diff --git a/draft/draft.tex b/draft/draft.tex index 19894dd..9f03924 100644 --- a/draft/draft.tex +++ b/draft/draft.tex @@ -292,7 +292,9 @@ \newcommand{\spto}{\mathrel{\tikz{\draw[-{Stealth}] (0.0,0) -- (0.4,0); \draw (0.17,0.07) -- (0.17,-0.07);}}} \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);}}} \newcommand{\powf}{\mathcal{P}} +\newcommand{\powfi}{\mathcal{P}_{\mathsf{inj}}} \newcommand{\sappr}{\sqsupseteq} +\newcommand{\Dom}{\mathsf{Dom}} \newcommand{\simeet}{% \mathbin{% @@ -401,6 +403,117 @@ We want to say that a natural order structure entails an order over a functor de \begin{proof} 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 \end{proof} + +\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{prop}\label{prop:lift-gen-func} + 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. +\end{prop} +\begin{proof} + 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 $\powf Fg(k')=h$. + + 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 $k'\subseteq k$. + + Assuming $z\in\powf Fg(k')$, then there exists $y'\in k'$ that $z=Fg(y')$, so $z\in h$ and $\powf Fg(k')\subseteq h$. + + 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 +\end{proof} + +\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{remark} + We give an example that justifies why the $g$ in~\autoref{def:coliftable-ord} should be surjective. + 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{remark} + +\begin{remark} + 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. +\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} + + \section{Coalgebraic Bisimulation}%\label{sec:} In this section, by $\spa(\BC)$ we refer to spans in a category $\BC$ that has products, and by $\rel(\BC)$ we refer to the category of relations in $\BC$, i.e.\ @@ -1091,7 +1204,7 @@ A big concern with this approach is that Comma Objects are defined in a 2-catego \subsection{Choosing a suitable order for our setting} 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. -\section{Symmetric Simulation is Bisimulation} +\section{Symmetric Simulation is a Bisimulation} \begin{definition}[Graph] In a category $\BC$ a graph is a tuple $(R,X)$ of the following form: \begin{equation*} @@ -2049,6 +2162,35 @@ We recall that in the above diagram $\sigma_3$ is a bisimulation, and the rest a % \end{gather*} % We have $(Fp_1)^\dagger\comp\sigma$. %\end{proof} +\begin{example} +And another counter-example!!! +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.\\ +We take $R=\{(1,1),(1,2),(2,1),(2,2)\}$, and $X=\{1,2,3\}$. $\alpha$ is defined as below: +\begin{gather*} + \alpha(x)= + \begin{cases} + \{1,2\} & x=1 \\ + \{2,3\} & x=2\\ + \{3\} & x=3 + \end{cases} +\end{gather*} +$\sigma$ is defined as below: +\begin{gather*} + \sigma(w)= + \begin{cases} + \{(1,2),(2,2)\} & w=(1,2) \\ + \{(2,1),(2,2)\} & w=(2,1) \\ + \{(1,2),(2,1)\} & w=(1,1) \\ + \{(2,2),(1,1)\} & w=(2,2) + \end{cases} +\end{gather*} +$R$ 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 $\alpha(2)$. + +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. + +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. +\end{example} + \subsection{The concrete proof} %\begin{lemma}\label{lem:sim-opsim-inc1}\ppnote{Actually, this lemma holds for every functor in an arbitrary category.} % 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: @@ -2192,18 +2334,26 @@ Now, we prove our main statement. \end{cor} 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. \subsection{Maybe Functor} -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 $\id_{X+1}\cup\{(\bot,x)\mid x\in X\}$.\\ -Now, we prove that the given order on the maybe functor is a liftable one. +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 $f$). +Now, we prove that the given order on the maybe functor is liftable and coliftable. \begin{lemma}\label{lem:maybe-lif} The order structure on the set-functor $FX=X+1$ is a liftable order. \end{lemma} \begin{proof} - 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: - \begin{itemize} - \item $h=\bot$: In this case we take $k'=\bot$, so we have $k'\appr k$, and then we have $Fg(k')=\bot=h$. - - \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 - \end{itemize} +% 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: +% \begin{itemize} +% \item $h=\bot$: In this case we take $k'=\bot$, so we have $k'\appr k$, and then we have $Fg(k')=\bot=h$. +% \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 +% \end{itemize} +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: +\begin{gather*} + k'(x)= + \begin{cases} + \bot&x\notin\Dom(h)\\ + k(x)&x\in\Dom(h) + \end{cases} +\end{gather*} +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 \end{proof} \begin{lemma}\label{lem:maybe-colif} The order structure on the set-functor $FX=X+1$ is a coliftable order. @@ -2286,41 +2436,79 @@ So, proven by Dubut, for every AM-simulation relation over a coalgebra $(X,\alph % 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 $\beta(x_1,x_2)\in R$. % \qed %\end{proof} - +%%%%%%Second proof: \begin{prop}\label{prop:sym-sim-bis-may} 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. \end{prop} \begin{proof} 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 - \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 $Fp_2\comp\sigma(x_1,x_2)=\bot$: 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 $\powf g(k')=h$. - - 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{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} + \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}