Denote by $\mathbf{Epic}(A)$ the full subcategory of $\mathbf{A}/A$ whose objects are the epic morphisms. Call a "quotient object" of $A$ an isomorphism class of objects in $\mathbf{Epic}(A)$.

Theorem: Let $A \xrightarrow{e} X$ and $A \xrightarrow{e'} X'$ be epics in $\mathbf{Set}$.

$e$ and $e'$ are isomorphic in $\mathbf{Epic}(A)$ if and only if they induce the same equivalence relation on $A$.

Proof: Suppose $e$ and $e'$ are isomorphic in $\mathbf{Epic}(A)$. This means there is an isomorphism $f : X \rightarrow X'$ such that $f \circ e = e'$. Let $R \subset A \times A$ be the equivalence relation induced by $e$, similarly for $R'$. Suppose that $(s, t) \in R'$. In particular $e'(s) = e'(t)$. Then, $(f \circ e)(s) = (f \circ e)(t)$. Since $f$ is in particular injective, we have $e(s) = e(t)$. So $R' \subseteq R$. The converse follows similarly by noting that we have an isomorphism $f^{-1}$ and $e = f^{-1} \circ e'$. So $R = R'$.

Suppose that $R = R'$. Define a map $f : X \rightarrow X'$ as follows. Given an $x \in X$, choose an $a \in A$ such that $e(a) = x$ (which must exist, since epics in $\mathbf{Set}$ are surjective), and let $f(x) = e'(a)$. This is well defined because, if we were to choose a different element $b$ with $f(b) = x$, we have $(a, b) \in R$ so $(a, b) \in R'$ and $e'(b) = e'(a)$. To see $f$ is injective, take $x \neq x'$. Then we must have an $a, b$ with $e(a) = x$ and $e(b) = x'$ but $(a, b) \not\in R$ (if it were, $e$ would be identical on $a$ and $b$) -- it follows that $(a,b) \not\in R'$ so $e'(a) \neq e'(b)$, so $f$ takes distinct values. It's easy to see that $f$ is surjective -- given an $x' \in X'$, take an $a$ with $e'(a) = x'$, and note that the element $e(a)$ maps onto $x'$. So $f$ is a bijection and thus an isomorphism. To check that it is an isomorphism in $\mathbf{Epic}(A)$, we must ensure that $f \circ e = e'$. Indeed, given $a \in A$, $(f \circ e)(a)$ is equal to $e'(b)$ for some element $b \in A$ with $(a, b) \in R$. But of course $e'(b) = e'(a)$ because $(a, b) \in R'$. So, $f$ is indeed an isomorphism of epics.

So we have the dual result to that of 5.1.40 -- the "quotient objects" of $A$ are in correspondence with the equivalence relations on $A$.