}
|
||||
% and since $R$ is symmetric, we have
|
||||
% % \begin{enumerate}
|
||||
% \item $\alpha(x_2)=(p_1+1)\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2},
|
||||
% \item $\alpha(x_1)\sappr (p_2+1)\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3},
|
||||
% \end{enumerate}
|
||||
% Followed by $\alpha(x_2)\sappr (p_2+1)\comp\sigma(x_1,x_2)$ we have the following cases:
|
||||
% \begin{itemize}
|
||||
% \item $\alpha(x_2)=(p_2+1)\comp\sigma(x_1,x_2)$: In this case, we already have $\sigma$ as a witness for $R$ to be a bisimulation.
|
||||
% \item $p_2+1\comp\sigma(x_1,x_2)=\bot$: In this case, we have
|
||||
% \begin{align*}
|
||||
% (p_2+1)\comp\sigma(x_1,x_2)=\bot,&\\
|
||||
% \Rightarrow&\sigma(x_1,x_2)=\bot,\\
|
||||
% \Rightarrow&(p_1+1)\comp\sigma(x_1,x_2)=\bot,\\
|
||||
% \Rightarrow&\alpha(x_1)=\bot,&\eqref{eq:maybe-func-set-1}\\
|
||||
% \Rightarrow&(p_2+1)\comp\sigma(x_2,x_1)=\bot,&\eqref{eq:maybe-func-set-3}\\
|
||||
% \Rightarrow&\sigma(x_2,x_1)=\bot,\\
|
||||
% \Rightarrow&(p_1+1)\comp\sigma(x_2,x_1)=\bot,\\
|
||||
% \Rightarrow&\alpha(x_2)=\bot,&\eqref{eq:maybe-func-set-2}\\
|
||||
% \Rightarrow&\alpha(x_2)=(p_1+1)\comp\sigma(x_1,x_2).
|
||||
% \end{align*}
|
||||
% \end{itemize}\qed
|
||||
\begin{enumerate}
|
||||
\item $\alpha\comp p_1=(p_1+1)\comp\sigma$\label{eq:maybe-func-set-1},
|
||||
\item $\alpha\comp p_2\sappr (p_2+1)\comp\sigma$\label{eq:maybe-func-set-2},\\
|
||||
and since $R$ is symmetric, we have
|
||||
\end{enumerate}
|
||||
Followed by~\eqref{eq:maybe-func-set-2} for every $(x_1,x_2)\in R$ we have $\alpha(x_2)=(p_2+1)\comp\sigma(x_1,x_2)$ and $\Dom((p_2+1)\comp\sigma)\subseteq\Dom(\alpha\comp p_2)$. We have
|
||||
\begin{align*}
|
||||
\Dom((p_2+1)\comp\sigma)&\\
|
||||
=&\Dom(\sigma)\\
|
||||
=&\Dom((p_1+1)\comp\sigma)\\
|
||||
=&\Dom(\alpha\comp p_1)&\by{\eqref{eq:maybe-func-set-1}}\\
|
||||
=&\Dom(\alpha\comp p_1\comp s)\\
|
||||
=&\Dom(\alpha\comp p_2)
|
||||
\end{align*}
|
||||
So, we have $\alpha\comp p_2= (p_2+1)\comp\sigma$ as well that means that $\sigma$ also serves as a witness for $R$ to be a bisimulation.
|
||||
\end{proof}
|
||||
|
||||
Now, we want to abstract the given proof for an extensive category that has terminal objects (so that we have the maybe functor). We assume a natural order structure $\appr$ for the maybe functor. %For every objects $X$ and $Y$, we define $\appr$ on each $\Hom(X,Y+1)$ by saying that for $f,g\in\Hom(X,Y+1)$ we have $f\appr g$ whenever either $f=g$ or $f=\bot$.\sgnote{This abstract order is coarser than (not the concretisation of) the pointwise Set order at \autoref{lem:set-ord-str}: here the \emph{whole} map must be $\bot$, whereas pointwise a map may be $\bot$ on some points and agree elsewhere. Please check the two agree on the hom-sets you actually use, or justify why the coarser order suffices.}
|
||||
Now, we want to abstract the given proof for an extensive category that has terminal objects (so that we have the maybe functor). We assume a natural order structure $\appr$ for the maybe functor. We use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$.%For every objects $X$ and $Y$, we define $\appr$ on each $\Hom(X,Y+1)$ by saying that for $f,g\in\Hom(X,Y+1)$ we have $f\appr g$ whenever either $f=g$ or $f=\bot$.\sgnote{This abstract order is coarser than (not the concretisation of) the pointwise Set order at \autoref{lem:set-ord-str}: here the \emph{whole} map must be $\bot$, whereas pointwise a map may be $\bot$ on some points and agree elsewhere. Please check the two agree on the hom-sets you actually use, or justify why the coarser order suffices.}
|
||||
%Although even in the context that we are at the moment, it does not seem plausible to prove that a symmetric simulation is a bisimulation without having an operator like $\join$ in~\autoref{prop:sym-sim-bis-maybe} that takes two morphisms of the same type and gives one.
