minor

draft/draft.tex

@@ -527,15 +527,7 @@ The order structure that we can define for this functor is that for sets $X$ and
 	%		\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}
 \begin{proof}
-<<<<<<< HEAD
-	$(I)$ 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 $t\mathrel{(Ff\comp\sappr)}s$.
-	
-	$(II)$ Basically, by definition of $\op$ and relation composition we have
-=======
 	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:
->>>>>>> 8c1054256a4ef67f109cc6d30f3d3bd050e936cb
 	\begin{gather*}
 		k'(x)=
 		\begin{cases}
@@ -571,7 +563,7 @@ The definition derives the ordering on morphisms of each hom-set $\Hom(X,\sub Y)
 	The pointwise ordering on $\sub$ is a liftable ordering.
 \end{prop}
 \begin{proof}
-	Assuming that $\mu\in SX$, $g\c X\to Y$ is surjective, $\nu\in SY$, and $\nu\appr \sub g(\mu)$, then there exists $\mu'\in \sub X$ such that $\mu'\appr \mu$ and $\sub g(\mu')=h$.\\ 
+	Assuming that $\mu\in SX$, $g\c X\to Y$ is surjective, $\nu\in SY$, and $\nu\appr \sub g(\mu)$, then there exists $\mu'\in \sub X$ such that $\mu'\appr \mu$ and $\sub g(\mu')=\nu$.\\ 
 	The assumption $\nu \appr Sg(\mu)$ means that for every $y \in Y$,
 	\begin{gather*}
 	\nu(y) \leq Sg(\mu)(y) = \sum_{x \in g^{\mone}(y)} \mu(x).