Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 75 / 7 / Solution

Interpret the given relation by a subobject $R\hookrightarrow A\times B$ and write its projections as $p:R\to A$, $q:R\to B$. Totality says that the image of $p$ is all of $A$, so $p$ is a regular epimorphism. Single-valuedness makes $p$ monic: if $pr=ps$ for two arrows into $R$, the functionality sequent forces $qr=qs$, and the inclusion into $A\times B$ then gives $r=s$.

A regular epimorphism that is monic is invertible. Indeed its kernel pair consists of equal projections, and its coequalizer universal property supplies an inverse. Therefore defineThe graph of this map is exactly the original relation $R$, since $(p,q)=(1_A,f)p$. The graph also makes the map unique. This proves total functional relations define morphisms, using regular-category semantics and no choice of individual witnesses.