minor

draft/draft.tex

@@ -2322,12 +2322,19 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 	\todo{Finish.}
 \end{proof}
 \begin{definition}[Natural Order Structure]\label{def:nat-ord}
-	A \emph{natural order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ such that if $\alpha\appr\beta$ in $\Hom(X,FY)$, $f\c X\to X'$, $g\c Y\to Y'$, then:
+	A \emph{natural order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ such that if $\alpha\appr\beta$ in $\Hom(X,FY)$, $f\c X'\to X$, $g\c Y\to Y'$, then:
 	\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
-		\item $\alpha\comp f\appr\beta\comp f$ in $\Hom(X',Y)$. \label{item:nat-ord:I}
-		\item $Fg\comp\alpha\appr Fg\comp\beta$ in $\Hom(X,Y')$. \label{item:nat-ord:II}
+		\item $\alpha\comp f\appr\beta\comp f$ in $\Hom(X',FY)$. \label{item:nat-ord: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}
@@ -2343,7 +2350,7 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
 	\end{enumerate}
 \end{lemma}
 \begin{proof}
-	$(I)$ Assuming $t\mathrel{Ff\comp\sappr} x$, 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 $t\mathrel{Ff\comp\sappr} x$\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 $t\mathrel{\sappr\comp Ff} x$.
 	
 	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 $t\mathrel{Ff\comp\appr}s$.
 	
@@ -2417,9 +2424,9 @@ 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 $h\subseteq\powf g\comp k$, there exists $k'\c X\to\powf Y$ that for every $x\in X$, $k'(x)=k(x)\setminus\{y\mid g(y)\notin h(x)\}$. 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. 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 $\powf g\comp k'(x)=h(x)$.
 	
-	Assuming $z\in\powf g(k'(x))$, then there exist $y'\in k(x)\setminus \{y\mid g(y)\notin h(x)\}$ 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 $z\in\powf g(k'(x))$, 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 $z\in h(x)$, 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
 \end{example}
@@ -2444,7 +2451,7 @@ 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 $r$, $r=\pi_2\comp\pi_1^\op$.
+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!}
 \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}