Questions about Simmons' construction of his first construction of coequalizer for the category of $\textbf{Sets}$. The following is from Harold Simmon's An introduction Category Theory which contains the definition for the concepts of both equalizer and coequalizer, along with the first of two construction for the coequalizer for the category of Sets. 

$\quad$ [Simmons, Definition 1]  Given a parallel pair $${\stackrel{\large{A}\stackrel{\small\small f}{\stackrel{\longrightarrow}{\longrightarrow}}\large{B}}{\small\small g}}$$ of arrows in a category $C$ $$\text{an equalizer} \quad \text{a coequalizer}$$ is an arrow $$S\xrightarrow{k}A \quad B\xrightarrow{k}S$$ which makes equal $f$ and $g$, and has the following universal property.
$\quad$For each arrow $$X\xrightarrow{h}A \quad B\xrightarrow{h}X$$ which makes equal the parallel pair, there is a unique arrow $$X\xrightarrow{m}S \quad S\xrightarrow{m}X$$ such that $$h=k\circ m\quad h=m\circ k$$ holds.  This $m$ is the mediating arrow (or mediator) for the arrow $h$.
$\quad$Let $S$ be an arbitrary set, and let $\sim$ be an equivalence relation on $S$.  This relation partition $S$ into blocks (equivalence classes).  For each $s\in S$ let $[s]$ be the block in which $s$ lives, and let $$S/\sim$$ be the set of all such blocks. Let $$S\xrightarrow{\sigma}S/\sim$$ $$S\mapsto [s]$$be the induced surjection.
$\quad$Let $$S\xrightarrow{\sigma}X$$ be any function.  $\color{red}{The \,kernel \,of \,h \,is \,the \,relation \,\approx \,on \,S \,given \,by \,}$ $$\color{red}{s_1\approx s_2 \Longleftrightarrow h(s_1)=h(s_2).}$$ $\color{red}{for\,s_1, s_2\in S.}$  Trivially, this is an equivalence relation.
$\quad$$\color{red}{Now\, suppose\, \approx\, includes\, \sim, \,that\, is \,S \,given \,by \,}$ $$\color{red}{s_1\sim s_2 \Longrightarrow h(s_1)=h(s_2)}$$ $\color{red}{for\,s_1, s_2\in S. }$   Under these conditions there is a commuting triangle 



for some unique function $h^{\sharp}$.  This function is given by $$h^{\sharp}([s])=h(s)$$ for $s\in S.$  To show that $h^{\sharp}$ is well defined; the function $\sigma$ is surjective, there can be at most one function $h^{\sharp}$ to make the diagram commute (the one in the screen shot above)  For $s_1, s_2 \in S$ we have $$[s_1]=[s_2] \Longrightarrow s_1\sim s_2\Longrightarrow h(s_1)=h(s_2)$$ to show that the suggested function $h^{\sharp}$ is well defined.  For $s\in S$ we have $$(h^{\sharp}\circ \sigma)(s)=h^{\sharp}([s])=h(s)$$ to show that the triangle commutes.

Questions:  The portion of the block text in red, I would like to know why in the construction, there needs to be defined two equivalence relations: $\approx$, $\sim$, and $\sim  \,\subset \,\approx$, and also why where both of the equivalence relation has to be the kernel of a function.  The other thing is, in $\color{red}{s_1\sim s_2 \Longrightarrow h(s_1)=h(s_2)}$, shouldn't the $\Longrightarrow$ be $\Longleftrightarrow$ instead?  Meaning, there is that a misprint?  Thank you in advance.
 A: When $\sim$ is first introduced in 2.6.6, Simmons is explaining how equivalence relations and quotients of sets relate to functions, before considering how to build the coequalizer in 2.6.7. I think your problem is that Simmons uses $S$ for an arbitrary set in 2.6.6, but in the proof of existence of coequalizers he is going to apply it to $B$, not $S$, and the $S$ from the definition of coequalizers is $B/{\sim}$.
Perhaps it is better to see the overall strategy for constructing $S$ and $k : B \rightarrow S$ to make a coequalizer.

*

*Define, using $f$ and $g$, a purpose-made equivalence relation $\sim$ on $B$.

*Take $S = B/{\sim}$ and $k$ to be the function mapping $b \in B$ to $[b]$, its equivalence class.

*Given some $h : B \rightarrow X$ such that $h \circ f = h \circ g$, we require an $m : S \rightarrow X$ such that $m \circ k = h$, unique with this property.

*$h$ defines an equivalence relation $\approx$ on $B$, its kernel. So in step 1, we need to arrange that, $\approx$ is a coarser equivalence than $\sim$, which is to say, for all $b_1,b_2 \in B$, we have $b_1 \sim b_2 \Rightarrow h(b_1) = h(b_2)$. This will not be $\Leftrightarrow$, because we need it to work for all $h$ satisfying the requirement in 3. Not a misprint.

*Therefore the function $m([b]) = h(b)$ is well-defined on $S = B/{\sim}$, and $m \circ k = h$ by definition, and the uniqueness follows from the surjectivity of $k$.

We are left with how to define $\sim$ in part 1 so that it satisfies the requirement when it is used in part 4. Define $R \subseteq B \times B$ by
$$
R = \{ (f(a),g(a)) \mid a \in A \}
$$
($R$ is called $\rightsquigarrow$ by Simmons). Then take $\sim$ to be the "equivalence closure" of $R$, the finest equivalence relation containing $R$. If $\approx$ is the kernel of some $h : B \rightarrow X$ with $h \circ f = h \circ g$, then for all $a \in A$, $h(f(a)) = h(f(b))$, so $\approx$ contains $R$ and therefore is coarser than $\sim$, the property required in 4.