|
||||
The following lemma is an abstraction of saying that in $\Set$ assuming $f\c X\to Y+1$ and $g\c Y\to Z$, we have $\Dom((g+1)\comp f)=\Dom(f)$ that we have already used multiple times in the concrete proofs.
|
||||
\begin{lemma}
|
||||
Assuming that $(X_Y,f_Y,i)$ and $(X_Y,p,i)$ are pullbacks in the following diagram, then $p=g\comp f_Y$.
|
||||
\begin{equation*}
|
||||
\begin{tikzcd}[ampersand replacement=\&]
|
||||
{X_Y} \& X \\
|
||||
Y \& {Y+Q} \\
|
||||
Z \& {Z+Q'}
|
||||
\arrow["i", tail, from=1-1, to=1-2]
|
||||
\arrow["{f_Y}"', from=1-1, to=2-1]
|
||||
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
||||
\arrow["p"', bend right=50, from=1-1, to=3-1]
|
||||
\arrow["f", from=1-2, to=2-2]
|
||||
\arrow["{\inl_Y}"', tail, from=2-1, to=2-2]
|
||||
\arrow["{g}"', from=2-1, to=3-1]
|
||||
\arrow["g+q", from=2-2, to=3-2]
|
||||
\arrow["{\inl_Z}"', tail, from=3-1, to=3-2]
|
||||
\end{tikzcd}
|
||||
\end{equation*}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
IT simply relies on the fact that injections in an extensive category are assumed to be monic, and that the pullback of a monomorphism gives a monomorphism leg in the pullback's span.
|
||||
\end{proof}
|
||||
Assuming $f,g\in\Hom(X,Y+1)$, and that $f'\c X_f\rightarrowtail X$ and $g'\c X_g\rightarrowtail X$, and $X_f$ and $X_g$ are pullbacks of $f$ and $\inl$, and $g$ and $\inl$ accordingly, then $f\appr g$ if there exists $h\c X_f\rightarrowtail X_g$ that commutes in the following diagram:
|
||||
\begin{equation*}
|
||||
\begin{tikzcd}[ampersand replacement=\&]
|
||||
@@ -2335,50 +2523,82 @@ Assuming $f,g\in\Hom(X,Y+1)$, and that $f'\c X_f\rightarrowtail X$ and $g'\c X_g
|
||||
The order structure on the functor $F\c\BC\to\BC$ defined as $FX=X+1$ is a liftable order.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
For morphisms $h\c X\to Z+1$, $g\c Y\to Z$, and $k\c X\to Y+1$, we assume $h\appr Fg\comp k$ that means we have two cases:
|
||||
\begin{itemize}
|
||||
\item $h=\bot$: In this case we take $k'=\bot$. Now, $k'\appr k$ and $Fg\comp k'=h$.
