effective equivalence relation
EffectiveEquivalenceRelation
2008-11-06 02:10:30
2008-11-06 02:21:45
Definition
relation on objects
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Recall that given an equivalence relation $R$ on a set $A$, we can form the quotient $A/R$ of $A$ by $R$. Elements of $A/R$ are the equivalence classes under $R$. There are two functional properties of $A/R$:
\begin{itemize}
\item If $p_1,p_2$ are projections of $R$ onto $A$, given by $p_1(a,b)=a$ and $p_2(a,b)=b$, then the canonical surjection $q:A\to A/R$ is the coequalizer of $p_1$ and $p_2$.
\begin{proof}
First, $q\circ p_1(a,b) = q(a)=[a]=[b]= q(b)=q\circ p_2$. Suppose that $r:A\to B$ is another function with $r\circ p_1=r\circ p_2$. Define $f:A/R\to B$ by $f([a])=r(a)$. This is a well-defined function because $[a]=[b]$ implies that $r(a)=r\circ p_1(a,b)=r\circ p_2(a,b)=r(b)$. This shows that $f\circ q=r$, which also implies that $f$ is uniquely determined.
\end{proof}
\item $p_1$ and $p_2$ form a kernel pair of $q$.
\begin{proof}
Again, $q\circ p_1 = q\circ p_2$, as was just shown previously. Now suppose $g,h:C\to A$ are functions with $q\circ g = q\circ h$. For any $c\in C$, we see that $[g(c)]=q(g(c))=q(h(c))=[h(c)]$, so that $(g(c),h(c))\in R$. Define $s:C\to R$ by $s(c)=(g(c),h(c))$. Then $p_1\circ s=g$ and $p_2\circ s=h$. It is again easy to see that $s$ is uniquely determined by $g$ and $h$. Hence, $p_1,p_2$ are a kernel pair of $g$.
\end{proof}
\end{itemize}
\textbf{Definition}. An equivalence relation object $(R,p_1,p_2)$ on an object $A$ in a category $\mathcal{C}$ is said to be an \emph{effective equivalence relation object} if
\begin{itemize}
\item the projections $p_1,p_2$ has a coequalizer $q:A\to A/R$, and
\item $p_1,p_2$ form the kernel pair of $q$.
\end{itemize}
In other words, $R$ is effective iff there is an exact fork
$$\xymatrix@+=2cm{R \ar@<0.75ex>[r]^-{p_1} \ar@<-0.75ex>[r]_-{p_2} & A \ar[r]^-q & A/R}$$
In \textbf{Set}, the category of sets, every equivalence relation object (which is just an equivalence relation on a set) is effective, as we have just shown above. However, this is not true in general. For example, not every equivalence relation object is effective in \textbf{Top}, the category of topological spaces (and continuous functions).
More to come...