Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 1 / iv / Solution

If $f:A\to B$ has a section $g:B\to A$, functoriality gives $H_f{H}_g=1_{H_B}$. Thus $H_f$ is a split epimorphism and in particular an epimorphism.

For the converse, justify the needed pointwise surjectivity directly. Let $I\subseteq H_B$ be the image subpresheaf of $H_f$. Form a new presheaf by taking, at every object, two copies of $H_B(U)$ and identifying their elements in $I(U)$. Restriction maps descend because $I$ is a subpresheaf. The two canonical transformations $j_1,j_2:H_B\to H_B{⨿}_I{H}_B$ agree after $H_f$. If any component of $I$ is proper, an element outside it distinguishes $j_1$ from $j_2$. Therefore epimorphicity forces every component of $H_f$ to be surjective.

In particular, its component at $B$ has $1_B$ in its image: there is $g:B\to A$ with $fg=1_B$. This proves the Yoneda embedding detects split epimorphisms criterionOrdinary epimorphicity of $f$ alone would not suffice.