Constructing coequalizers via pushouts and initial objects Is it possible to construct a coequalizer in terms of a pushout and an initial object? How can this be done for example in Set? 
 A: Yes, it is possible. 
Let $A$ be a category with pushouts and initial object $i\in A$. First observe that coproducts can be constructed in terms of pushouts and initial object.
Let $a,b\in A$ be objects of $A$. Then their coproduct $a\sqcup b$ is the pushout of the pair $(i_a,i_b)$:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
i & \ra{i_a} & a & \\
\da{i_b} & & \da{s_a} &\\
b & \ra{s_b} & a\sqcup b & \\
\end{array}
$$
If $f\colon a\to c$ and $g\colon b\to c$ are two arrows in $A$, then denote by $(f,g)\colon a\sqcup b\to c$ the unique arrow such that $(f,g)\circ s_a=f$ and $(f,g)\circ s_b=g$.
Let $f,g\colon a\to b$ be two arrows in $A$. Then their coequalizer is the pushout of the pair $((f,id(b)),(g,id(b)))$:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
a\sqcup b & \ra{(f,id(b))} & b & \\
\da{(g,id(b))} & & \da{p} &\\
b & \ra{q} & Coeq(f,g) & \\
\end{array}
$$
Indeed, let $c$ be an object of $A$ and $h\colon b\to c$ be an arrow, such that $h\circ f=h\circ g$. Then $h\circ(f,id(b))=h\circ(g,id(b))$ by the universal property of coproduct $a\sqcup b$. Thus we have the unique arrow $x\colon Coeq(f,g)\to c$, such that $x\circ p=x\circ q=h$ by the universal property of pushout. Conversely, if $y\colon Coeq(f,g)\to c$ satisfies $y\circ p=h$, then
$$
y\circ q=y\circ q\circ id(b)=y\circ q\circ (g,id(b))\circ s_b=y\circ p\circ (f,id(b))\circ s_b=
$$
$$
=h\circ(f,id(b))\circ s_b=h, 
$$
therefore $x=y$.
Of course, you can easily apply this construction to $\mathbf{Set}$.
