some more

draft/draft.tex

@@ -2328,13 +2328,7 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 		\item $Fg\comp\alpha\appr Fg\comp\beta$ in $\Hom(X,FY')$. \label{item:nat-ord:II}
 	\end{enumerate}
 \end{definition}
-\begin{prop}
-	In $\Set$, assuming that $Ff\comp g\appr h$ where $g,h\in\Hom(1,FY)$, generalizes to the case where $g,h\in\Hom(X,FY)$, where $X$ is an arbitrary set.  
-\end{prop}
-\begin{proof}
-	We assume that for every set $Y$ we have a preorder structure on $\Hom(1,FY)$. 
-	\todo{Prove it and then edit the examples accordingly!}
-\end{proof}
+
 \begin{definition}[Good Order Structure]\label{def:good-ord}
 	A \emph{good 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}
@@ -2342,6 +2336,20 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 	A \emph{cogood 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 $\Hom(X,FZ)$, 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 good 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 cogood order structures.
+\end{remark}
 \begin{lemma}\label{lem:good}
 	Assuming that a a functor $F$ has an order structure $\appr$ that is good, then for every $f\in\Hom(X,FY)$ we have:
 	\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
@@ -2350,9 +2358,9 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 	\end{enumerate}
 \end{lemma}
 \begin{proof}
-	$(I)$ Assuming $t\mathrel{Ff\comp\sappr} x$\footnote{add parenthesis}, there exists $s$ such that $t\sappr s$ and $Ff(s)=x$. Since $\appr$ is good, and thus natural, by~\autoref{def:nat-ord}, from $s\appr t$ we get $x\appr Ff(t)$ that is $t\mathrel{\sappr\comp Ff} x$.
+	$(I)$ Assuming $t\mathrel{(Ff\comp\sappr)} x$, there exists $s$ such that $t\sappr s$ and $Ff(s)=x$. Since $\appr$ is good, 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:good-ord} since $Ff(t)\sappr x$ there exists $s$ that $t\sappr s$ and $Ff(s)=x$ that is $t\mathrel{Ff\comp\appr}s$.
+	Assuming $t\mathrel{(\sappr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\sappr x$. By~\autoref{def:good-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*}
@@ -2424,11 +2432,11 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 	They all follow in an obvious way from~\autoref{lem:good} and~\autoref{lem:cogood}. 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 good order. For every $h\c X\to\powf Z$, $k\c X\to \powf Y$, and $g\c Y\to Z$, such that $h\subseteq\powf g\comp k$, we define $k'\c X\to\powf Y$ that for every $x\in X$, $k'(x)=k(x)\setminus\{y\mid g(y)\notin h(x)\}$\footnote{Change to $\{y\mid g(y)\in h(x)\}$}. We show that for every $x\in X$, we have $\powf g\comp k'(x)=h(x)$.
+	In the category of sets, subset over the powerset functor is an example of a good 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 $z\in\powf g(k'(x))$, then there exists $y'\in k(x)\setminus \{y\mid g(y)\notin h(x)\}$ that $g(y')=z$, so $y'\in k(x)$, $y'\notin\{y\mid g(y)\notin h(x)\}$. $y'\notin\{y\mid g(y)\notin h(x)\}$ means that $g(y')\in h(x)$, thus $\powf g\comp k'\subseteq h$.
+	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 $z\in h(x)$, since $h\subseteq \powf g\comp k$, then $z\in\powf g\comp k(x)$. So, there exists $y'$ such that $y'\in k(x)$ and $g(y')=z$. Since $g(y')\in h(x)$ then $y'\notin\{y\mid g(y)\notin h(x)\}$ meaning that $y'\in k'(x)$ that means $z\in \powf g(k'(x))$, thus $h\subseteq \powf g\comp k'$.\qed
+	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{example}
 	In the category of sets, subset over the powerset functor is NOT an example of a cogood 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 $z\in\mathbb{Z}$, 
@@ -2451,7 +2459,13 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 	\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}
-The previous example suggests that cogoodness is a strong condition. Perhaps it can be limited to be meaningful. Maybe the morphism $g$ can be limited to legs of a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, where for a relation $r$, $r=\pi_2\comp\pi_1^\op$.\footnote{Or maybe because the $g$ in the counter-example is not surjective!}
+The previous example suggests that cogoodness is a strong condition. It can be limited to be meaningful. If we limit it to the case where the $g$ in~\autoref{def:cogood-ord} is a surjection, we can use if for the scenario, where we have a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, where for a relation $r$, $r=\pi_2\comp\pi_1^\op$. Then we have the following:
+\begin{example}
+	In the category of sets, subset over the powerset functor is an example of a cogood order structure if the $g$ in~\autoref{def:cogood-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}
+\todo{Pedro!}
 \begin{prop}
 	For a natural order structure $\appr$ on a set-functor $F$, if for every $f\in\Hom(X,FY)$ we have $Ff\comp\appr=\appr\comp Ff$, then $\appr$ is cogood.
 \end{prop}