rel

draft/draft.tex

@@ -815,7 +815,6 @@ As mentioned in the previous section, there is a way to define morphisms in $\sp
 			\arrow["{p_2}"', from=2-1, to=2-2]
 		\end{tikzcd}
 	\end{gather*}
-\end{example}
 For morphisms
 \begin{gather*}
 	(f,g,w)\c(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(Y \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Z)
@@ -873,7 +872,7 @@ As mentioned in the previous section, there is a way to define morphisms in $\sp
 		\arrow["{c_V}"', from=2-2, to=2-3]
 	\end{tikzcd}
 \end{equation*}
-	So, $(w,v,u)$ is a morphism of type $(R \stackrel{c_R}{\leftarrow} R\odot W \stackrel{c_W}{\to}W)\to(S \stackrel{c_S}{\leftarrow} S\odot V \stackrel{c_V}{\to}V)$. So, we define $(f,g,w)\odot(g,h,v)=(w,v,u)$. To define the natural isomorphisms is a cumbersome task, but it is known in the literature.
+So, $(w,v,u)$ is a morphism of type $(R \stackrel{c_R}{\leftarrow} R\odot W \stackrel{c_W}{\to}W)\to(S \stackrel{c_S}{\leftarrow} S\odot V \stackrel{c_V}{\to}V)$. So, we define $(f,g,w)\odot(g,h,v)=(w,v,u)$. To define the natural isomorphisms is a cumbersome task, but it is known in the literature. Additionally, showing that $\rel(\BC)\rightrightarrows\BC$ is also a double category is almost the same. The functors are defined the same, except that the functor $\odot$ takes the image of the pullback over its legs, instead of the pullback itself.
 %	The following diagrams help to see why the defined $\odot$ is a functor.
 %	\begin{equation*}
 	%		\begin{tikzcd}[ampersand replacement=\&]
@@ -958,6 +957,10 @@ As mentioned in the previous section, there is a way to define morphisms in $\sp
 		%			\arrow["{c_{V'}}"', from=4-8, to=3-9]
 		%		\end{tikzcd}
 	%	\end{equation*}
+\end{example}
+\begin{example}
+	\todo{Talk about $\spa_a$ and $\rel_a$.}
+\end{example}
 \section{Coalgebraic Bisimulation}%\label{sec:}
 %
 %\begin{definition}[Relation Lifting]