Constructing an ordering on a set P given a coequalizer Suppose that $f, g : (A, \leq) \to (B, \leq)$ are monotone maps in the category of preordered sets, Ord and suppose that $e : B \to Q$ is a coequalizer in Set for the functions $f, g : A \to B$.
How can we define a preorder on the set $Q$ so that $e$ is a monotone coequalizer in the category Ord?
Construction: Suppose $x, y \in A$. If $f(x) \leq f(y)$ AND $g(x) \leq g(y)$, then we say $e(f(x)) \leq e(f(y))$ in $Q$. Since $e$ is a coequalizer, this is the same as saying $e(g(x)) \leq e(g(y))$. It is pretty clear that this construct makes $e$ a monotone map. I'm not sure how to see if $e$ is a coequalizer with this construction.
 A: $$ \newcommand{\catname}[1]{{\textbf{#1}}} $$
Unfortunately, your construction fails since $e$ might fail to be a morphism in $\catname{PreOrd}$. Consider $A := \{a_1\}$, $B := \{b_1, b_2, b_3\}$, $f(a_1) := b_1$, $g(a_1) := b_2$ and $b_1 \leq b_3$. The coequalizer in $\catname{Set}$ would be $Q := \{q_{12}, q_3\}$ with $e(b_1) = e(b_2) = q_{12}$ and $e(b_3) = q_3$. According to your construction, you would add no other order to $Q$ than already required from a preorder. Now $b_1 \leq b_3$, but $e(b_1) = q_{12} \not\leq q_3 = e(b_3)$!

To salvage your construction, you have to endow $Q$ with the smallest preorder that makes $e$ order-preserving, see also ncatlab on coequalizers in Pos.
Now for that $(Q, e)$ being a coequalizer in $\catname{PreOrd}$ you have to show two things:


*

*$(Q, e)$ is a cocone

*$(Q, e)$ is in fact a coequalizer (or in general: colimit)


ad 1: Trivially due to $(UQ, Ue)$ being already a coequalizer in $\catname{Set}$ where $U: \catname{PreOrd} \to \catname{Set}$ is the forgetful functor.
ad 2: Let $(P, p)$ be another cocone. We have to show: $\exists !\varphi: Q \to P: \varphi = \varphi \circ e$. Certainly, $(UP, Up)$ is a cocone of the same diagram (under $U$) in $\catname{Set}$. Hence we have $\exists !\varphi: Q \to P: \varphi = \varphi \circ e$ — however with all symbols refering to $\catname{Set}$! Most symbols can be carried over to $\catname{PreOrd}$ seamlessly, the existential quantifier as well: if we had two such morphisms in $\catname{PreOrd}$, we would also have both of them (under $U$) in $\catname{Set}$. The only thing left to verify is that $\varphi$ is indeed a morphism in $\catname{PreOrd}$:
With $q_1 \leq q_2$, we need to show $\varphi(q_1) \leq \varphi(q_2)$. Since we endowed $Q$ with the smallest possible preorder that made $e$ order-preserving, the first inequality tells us that there must be $b_1, b_2$ with $e(b_1) = q_1$ and $e(b_2) = q_2$ (*). But then we have $\varphi(q_1) = \varphi(e(b_1)) = p(b_1)$ due to the universal property in $\catname{Set}$. The same holds with $\varphi(q_2) = p(b_2)$. Since we know that $p$ is order-preserving as a $\catname{PreOrd}$ morphisim, we have $\varphi(q_1) = p(b_1) \leq p(b_2) = \varphi(q_2)$, which was our proof goal.
(*) might be a bit subtle. Precisely speaking, $Q$ having such a smallest preorder and $q_1 \leq q_2$ imply that either what is assumed above ($\exists b_1, b_2: \ldots$), $q_1 = q_2$ (reflexivity) or $q_1 \leq y_1 \leq \ldots \leq y_n \leq q_2$ (transitivity). The first case is handled above. The second case trivially yields $\varphi(q_1) \leq \varphi(q_2)$. For the third case we continue the same case analysis until we reach to $q_1 \leq z_1 \leq \ldots \leq z_m \leq q_2$ where each inequality is case 1, i.e. for $q_1$ and $z_1$ there are $b_1, b_2$, for $z_1$ and $z_2$ there are $b_3, b_4$ and so on. We then apply the proof we did in the first case and get the inequality $\varphi(q_1) \leq \varphi(z_1) \leq \ldots \leq \varphi(z_m) \leq \varphi(q_2)$ in $P$. Due to transitivity in $P$ we then have $\varphi(q_1) \leq \varphi(q_2)$.
If you're interested in stating this even more formally, I'd say this is applying transfinite induction on the very construction of the "smallest preorder on Q". Maybe it goes along the way of the following wellfoundded binary relation $R$ on $Q$:
$(q_1 \leq q_2) \sqsubseteq (q_3 \leq q_4)$ iff. $q_1 \leq q_2$ was (one of) the reasons why $q_3 \leq q_4$ has been brought to existence.
Here is an exemplary excerpt of that relation as a Hasse diagram. The leaf nodes are either due to reflexivity (on the left) or due to being required by order-preservation of $e$:
$R$ depicting three disconnected nodes of the form $q \leq q$ and one tree with root $q_1 \leq q_2$ and leaves $q_1 \leq z_1$, $z_1 \leq z_2$, ..., $z_m \leq q_2$.">