|
||||
\item $h=Fg\comp k$: In this case we take $k'=k$. Now, $k'\appr k$ and $Fg\comp k'=h$.\qed
|
||||
\end{itemize}
|
||||
\begin{equation*}
|
||||
\begin{tikzcd}[ampersand replacement=\&]
|
||||
\& Z \& \\
|
||||
{X_h} \&\& {Z+1} \\
|
||||
{X_k} \& X \\
|
||||
Y \& {Y+1} \\
|
||||
Z \& {Z+1}
|
||||
\arrow["{\inl_Z}", tail, from=1-2, to=2-3]
|
||||
\arrow["q", from=2-1, to=1-2]
|
||||
\arrow["\lrcorner"{anchor=center, pos=0.125, rotate=45}, draw=none, from=2-1, to=2-3]
|
||||
\arrow["e"', dashed, from=2-1, to=3-1]
|
||||
\arrow["{h_d}", tail, from=2-1, to=3-2]
|
||||
\arrow["{k_d}", tail, from=3-1, to=3-2]
|
||||
\arrow["{k_Y}"', from=3-1, to=4-1]
|
||||
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-1, to=4-2]
|
||||
\arrow["p"', bend right=50, from=3-1, to=5-1]
|
||||
\arrow["h"', from=3-2, to=2-3]
|
||||
\arrow["k"', from=3-2, to=4-2]
|
||||
\arrow["{\inl_Y}"', tail, from=4-1, to=4-2]
|
||||
\arrow["g"', from=4-1, to=5-1]
|
||||
\arrow["{g+1}"', from=4-2, to=5-2]
|
||||
\arrow["{\inl_Z}"', tail, from=5-1, to=5-2]
|
||||
\end{tikzcd}
|
||||
\end{equation*}
|
||||
We define $k'=k\comp\mathsf{iso}\comp(k_d+1)\comp(e+1)\comp\mathsf{iso}$. We have:
|
||||
\begin{align*}
|
||||
(g+1)\comp k\comp\mathsf{iso}\comp(k_d+1)\comp(e+1)\comp\mathsf{iso}\\
|
||||
(g+1)\comp k\comp\mathsf{iso}\comp(h_d+1)\comp\mathsf{iso}
|
||||
\end{align*}
|
||||
\todo{Finish! Perhaps, $k'=k\comp\mathsf{iso}\comp(k_d+1)\comp(e+1)\comp\mathsf{iso}$.}
|
||||
\end{proof}
|
||||
To have an abstraction of~\autoref{lem:maybe-func-set} recalling that our category is extensive, we use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$. We take $\brks{\alpha\comp p_1,\alpha\comp p_2}\c R\to (X+1)\times(X+1)$, where $(X+1)\times (X+1)\iso X^2+(2\times X)+1$, and we assume the following pullbacks exist:
|
||||
\begin{equation*}
|
||||
\begin{tikzcd}[ampersand replacement=\&]
|
||||
{R_{X^2}} \& R \& {R_{2\times X}} \& R \\
|
||||
{X^2} \& {X^2+(2\times X)+1} \& {(2\times X)} \& {X^2+(2\times X)+1} \\
|
||||
\& {R_1} \& R \\
|
||||
\& 1 \& {X^2+(2\times X)+1}
|
||||
\arrow["{q_1}", from=1-1, to=1-2]
|
||||
\arrow["{q_2}"', from=1-1, to=2-1]
|
||||
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
||||
\arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=1-2, to=2-2]
|
||||
\arrow["{r_1}", from=1-3, to=1-4]
|
||||
\arrow["{r_2}"', from=1-3, to=2-3]
|
||||
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=2-4]
|
||||
\arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=1-4, to=2-4]
|
||||
\arrow["{\mathsf{in}_1}"', from=2-1, to=2-2]
|
||||
\arrow["{\mathsf{in}_2}"', from=2-3, to=2-4]
|
||||
\arrow["{s_1}", from=3-2, to=3-3]
|
||||
\arrow["{s_2}"', from=3-2, to=4-2]
|
||||
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-2, to=4-3]
|
||||
\arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=3-3, to=4-3]
|
||||
\arrow["{\mathsf{in}_3}"', from=4-2, to=4-3]
|
||||
\end{tikzcd}
|
||||
\end{equation*}
|
||||
%\begin{proof}
|
||||
% For morphisms $h\c X\to Z+1$, $g\c Y\to Z$, and $k\c X\to Y+1$, we assume $h\appr Fg\comp k$ that means we have two cases:
|
||||
% \begin{itemize}
|
||||
% \item $h=\bot$: In this case we take $k'=\bot$. Now, $k'\appr k$ and $Fg\comp k'=h$.
