Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 119 / 1 / c / Solution

Assume (c), and take an epimorphism $p:M\twoheadrightarrow h_A$ with $M$ a monofunctor. Since $h_A$ is a projective object in a category, there is a natural transformation $s:h_A\to M$ with $ps=1_{h_A}$. A retract in a category of a monofunctor is a monofunctor: if $h_A(f)(u)=h_A(f)(v)$, then naturality of $s$ gives $M(f)s(u)=M(f)s(v)$, injectivity of $M(f)$ gives $s(u)=s(v)$, and applying $p$ gives $u=v$. Thus (c) implies (b), completing

The stronger assertion that every set-valued functor is a monofunctor holds exactly when $\mathcal{C}$ is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.

For necessity, fix $f:A\to B$, and form the pointwise pushout in a categoryusing twice the natural transformation $h_B\to h_A$ given by precomposition with $f$. At an object $X$, $P(X)$ consists of two copies of $\mathcal{C}(A,X)$, with the two copies of $kf$ identified for every $k:B\to X$. No different underlying elements are identified: each generating relation simply joins the two copies of one element. In $P(B)$, the two copies of $f$ coincide, since $f=1_{B}f$. Hence $P(f)$ sends the two copies of $1_A$ in $P(A)$ to the same element. If $P$ is a monofunctor, those copies of $1_A$ coincide. The description of the pushout implies $rf=1_A$ for some $r:B\to A$. Thus every morphism is a split monomorphism.

Apply the same conclusion to $r$: there is $t:A\to B$ with $tr=1_B$. Then $t=trf=f$, and so $fr=1_B$ as well. Every arrow is invertible, proving