opposite internal relation




Let CC be a finitely complete category. This means that CC is symmetric, and for objects XX and YY there exist isomorphisms B X,Y:X×YY×XB_{X, Y}: X \times Y \cong Y \times X and B Y,X:Y×XX×YB_{Y, X}: Y \times X \cong X \times Y such that B X,YB Y,X=id X×YB_{X, Y} \circ B_{Y, X} = id_{X \times Y} and B Y,XB X,Y=id Y×XB_{Y, X} \circ B_{X, Y} = id_{Y \times X}. Given an internal relation R(s,t)X×YR\stackrel{(s,t)}\hookrightarrow X \times Y, the pullback of the internal relation (s,t)(s,t) along the braiding B Y,XB_{Y, X} is the opposite internal relation span

R R op,RR op(t,s)Y×XR \stackrel{\dagger_{R^\op,R}}\leftarrow R^\op\stackrel{(t,s)}\hookrightarrow Y \times X

with a morphism R,R op:RR op\dagger_{R,R^\op}:R \to R^\op such that R,R op R op,R=id R op\dagger_{R,R^\op} \circ \dagger_{R^\op,R} = id_{R^\op} and R op,R R,R op=id R\dagger_{R^\op,R} \circ \dagger_{R,R^\op} = id_{R}.

See also

Last revised on May 14, 2022 at 12:46:12. See the history of this page for a list of all contributions to it.