|
||||
% \item $h=Fg\comp k$: In this case we take $k'=k$. Now, $k'\appr k$ and $Fg\comp k'=h$.\qed
|
||||
% \end{itemize}
|
||||
%\end{proof}
|
||||
%To have an abstraction of~\autoref{lem:maybe-func-set} recalling that our category is extensive, we use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$. We take $\brks{\alpha\comp p_1,\alpha\comp p_2}\c R\to (X+1)\times(X+1)$, where $(X+1)\times (X+1)\iso X^2+(2\times X)+1$, and we assume the following pullbacks exist:
|
||||
%\begin{equation*}
|
||||
% \begin{tikzcd}[ampersand replacement=\&]
|
||||
% R \& I \& {X^2+(2\times X)+1}
|
||||
% \arrow["e", two heads, from=1-1, to=1-2]
|
||||
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}"', bend right=20, from=1-1, to=1-3]
|
||||
% \arrow["m", tail, from=1-2, to=1-3]
|
||||
% {R_{X^2}} \& R \& {R_{2\times X}} \& R \\
|
||||
% {X^2} \& {X^2+(2\times X)+1} \& {(2\times X)} \& {X^2+(2\times X)+1} \\
|
||||
% \& {R_1} \& R \\
|
||||
% \& 1 \& {X^2+(2\times X)+1}
|
||||
% \arrow["{q_1}", from=1-1, to=1-2]
|
||||
% \arrow["{q_2}"', from=1-1, to=2-1]
|
||||
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
||||
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=1-2, to=2-2]
|
||||
% \arrow["{r_1}", from=1-3, to=1-4]
|
||||
% \arrow["{r_2}"', from=1-3, to=2-3]
|
||||
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=2-4]
|
||||
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=1-4, to=2-4]
|
||||
% \arrow["{\mathsf{in}_1}"', from=2-1, to=2-2]
|
||||
% \arrow["{\mathsf{in}_2}"', from=2-3, to=2-4]
|
||||
% \arrow["{s_1}", from=3-2, to=3-3]
|
||||
% \arrow["{s_2}"', from=3-2, to=4-2]
|
||||
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-2, to=4-3]
|
||||
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=3-3, to=4-3]
|
||||
% \arrow["{\mathsf{in}_3}"', from=4-2, to=4-3]
|
||||
% \end{tikzcd}
|
||||
%\end{equation*}
|
||||
So, we have the following:
|
||||
\begin{gather*}
|
||||
R\iso R_{X^2}+R_{2\times X}+R_{1}
|
||||
\end{gather*}
|
||||
And now we can use $q_2+r_2+s_2\c R_{X^2}+R_{2\times X}+R_{1}\to X^2+(2\times X)+1$ instead of $\brks{\alpha\comp p_1,\alpha\comp p_2}$. To prove~\autoref{lem:maybe-func-set} abstractly is to prove that $q_2+r_2+s_2$ factors through $X^2+1$. To achieve this, we need to show that $R_{2\times X}\iso 0$.
|
||||
\sgnote{This abstraction is unfinished: the crux $R_{2\times X}\iso 0$ (the ``no mixed pairs'' content of \autoref{lem:maybe-func-set}) is stated but not proved, and the subsection ends here. Either complete the argument or mark it clearly as work in progress.}
|
||||
%%\begin{equation*}
|
||||
%% \begin{tikzcd}[ampersand replacement=\&]
|
||||
%% R \& I \& {X^2+(2\times X)+1}
|
||||
%% \arrow["e", two heads, from=1-1, to=1-2]
|
||||
%% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}"', bend right=20, from=1-1, to=1-3]
|
||||
%% \arrow["m", tail, from=1-2, to=1-3]
|
||||
%% \end{tikzcd}
|
||||
%%\end{equation*}
|
||||
%So, we have the following:
|
||||
%\begin{gather*}
|
||||
% R\iso R_{X^2}+R_{2\times X}+R_{1}
|
||||
%\end{gather*}
|
||||
%And now we can use $q_2+r_2+s_2\c R_{X^2}+R_{2\times X}+R_{1}\to X^2+(2\times X)+1$ instead of $\brks{\alpha\comp p_1,\alpha\comp p_2}$. To prove~\autoref{lem:maybe-func-set} abstractly is to prove that $q_2+r_2+s_2$ factors through $X^2+1$. To achieve this, we need to show that $R_{2\times X}\iso 0$.
