The following question is taken from $\textit{Arrows, Structures and Functors the categorical imperative}$ by Arbib and Manes
Let $E$ be the equivalence relation on $\mathbb{N}$ whose two equivalence classes are "even" and "odd". Note that $E$ is a poset via $(n, m)\leq (n',m')$ iff $n\leq n'$ and $m\leq m'$ such that the restricted projections $p,q: E\rightarrow \mathbb{N}$ are order-preserving. Show that $coeq(p,q)$ exists in $\textbf{Poset,}$ and is not the same as in $\textbf{Set}$.
$\textbf{What I would like to know are the following:}$
$\textbf{(1)}$ Since the equivalence relation consists of only two equivalence classes: even and odd; does that mean in either of the two equivalence classes, say "even", all of $m,n, m',n'$ have all to be even, similarly for the case odd.
$\textbf{(2)}$ How do the restricted projections $p,q$ work in practice? Is it like $p_1(n,m)=n$, $p_2(n',m')=n'$ and $q_1(n,m)=m$ $q_2(n',m')=m'$ with $(n,m)\leq(n',m')$ implying $p_1(n,m)\leq p_2(n',m')$ and $q_1(n,m)\leq q_2(n',m'),$ where $\leq$ denotes usual inequality and only applies to pairs of order pairs, $(n, m), (n',m')$, iff all $m,n, m',n'$ have the same parity. Obviously this is order preserving. But is this the correct way to construct the projection maps?
$\textbf{(3)}$ To construct the coequalizer for this pair of projection, do I use the solution from the following post, (I actually don't know how to apply it) or do I just do something like the following:
Let the map $h:\mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}\times \mathbb{N}/E$. As an example if $n\leq n'$, then $h(p_1(n,m), p_2(n',m'))=0$ if $n, n'$ are both even else, $h(p_1(n,m), p_2(n',m'))=1$ if $n, n'$ are both odd. Similar construction for the case where $h(q_1,q_2).$
$\textbf{(4)}$ Lastly what does it mean that the coequalizer for this particular $\textbf{Poset}$ is not the same as in $\textbf{Set}$?
Can someone explain some of my confusions and also check on my ideas about how to go about solving this problem and if my notations and understanding of the problems are correct. Thank you in advance.
(1) No: $(n,m)\in E$ means $n$ and $m$ have the same parity. So $(n,m)\leq(n',m')$ means that $n$ and $m$ have the same parity, $n'$ and $m'$ have the same parity (possibly different from that of $n$ and $m$, and that $n\leq n'$ and $m\leq m'$.
(2) This is the correct way to construct the restricted projection maps. They are the compositions of $\mathbb N\times\mathbb N\to\mathbb N$ with the inclusion $E\hookrightarrow\mathbb N\times\mathbb N$. Note that, in accordance with my correction of (1), the inequality only applies to pairs of ordered pairs $(n,m)$ and $(m',n')$ where $n$ and $m$ have the same parity, and $n'$ and $m'$ have the same parity, but possibly different from that of $n$ and $m$.
For (3) and (4), note that a function $f\colon N\to X$ to a (po)set $X$ coequalizes the projection maps $E\rightrightarrows N$ if $f(n)=f(m)$ for every pair of numbers $(n,m)$ that are both odd or both even.
In the case of sets, this is the only condition, and thus such functions $f$ correspond to functions $g\colon N/E\to X$ where $N/E$ is the set of equivalence classes of $E$, i.e. the set of two elements: the set of even numbers, and the set of odd numbers. More precisely, if $q\colon N\to N/E$ is the function sending any even number to the subset of all even numbers, and any odd number to the subset of all odd numbers, then $f\colon N\to X$ coequalizes the pair of restricted projections $E\rightrightarrows N$ if and only if $f$ factors as $f=g\circ q$ for a unique function $g\colon N/E\to X$. Thus, $q\colon N\to N/E$ is the coequalizer of $E\rightrightarrows N$ in the category of sets.
In the case of posets, the fact that $f\colon N\to X$ is a morphism of posets means that $f(x)\leq f(y)$ if $x\leq y$. In partuclar $f(x)\leq f(x+1)\leq f(x+2)=f(x)$, where the last equality holds because $x$ and $x+2$ have the same parity, i.e. $(x,x+2)\in E$. Thus, we have $f(x)=f(x+1)$ for all $N$, i.e. $f\colon N\to X$ is constant. Since any constant function is a morphism of posets, and since constant functions are the ones that factor (necessarily uniquely) through singletons, we can conclude that any function $\top\colon N\to\{*\}$ to a singleton $\{*\}$ poset is the coequalizer of $E\rightrightarrows N$ in the category of posets.