## (Least) upper bounds for order-preserving maps

A map $f:P\to Q$ between non-empty partially ordered sets is said to be order-preserving if $x, y \in P, x\leq_P y$ entails $f(x) \leq_Q f(y)$. The collection $Mor(P,Q)$ of order-preserving maps $f:P\to Q$ can be endowed with an ordering relation in a natural way by setting

$f \leq g$ in $Mor(P,Q)$ iff $f(x) \leq g(x)$ for all $x\in P$.

When considering any partially ordered set, or poset, for short, a natural question arising is that of the existence of (least) upper bounds, and dually, (greatest) lower bounds. A subset $S$ of a poset $(P,\leq)$ is said to have an upper bound if there exists $z \in P$ such that $z \geq s$ for all $s\in S$. The collection of upper bounds of $S$ is denoted by $S^u$.

The basic question of this post is: When do $f, g \in Mor(P,Q)$ have a (least) upper bound?

When you start tinkering with $Mor(P,Q)$ and its elements, you notice that the existence of bounds in the codomain $Q$ is more important than in the domain $P$. This can be illustrated by two examples:

(1) If $f(x)$ and $g(x)$ have no upper bound in $Q$ for some $x \in P$, then $f,g$ are not bounded in $Mor(P,Q)$.

(2) If $Q$ is a lattice, then so is $Mor(P,Q)$.

As for the proof of statement (2), let us denote the supremum of $a,b \in Q$ by $a\vee b$ and the infimum by $a\wedge b$. Let $f,g \in Mor(P, Q)$. Then it is easy to verify that the map $h:P \to Q$ given by

$h(x) = f(x) \vee g(x)$ for all $x \in P$

is the supremum of $f,g$; the infimum is constructed in a similar manner.

Let’s have a look at the converse statements of (1) and (2).

The converse of (1) would say that if $\{f(x), g(x)\}^u \neq \emptyset$ for all $x\in P$, then $f, g$ have an upper bound in $Mor(P,Q)$. We can construct a finite counterexample to this statement by letting $V = \{b, 0, 1\}$ ordered by $b < 0, b < 1$ and $Q = \{0,1\}^2$ ordered by $(0, j) < (1,k)$ for $j,k \in \{0,1\}$. So if we draw $V$ we get a v-shaped figure, and $Q$ looks like a butterfly.

Let $f,g: V\to Q$ be the maps given by $f(b) = (0,0)$ and $g(b) = (0,1)$ and $f(0) = g(0) = (1,0)$ and $f(1) = g(1) = (1,1)$. Then $\{f(x), g(x)\}^u \neq \emptyset$ for all $x\in V$, but it is easily seen that there is no order-preserving map that is an upper bound of $f, g$.

The converse of (2), on the other hand, is true. In order to tackle this, we need the concept of a retract. We say that a poset $S$ is a retract of a poset $T$ if there are order-preserving maps $i:S\to T$ and $r:T \to S$ such that $r\circ i: S\to S$ is the identity map.

Lemma 1. $Q$ is a retract of $Mor(P,Q)$.
Proof. Let $i:Q \to Mor(P,Q)$ be defined by $i:q \mapsto c_q$ where $c_q:P\to Q$ is the constant map that maps everything from $P$ to $q\in Q$. For defining $r:Mor(P,Q) \to Q$, fix $a_0 \in P$ and for $f\in Mor(P,Q)$ let $r(f) = f(a_0)$. A routine check shows that $i, r$ are order-preserving and $r\circ i = \textrm{id}_Q$.

Lemma 2. Any retract of a lattice is a lattice.
Proof. Let $L$ be a lattice and $P$ a poset and $i:P\to L$ and $r:L \to P$ such that $r\circ i: P\to P$ is the identity map. For $a, b \in P$ it suffices to check that $r(i(a) \vee i(b))$ is the least upper bound of $a,b$. It clearly is an upper bound. Suppose $v \in \{a,b\}^u$. Then $i(v) \in \{i(a), i(b)\}^u$ since $i$ is order-preserving, therefore $i(a)\vee i(b) \leq i(u)$. Using the fact that $r$ is order-preserving we get that $r(i(a) \vee i(b)) \leq r(i(u)) = u$, so $r(i(a) \vee i(b))$ is the least upper bound.

In summary, we get that $Mor(P,Q)$ is a lattice if and only if $Q$ is a lattice. (Note that this equivalence only works for $P \neq \emptyset$).

We gave partial answers, but are still lacking a full one, to the main question of this post: When do $f, g \in Mor(P,Q)$ have a (least) upper bound?