|
||||
%\sgnote{This abstraction is unfinished: the crux $R_{2\times X}\iso 0$ (the ``no mixed pairs'' content of \autoref{lem:maybe-func-set}) is stated but not proved, and the subsection ends here. Either complete the argument or mark it clearly as work in progress.}
|
||||
\section{Relators}
|
||||
\subsection{Two-way similarity in Hughes-Jacobs}
|
||||
Hughes and Jacobs define two-way similarity as $\leq\cap\leq^\op$. They give a sufficient condition for the two-way similarity to be the bisimilarity. We discuss that this condition does not allow us to say that a symmetric simularity is a bisimilarity. The condition is:
|
||||
@@ -2874,65 +3094,7 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
|
||||
% \end{enumerate}
|
||||
%\end{definition}
|
||||
|
||||
\begin{definition}[Liftable Order Structure]\label{def:liftable-ord}
|
||||
A \emph{liftable order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ that is a natural order structure, and if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $h\appr Fg\comp k$ in $\Hom(X,FZ)$, then there is $k'\c X\to FY$ such that $k'\appr k$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
|
||||
\end{definition}
|
||||
\begin{definition}[Coliftable Order Structure]\label{def:coliftable-ord}
|
||||
A \emph{coliftable order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ that is a natural order structure, and if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $Fg\comp k\appr h $ in $\Hom(X,FZ)$, and $g$ is surjective, then there is $k'\c X\to FY$ such that $k\appr k'$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}\label{lem:set-ord-str}
|
||||
In $\Set$, assuming that for every set $Y$ we have an order $\leq$ on $\Hom(1,FY)$ that satisfies \autoref{def:nat-ord}.\eqref{item:nat-ord:II}, and if for $A\in\Hom(1,FZ)$, $B\in\Hom(1,FY)$, and $g\c Y\to Z$, $h\leq Fg(B)$ in $\Hom(1,FZ)$ there exists $B'\in\Hom(1,FY)$ such that $B'\leq B$ and $B\in\Hom(1,FY)$ and $A=Fg(B')$, then there exists an order structure $\appr$ on $F$ that is a liftable order structure.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
For every $f,g\in\Hom(X,FY)$, we define $f\appr g$ whenever for all $x\in X$, $f(x)\leq g(x)$.
|
||||
|
||||
Now, assuming $f\c X'\to X$, $\alpha,\beta\c X\to FY$, and $\alpha\appr\beta$, then for every $x'\in X'$, we have $\alpha(f(x'))\leq\beta(f(x'))$, so we have $\alpha\comp f\appr \beta\comp f$. Additionally, if we assume $g\c Y\to Y'$, then since $\leq$ satisfies \autoref{def:nat-ord}.\eqref{item:nat-ord:II}, for every $x\in X$ we have $Fg\comp\alpha(x)\leq Fg\comp\beta(x)$.
|
||||
|
||||
Furthermore, if we have $g\c Y\to Z$, $k\c X\to FY$, $h\c X\to FZ$, and $Fg\comp k\appr h$, then for every $x\in X$ we have $Fg\comp k(x)\leq h(x)$, thus by the assumption of the lemma, for every $x$, there exists a set that we call $k'(x)$, for which we have $Fg(k'(x))\leq h(x)$. So, we have a function $k'\c X\to FY$, such that for every $x$ we have $Fg\comp k'(x)\leq h(x)$ that means $Fg\comp k'\appr h$.
|
||||
\qed
|
||||
\end{proof}
|
||||
\begin{remark}\label{rem:set-ord-str-co}
|
||||
The exact same argument in~\autoref{lem:set-ord-str} is true for coliftable order structures.
|
||||
\end{remark}
|
||||
\begin{lemma}\label{lem:liftable}
|
||||
Assuming that a a functor $F$ has an order structure $\appr$ that is liftable, then for every $f\in\Hom(X,FY)$ we have:
|
||||
\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
||||
\item $Ff\comp\sappr\quad=\quad\sappr\comp Ff$
|
||||
\item $(Ff)^\op\comp\appr\quad=\quad\appr\comp (Ff)^\op$
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
$(I)$ Assuming $t\mathrel{(Ff\comp\sappr)} x$, there exists $s$ such that $t\sappr s$ and $Ff(s)=x$. Since $\appr$ is liftable, and thus natural, by~\autoref{def:nat-ord}, from $s\appr t$ we get $x\appr Ff(t)$ that is $t\mathrel{(\sappr\comp Ff)} x$.
