In Example 1.3.21 Leinster says that the functor $F: \mathscr C\to \mathbf{Mon}$ sending a one-object category to the monoid of arrows from the unique object to itself is full, faithful, and essentially surjective on objects.
He says that fullness and faithfullness follows from Example 1.2.7 which says that a functor between categories corresponding to monoids is the same as a homomorphism of monoids. How does this imply that $F$ is full and faithful?
I also don't understand why $F$ is essentially surjective on objects. This would mean that every monoid is a set of arrows from some object to itself under composition. Why does this hold?
Basically the author is defining the functor $F$ as follows:
Each one-object category $A$ is sending into $FA=\operatorname{hom}(A, A)$, the set of all morphisms from $A$ to itself, endowed with the composition (this turns the set into a monoid).
A morphism $f\colon A\to B$ in $\mathcal{C}$ is a functor that is completely determined by its action over the morphisms of $A$. Then $F$ maps $f$ into this restriction: the way $f$ acts over the monoid $FA=\operatorname{hom}(A, A)$. This identification is injective and surjective: if $Ff = Fg$ then $Ff(a)=Fg(a)$ for all $a\in FA$, that is, $f(a)=g(a)$ (because of the construction of $F$ over morphisms). Then the observation of the author applies as $f$ and $g$ 'are homomorphisms'.
Can you conclude the surjectivity (full) of the identification (functor) and that $F$ is essentially surjective?
EDIT: Lord Shark already proved that $F$ is essentially surjective on objects, at the comments.