Question Involving Universal Elements From Categories for the Working Mathematician pg. 57:

Question: For this $H$ mentioned in the context of the second paragraph, what does it send the arrows of $\mathbf{Set}$ to? In particular, what does it send an arrow $f : S/E \rightarrow X$ to?
Does it send it to the arrow $g$ s.t $g(p) = f \circ p \in H(X)$?
 A: Let $g:X\to Y$ be any morphism of sets, then
$$H(g):H(X)\longrightarrow H(Y)$$
is given by the precomposition
$$H(g)f := g\circ f\ .$$
Notice that the unique function $h'$ such that $g\circ f = h\circ p$ is simply $g\circ f'$.
A: $$H(X) \equiv \{f\in\text{Hom}(S,X)\mid\forall s,s'\in S.s\sim s' \implies f(s)=f(s') \}$$ using $\sim$ instead of $E$ because it looks nicer.
The action is the same action of the $\text{Hom}$ functor in its second argument, namely post-composition, as you wrote.  That is, what you call $g$ is indeed $H(f)$.  You do have to check that it actually is a functor though, i.e. post-composing by an arbitrary function still satisfies the well-definedness condition.
You can verify that $p\in H(S/\!\!\sim)$ and for any $f : S \to X$, there is a unique $f' : S/\!\!\sim\ \!\!\to X$ such that $H(f')(p) = f$ which is to say there is for every function $f : S \to X$ such that $\forall s,s'\in S.s\sim s' \implies f(s) = f(s')$ a unique function $f' : S/\!\!\sim\ \!\!\to X$ such that $f' \circ p = f$.