|
||||
|
||||
Assuming $t\mathrel{(\sappr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\sappr x$. By~\autoref{def:liftable-ord} since $Ff(t)\sappr x$ there exists $s$ that $t\sappr s$ and $Ff(s)=x$ that is $t\mathrel{(Ff\comp\appr)}s$.
|
||||
|
||||
$(II)$ Basically, by definition of $\op$ and relation composition we have
|
||||
\begin{gather*}
|
||||
(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
|
||||
(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
|
||||
\end{gather*}
|
||||
So it follows directly from applying $\op$ on both sides of $(I)$.\qed
|
||||
\end{proof}
|
||||
\begin{lemma}\label{lem:coliftable}
|
||||
Assuming that a a functor $F$ has an order structure $\appr$ that is coliftable, then for every surjective $f\in\Hom(X,FY)$ we have:
|
||||
\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
||||
\item $Ff\comp\appr\quad=\quad\appr\comp Ff$
|
||||
\item $(Ff)^\op\comp\sappr\quad=\quad\sappr\comp (Ff)^\op$
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
$(I)$ Assuming $t\mathrel{(Ff\comp\appr)} x$, there exists $s$ such that $t\appr s$ and $Ff(s)=x$. Since $\appr$ is coliftable, and thus natural, by~\autoref{def:nat-ord}, from $t\appr s$ we get $Ff(t)\appr x$ that is $t\mathrel{(\appr\comp Ff)} x$.
|
||||
|
||||
Assuming $t\mathrel{(\appr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\appr x$. By~\autoref{def:coliftable-ord} since $Ff(t)\appr x$ there exists $s$ that $t\appr s$ and $Ff(s)=x$ that is $t\mathrel{(Ff\comp\appr)}s$.
|
||||
|
||||
$(II)$ Basically, by definition of $\op$ and relation composition we have
|
||||
\begin{gather*}
|
||||
(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
|
||||
(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
|
||||
\end{gather*}
|
||||
So it follows directly from applying $\op$ on both sides of $(I)$.\qed
|
||||
\end{proof}
|
||||
%\begin{prop}
|
||||
% For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has a coliftable order structure $\appr$, the following propositions hold:
|
||||
% \begin{enumerate}
|
||||
@@ -2976,40 +3138,14 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
|
||||
\begin{proof}
|
||||
They all follow in an obvious way from~\autoref{lem:liftable} and~\autoref{lem:coliftable}. The last one needs $\appr\comp\appr=\appr$ that comes from transitivity of $\appr$. \qed
|
||||
\end{proof}
|
||||
\begin{example}
|
||||
In the category of sets, subset over the powerset functor is an example of a liftable order. Using~\autoref{lem:set-ord-str} we only prove the case for every $h\in\Hom(1,\powf Z)$, $k\in\Hom(1,\powf Y)$. Additionally, for every $g\c Y\to Z$, such that $h\subseteq\powf g(k)$, we define $k'\in\powf Y$ that that $k'=\{y\mid g(y)\in h\}$. We show that $\powf g(k')=h$.
|
||||
%\begin{example}
|
||||
% In the category of sets, subset over the powerset functor is an example of a liftable order. Using~\autoref{lem:set-ord-str} we only prove the case for every $h\in\Hom(1,\powf Z)$, $k\in\Hom(1,\powf Y)$. Additionally, for every $g\c Y\to Z$, such that $h\subseteq\powf g(k)$, we define $k'\in\powf Y$ that that $k'=\{y\mid g(y)\in h\}$. We show that $\powf g(k')=h$.
|
||||
%
|
||||
% Assuming $z\in\powf g(k')$, then there exists $y'\in \{y\mid g(y)\in h\}$ that $z=g(y')$, so $z\in h$ and $\powf g(k')\subseteq h$.
|
||||
%
|
||||
% Assuming $z\in h$, since $h\subseteq \powf g(k)$, then $z\in\powf g(k)$. So, there exists $y'\in k$ such that $g(y')=z$. So, by the definition of $k'$ we have $y'\in k'$ that means $z\in\powf g(k')$.\qed
|
||||
%\end{example}
|
||||
|
||||
Assuming $z\in\powf g(k')$, then there exists $y'\in \{y\mid g(y)\in h\}$ that $z=g(y')$, so $z\in h$ and $\powf g(k')\subseteq h$.
