The following is taken from "Categories" by T.S.Blyth
$\color{green}{Background}$
Let $E$ be a non-empty set and let $R$ be an equivalence relation on $E.$ Consider the category $C$ whose objects are pairs $(X,f)$ where $X$ is a non-empty set and $f:E\to X$ is a mapping such that $xRy \Longrightarrow f(x)=f(y)$ (in other words, such that $R$ implies $R_f$ where $R_f$ is the equivalence relation on $E$ associated with $f$). Define $\text{Mor}_C((X,f),(Y,g))$ to be the set of all mappings $h:X\to Y$ such that $h\circ f=g:$
$\textbf{Example 1:}$
We claim that the set $E/R$ of $R-$classes together with the natural map $\sharp_R:E\to E/R,$ described by $\sharp_R(x)=[x]_R,$ is an initial object in $C.$ To see this; let $g:E\to Y$ be such that $xRy\Longrightarrow g(x)=g(y).$ This means that we have $\sharp_R(x)=\sharp_R(y)\Longrightarrow g(x)=g(y)$ and so we can define a mapping $h:E/R\to Y$ by setting $h([x]_R)=g(x).$ Clearly, we have $h\circ \sharp_R=g.$ That $h$ is unique with respect to this property follows from the fact that if $k$ is also such that $k\circ \sharp_R=g$ then, since $\sharp_R$ is surjective, hence epic, we obtain $k=h$ by right cancellation.
$\textbf{Example 2:}$
Given $f,g:A\to B$ in $\textbf{Set}, \textbf{Grp}$ or $_R\textbf{Mod}$ let $Q$ be the smallest equivalence relation (respectively, congruence) on $B$ such that $$(\forall a\in A)(f(a)\equiv g(a)\quad(Q).$$
Consider $B/Q$ endowed with the appropriate structure. Then the natural morphism $\sharp_Q:B\to B/Q$ is a coequaliser of $f,g.$ To see this, observe first from the definition of $Q$ that $\sharp_Q\circ f=\sharp_Q\circ g.$ Suppose now that $j:B\to X$ is such that $j\circ f\equiv j\circ g.$ If $R_j$ is the associated equivalence relation (respectively, congruence) we have that $$(\forall a\in A)f(a)\equiv g(a)\quad (R_f)$$ so that, by definition, $Q$ implies $R_j.$ It now follows from the universal property of $B/Q$ described in $\textbf{Example 1}$ that there is a unique morphism $k:B/Q\to X$ such that the diagram
is commutative. Consequently $\sharp_Q:B\to B/Q$ is a coequaliser of $f,g.$
$\color{blue}{Questions}$
What I would like to know is how does $Q$ implies $R_j.$ Am I correct to understand that $Q$ and $R_j$ are separate equivalence relations being defined on $B.$ If given two separate equivalence relations, to show that $k\circ \sharp_Q=j$ is well defined. Then we need the assumption $b_1Qb_2\Longrightarrow j(b_1)=j(b_2)?$
Thank you in advance.