= Solution
For a <locally small category> $\mathcal C$, an object $A$, and a functor $F:\mathcal C\to\mathbf{Set}$, the covariant <Yoneda lemma> is the natural bijection
$$
\operatorname{Nat}(\mathcal C(A,-),F)\longrightarrow F(A),
\qquad \alpha\longmapsto\alpha_A(1_A).
$$
Its inverse sends $x\in F(A)$ to the <natural transformation> whose component at $B$ maps $f:A\to B$ to $F(f)(x)$.
Suppose now that $\mathcal C$ is a <small category>. For $F:\mathcal C\to\mathbf{Set}$, form the <coproduct in a category>
$$
P=\coprod_{(A,x),\ x\in F(A)}\mathcal C(A,-).
$$
The <Yoneda lemma> associates to every summand the natural transformation determined by $x$, and these transformations combine to a map $p:P\to F$. At an object $B$, the element $y\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,y)$, so $p$ is a <pointwise epimorphism in a functor category>. Each <representable functor> $\mathcal C(A,-)$ is a <projective object in a category>, since
$$
\operatorname{Nat}(\mathcal C(A,-),- )\cong\operatorname{ev}_A
$$
and evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so $P$ is the required projective object. This is the <projective cover of a set-valued functor by representables>.
We next prove the three equivalent conditions. If every <morphism> of $\mathcal C$ is a <monomorphism>, then for $f:B\to C$ and every $A$, postcomposition
$$
\mathcal C(A,B)\longrightarrow\mathcal C(A,C),
\qquad g\longmapsto fg
$$
is injective. Thus every covariant <representable functor> is a <monofunctor>. Conversely, taking $A=B$ shows that injectivity for every representable implies that $fg=fh$ forces $g=h$, so every $f$ is monic.
If all representables are monofunctors, the object $P$ above is a monofunctor because a coproduct of injective functions is injective. Hence every $F$ is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable $R$. Choose an epimorphism $q:M\twoheadrightarrow R$ with $M$ a monofunctor. Since $R$ is projective, $1_R$ lifts to $s:R\to M$ with $qs=1_R$. Thus $R$ is a <retract in a category> of $M$. Every retract of a monofunctor is a monofunctor: if $R(f)x=R(f)y$, then injectivity of $M(f)$ applied to $s(x)$ and $s(y)$ gives $s(x)=s(y)$, and applying $q$ gives $x=y$. This completes the equivalence.
Finally, every functor $\mathcal C\to\mathbf{Set}$ is a monofunctor exactly when every morphism of $\mathcal C$ is a <split monomorphism>. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix $f:A\to B$ and form a quotient of $\mathcal C(A,-)\sqcup1$ by identifying the distinguished point with every arrow of the form $kf:A\to X$, where $k:B\to X$. In the resulting functor $Q$, the two elements $[1_A]$ and $*$ of $Q(A)$ have equal images under $Q(f)$ because $[f]=*$. If every functor is a monofunctor, $Q(f)$ is injective, so $[1_A]=*$. By construction this means $1_A=rf$ for some $r:B\to A$. Thus $f$ is split monic. Equivalently, every morphism of $\mathcal C$ must be an <absolute monomorphism>.
Back to article page
