# Construction of a partial order on a quotient of a coproduct

[NB: Throughout this post, let the subscript $i$ range over the set $\unicode{x1D7DA} \equiv \{0, 1\}$.]

Let $(Y, \leqslant)$ be a poset, and $X\subseteq Y$. Let $\iota_i$ be the canonical inclusions $Y \hookrightarrow (Y \amalg Y)$. Define $Z = (Y \amalg Y)/\sim$, where $\sim$ is the smallest equivalence relation on $Y\amalg Y$ that identifies $\iota_0(x)$ and $\iota_1(x)$, for all $x \in X \subseteq Y$. Finally, define functions $f_i = \pi\;{\scriptstyle \circ}\;\iota_i$, where $\pi$ is the canonical projection $(Y \amalg Y) \to Z = (Y \amalg Y)/\sim$.

I'm looking for a construction of a partial order $\leqslant_Z$ on $Z = (Y \amalg Y)/\sim$ such that both $f_0$ and $f_1$ are order-preserving wrt $\leqslant_Z$.

This is what I have so far:

Since the $f_i$ are injections, the $\leqslant_i$ given by

$$\leqslant_i \;\;=\;\; \{(f_i(y), f_i(y\,')) \;\;|\;\; (y, y\,' \in Y) \wedge (y \leqslant y\,')\} \;\;\cup\;\; I_Z.$$

...(where $I_Z$ is the identity on $Z$) are well-defined partial orders on $Z$.

Now, let $T$ be the transitive closure of the relation on $Z$ given by $\leqslant_0 \cup \leqslant_1$. I.e. $T = \bigcap_{V \in \mathscr{T}} V$, where $\mathscr{T} \neq \varnothing$ is the family of transitive relations containing $\leqslant_0 \cup \leqslant_1$. $T$ is obviously transitive, and it is also reflexive, since $I_Z \subseteq V, \forall \; V \in \mathscr{T}$.

The question therefore reduces to whether $T$ is antisymmetric (i.e. $T\cap T^{-1} = I_Z$).

I'm having a hard time finding a halfway reasonable-looking approach to this last point.

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In the following, the subscript index $i$ ranges over $\mathbb{Z}/2$.

Let $<_Y \;=\; \leqslant_Y - I_Y$. Likewise, let $<_i \;=\; \leqslant_i - I_Z = \{(f_i(x), f_i(y))|(x, y) \in <_Y\}$, $<^* \;=\; <_0 \;\cup\; <_1$, and $<_* \;=\; <_0 \;\cap\; <_1$. Also, define $\tilde{X} = f_0[Y] \cap f_1[Y] = f_0[X] = f_1[X]$, and $\tilde{E}_i = f_i[Y - X]$. Hence $Z = \tilde{X} \cup \tilde{E}_0 \cup \tilde{E}_1$ and $\tilde{X} \cap \tilde{E}_i = \tilde{E}_0 \cap \tilde{E}_1 = \varnothing$.

First, note that $\text{dom}(<_*) \subseteq \text{dom}(<_0) \cap \text{dom}(<_1) \subseteq f_0[Y] \cap f_1[Y] = \tilde{X}$, and similarly, $\text{ran}(<_*) \subseteq \tilde{X}$. Therefore $<_* \;\subseteq\;\tilde{X}\times\tilde{X}$, and thus, $\forall x, x' \in \tilde{X}$, we have that $x <_i x' \Leftrightarrow x <_* x'$.

Now, denote by $\mathscr{C}_p$ the class of all cycles in $<^*$ with basepoint $p\in Z$. Suppose that $\mathscr{C}_p \neq \varnothing$, for some $p \in Z$. Let $C = (p = p_0, p_1, \cdots, p_n = p) \in \mathscr{C}_p$ be of minimal length. From this last property, and the transitivity of the $<_i$, it follows that the pairs $(p_j, p_{j+1}), \forall j \in \mathbb{Z}/n$ that make up $C$ must belong alternately to $<_i$ and $<_{i+1}$. Abusing notation slightly, this situation can be summarized like this: $$p_0 <_i p_1 <_{i+1} p_2 <_i \cdots <_i p_{n-1} <_{i+1} p_n = p = p_0$$

...for some $i\in \mathbb{Z}/2$. Now, for every $j\in \mathbb{Z}/n$, it follows from $p_{j-1} <_{i_j} p_j$ and $p_j <_{{i_j}+1} p_{j+1}$ that $p_j \in \tilde{X}$, and from this, and the fact (shown above) that the restrictions of the $<_i$ to $\tilde{X}$ are equal to $<_*$, we conclude that

$$p_0 <_* p_1 <_* p_2 <_* \cdots <_* p_{n-1} <_* p_n = p = p_0.$$

Since this contradicts the asymmetry of $<_*$, we conclude that $\mathscr{C}_p = \varnothing, \forall p\in Z$. From this it follows that $<^*$ is acyclic, and so its transitive closure $\overline{<^*}$ must be irreflexive. So we have that $\overline{<^*} \;\cup\; I_Z$ is a partial order on $Z$.

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Suppose that $z\leq w$ and $w\leq z$, this means that there are finite sequences so that $z=z_0\leq z_1\leq ...\leq z_n=w$ and $w=w_0\leq w_1\leq ...\leq w_m=z$ where for all $i< n$ $(z_i,z_{i+1})\in (\leq_0\cup \leq_1\cup I_Z)$ and for all $j<m$ $(w_j,w_{j+1})\in (\leq_0\cup \leq_1\cup I_Z)$. This implies $z$ equivalent to $w$.

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