# Monotone function

Let $$\langle P, \leq_P \rangle$$ and $$\langle Q, \leq_Q \rangle$$ be posets. Then a func­tion $$\phi : P \rightarrow Q$$ is said to be mono­tone (al­ter­na­tively, or­der-pre­serv­ing) if for all $$s, t \in P$$, $$s \le_P t$$ im­plies $$\phi(s) \le_Q \phi(t)$$.

## Pos­i­tive example

com­ment: dot source:

di­graph G { node = 0.1, height = 0.1 rankdir = BT; rank = same; com­pound = true; font­name=”MathJax_Main”;

sub­graph cluster_P { node style=filled,color=white; edge = “none”; style = filled; color = light­grey; font­color = black; la­bel = “P”; la­bel­loc = b; b → a; c → a;

} sub­graph cluster_Q { node style=filled; edge = “none”; color = black; font­color = black; la­bel= “Q”; la­bel­loc = b; u → t; } edge = blue, style = dashed font­color = blue; la­bel = “φ”;
la­bel­loc = t; b → t = false; a → t = false; c → u = false; }

<div>

Here is an ex­am­ple of a mono­tone map $$\phi$$ from a poset $$P$$ to an­other poset $$Q$$. Since $$\le_P$$ has two com­pa­rable pairs of el­e­ments, $$(c,a)$$ and $$(b,a)$$, there are two con­straints that $$\phi$$ must satisfy to be con­sid­ered mono­tone. Since $$c \leq_P a$$, we need $$\phi(c) = u \leq_Q t = \phi(a)$$. This is, in fact, the case. Also, since $$b \leq_P a$$, we need $$\phi(b) = t \leq_Q t = \phi(a)$$. This is also true.

## Nega­tive example

com­ment: dot source:

di­graph G { node = 0.1, height = 0.1 rankdir = BT; rank = same; com­pound = true; font­name=”MathJax_Main”;

sub­graph cluster_P { node style=filled,color=white; edge = “none”; style = filled; color = light­grey; font­color = black; la­bel = “P”; la­bel­loc = b; a → b; }

sub­graph cluster_Q { node style=filled; edge = “none”; color = black; font­color = black; la­bel= “Q”; la­bel­loc = b; w → u; w → v; u → t; v → t; } edge = blue, style = dashed font­color = blue; la­bel = “φ”;
la­bel­loc = t; b → u = false; a → v = false; } <div>

Here is an ex­am­ple of an­other map $$\phi$$ be­tween two other posets $$P$$ and $$Q$$. This map is not mono­tone, be­cause $$a \leq_P b$$ while $$\phi(a) = v \parallel_Q u = \phi(b)$$.