homework in progress

draft/draft.tex

@@ -614,8 +614,7 @@ Now, we prove $Sg(\mu') = \nu$.
 	\end{itemize}\qed
 \end{proof}
 
-
+\section{Spans and Relations}
-\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.\
 such spans $(X \stackrel{p_1}{\leftarrow} R
@@ -692,7 +691,37 @@ For the time being, we limit the discussion to the case $\BC=\Set$. For simplici
 We can define morphisms in $\rel$ differently by only requesting such functions $g_1$ and $g_2$ that $x\mathrel{R}y$ entails $g_1(x)\mathrel{S}g_2(y)$. Let us 
 call such morphisms \emph{anonymous} (because they omit the witnessing part $f$, which is unique for relations but not for general spans). This however yields 
 an equivalent definition. \todo{Add a proof.} \todo{Do the same for spans; prove equivalence under the axiom of choice.}
+\begin{prop}
+	Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects in $\spa$, there is an anonymous morphism $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\spa$ iff there is an onymous $(g_1,g_2,w)$ of the same type, with the witness $w\c R\to S$ in $\spa$.
+\end{prop}
+\begin{proof}
+	$(\Rightarrow):$ We define $w$ as $w(x_1,x_2)=(g_1(x_1),g_2(x_2))$. We have:
+	\begin{align*}
+		g_1\comp p_1(x_1,x_2)&\\
+		=&g_1(x_1)\\
+		=&q_1(g(x_1),g(x_2))\\
+		=&q_1\comp w(x_1,x_2)
+	\end{align*}
+	Similarly, we have $g_2\comp p_2(x_1,x_2)=q_2\comp w(x_1,x_2)$.
 	
+	$(\Leftarrow):$ Assuming $x_1\mathrel{R}x_2$, then $w(x_1,x_2)\in S$ as well, and by the definition of onymous morphisms we have $w(x_1,x_2)=(g_1(x_1),g_2(x_2))$, so we have $g_1(x_1)\mathrel{S}g_2(x_2)$. \qed
+\end{proof}
+
+\begin{prop}
+	Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects in $\rel$, there is an anonymous morphism $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\rel$ iff there is an onymous $(g_1,g_2,w)$ of the same type, with the witness $w\c R\to S$ in $\rel$.
+\end{prop}
+\begin{proof}
+	\todo{Finish}
+\end{proof}
+
+\begin{prop}
+	Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects in $\rel$, there is an anonymous morphism $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ iff there is an onymous $(g_1,g_2,w)$ of the same type, with the witness $w\c R\to S$.
+\end{prop}
+\begin{proof}
+	\todo{Finish}
+\end{proof}
+
+\section{Coalgebraic Bisimulation}%\label{sec:}
 By varying from anonymous to non-anonymous morphisms and from $\rel$ to $\spa$ we 
 can obtain for flavors of bisimulation (\autoref{eq:acz-mend-diag}--\autoref{def:vanila}) -- 
 \autoref{fig:anonymous_onymous} contains a comprehensible summary.