|
||||
|
||||
Assuming $z\in h$, since $h\subseteq \powf g(k)$, then $z\in\powf g(k)$. So, there exists $y'\in k$ such that $g(y')=z$. So, by the definition of $k'$ we have $y'\in k'$ that means $z\in\powf g(k')$.\qed
|
||||
\end{example}
|
||||
The following example justifies why the $g$ in~\autoref{def:coliftable-ord} should be surjective:
|
||||
\begin{example}
|
||||
If we take the surjectivity of $g$ out of~\autoref{def:coliftable-ord}, subset relation over the powerset functor is NOT an example of a coliftable order! For some set $X$ we take $h\c X\to\powf\mathbb{Z}$, for every $x\in X$, $h(x)=\mathbb{Z}$, $g\c\mathbb{Z}\to\mathbb{Z}$, and for every $z\in\mathbb{Z}$,
|
||||
\begin{gather*}
|
||||
g(z)=
|
||||
\begin{cases}
|
||||
0 & z\in\mathbb{Z}^+ \\
|
||||
1 & \mathsf{otherwise}
|
||||
\end{cases}
|
||||
\end{gather*}
|
||||
then
|
||||
\begin{gather*}
|
||||
\powf g(A)=
|
||||
\begin{cases}
|
||||
\emptyset & A=\emptyset \\
|
||||
\{0\} & A\subseteq \mathbb{Z}^+,A\neq\emptyset\\
|
||||
\{1\} & A\subseteq \mathbb{Z}^-\cup\{0\},A\neq\emptyset\\
|
||||
\{0,1\} & \mathsf{otherwise}
|
||||
\end{cases}
|
||||
\end{gather*}
|
||||
then no matter what $k\c X\to\powf\mathbb{Z}$ is there will be no $k'\c X\to\powf\mathbb{Z}$, for every $x\in X$, $\powf g(k'(x))=h(x)$, as $\powf g(k'(x))\subseteq\{0,1\}$, while $h(x)=\mathbb{Z}$, so $\powf g(k'(x))\subset h(x)$.
|
||||
\end{example}
|
||||
\begin{example}
|
||||
In the category of sets, subset over the powerset functor is an example of a coliftable order structure if the $g$ in~\autoref{def:coliftable-ord} is a surjective. Using~\autoref{rem:set-ord-str-co} we only prove the case for every $h\in\Hom(1,\powf Z)$, $k\in\Hom(1,\powf Y)$. Additionally, $g\c Y\to Z$, such that $\powf g(k)\subseteq h$. We define $k'=k\cup\{y\mid g(y)\in h\}$, and we show that $\powf g(k')=h$.
|
||||
|
||||
Obviously, $\powf g(k')\subseteq h$. Now, assuming $z\in h$ we prove that $z\in \powf g(k')$. Since $g$ is surjective, then exists $A\subseteq Y$ such that $\powf g(A)=h$. By the definition of $k'$, $A\subseteq k'$. So from $z\in h$ we have $z\in \powf g(A)$ that means that exists $a\in A$, such that $g(a)=z$. Now, since $A\subseteq k'$, then $a\in k'$. So, we have $z\in \powf g(k')$.\qed
|
||||
\end{example}
|
||||
\begin{definition}[Natural Relator]
|
||||
An $F$-relator $\relar$ is called \emph{natural}, whenever for every relation $r\c X\rto Y$, and all functions $f\c A\to X$, and $g\c B\to Y$, we have $\relar (g^\op\comp r\comp f)=(Fg)^\op\comp\relar r\comp Ff$.
|
||||
\end{definition}
|
||||
@@ -3101,7 +3237,7 @@ Perhaps if we can relax the definition of liftable by allowing $g$ to be a relat
|
||||
\begin{proof}
|
||||
\todo{Finish.}
|
||||
\end{proof}
|
||||
\subsection{Symmetrization of a Barr relator}
|
||||
\subsection{Symmetrization of a lax Barr relator}
|
||||
\begin{prop}
|
||||
Assuming $F$ is the maybe functor $FX=X+1$, then the syemmetrization of the left-lax Barr relator of $F$ is a Barr-relator.
|
||||
\end{prop}
|
||||
|
||||
Reference in New Issue
Block a user