Question about an exercise in Arbib and Manes’ text concerning coequalizer and a possible misprint. The following question, is taken from Arbib and Manes’ Arrows, structures and functors text:

Definition 1:  Given an equivalence relation $E$ on a set $A$, we define the equivalence class of an element $a$ of $A$ with respect to $E$ to be
$$
  [a]_E = \{ a' \mid \text{$a'\in A$ and $(a, a') \in E$} \}
$$
(when no ambiguity can arise, we write $[a]$ for $[a]_E$.)  The factor set or quotient set of $A$ with respect to $E$ is then the set of equivalence classes
$$A/E = \{ [a] \mid a \in A \} \,.$$


Definition 2: Given any function $f \colon A \rightarrow B$, we define the equivalence relation $E(f)$ of $f$ on $A$ by
$$
  E(f) = \{ (a, a') \in A \times A \mid f(a) = f(a') \} \,.
$$
[…]
Given any relation $R$ on $A$ (i.e. any subset of $A\times A$) we may associate with it the two projection maps
$$
  p_k
  \colon
  R \rightarrow A \,,
  \quad
  (a_1, a_2) \mapsto a_k \,,
  \quad
  k = 1, 2 \,.
$$
On the other hand, given any pair of maps
$$
  p_1, p_2 \colon R \rightarrow A
$$
with common domain, and with codomain $A$, we may define on $A$ the corresponding relation
$$
  E_R = \{ (p_1(r), p_2(r)) : r \in R \subset A \times A
  \tag{$1$}
$$
(where we forbear writing $E_{(R, p_1, p_2)}$ for brevity).


Definition 3: We say a map $A \xrightarrow{h} B$  is a coequalizer if and only if there exists a pair $p_1, p_2 \colon R \rightarrow A$ of maps such that $h \circ p_1 = h \circ p_2$, and such that whenever $A \xrightarrow{h'} B'$ satisfies $h' \circ p_1 = h'\circ p_2$, there is a unique map $B \xrightarrow{\psi} B'$ such that $\psi \circ h = h'$.

In this situation, we call $h$ the coequalizer of $p_1$ and $p_2$ and write $h = \mathrm{coeq}(p_1, p_2).$


Exercise Let $p_1, p_2 \colon R \rightarrow A$ be functions with disjoint images, i.e., $p_1(R) \cap p_2(R) = \emptyset$. Show that if $B$ is a $1$-element set and $h \colon A \rightarrow B$ is the unique function then $h = \mathrm{coeq}(p_1, p_2)$.  [Use Definition 3.]

For the Exercise above, I would like to know how does the unique map $h$ in the question is related to the unique map $\psi$ in Definition 3. Also, since $B$ is a singleton set, say $B = \{ b \}$ then, I have to make it into a relation and in turn into an equivalence relation by defining the following binary relation: $\{ (b, b) \in E \mid (b, b) \in B\}$. Also as stated in the conclusion of the exercise, if I need to show that $h = \mathrm{coeq}(p_1, p_2)$, then $h \circ p_1 = h \circ p_2$.
Thank you in advance.
 A: The coequaliser arrow $h$ is not supposed to be the arrow $\psi$. They have different domains (in general) for one thing. They only say ‘unique’ because $B$ is terminal, it is not to be confused with the uniqueness of $\psi$.
You also don’t need to find a relation on $B$ (there are only two anyway, the empty one and the trivial one): notice that the descriptive definition of coequaliser given by your text makes no reference at all to relations. However when your text goes onto to proving that $\mathsf{Set}$ has all coequalisers (at least, it should do…) you will find a constructive definition involving relations.
You just need to check $h$ satisfies the universal property; $h\circ p_1=h\circ p_2$ and, for any $h’$, a unique $\psi$ is borne. ($\psi$ is inherently related to $h’$, but not really to $h$)
By the way, the exercise is false. Let $R=\{\ast\}$, $A=\{0,1,2\}$ and $p_{1,2}$ defined by $p_{1,2}(\ast)=1,2$ respectively. These have disjoint images. Suppose $h:A\to\{b\}$ is really a coequaliser. Consider $h’:A\to\{0,1\}$ the function $0\mapsto0,1\mapsto1,2\mapsto1$. Then $h’\circ p_1=h’\circ p_2$ but, since $h’$ is not constant, it is completely impossible for $h’$ to factor through $B=\{b\}$.
