Monotonic function which is not an order-embedding

Question

I wrote down this definition during the lesson:

Let $\langle P, \leq_P\rangle$ and $\langle Q, \leq_Q\rangle$ be two posets.

A function $f: P \rightarrow Q$ is

• monotonic, if $x \leq_P y \implies f(x) \leq_Q f(y)$;
• order-embedding, if $x \leq_P y \iff f(x) \leq_Q f(y)$.

Of course, an order-embedding is also a monotonic function but I'm trying to show a monotonic function which is not an order-embedding (and so far, I wasn't able to find one).

Answer 1

Let $P$ consist of two non-comparable elements, and let $Q$ consist of two comparable elements. Each of the bijections from $P$ to $Q$ is monotonic, and neither is an order-embedding. Thus, the two are not equivalent even for bijections.

Answer 2

Round reals to integers. Any of the usual rounding functions (down, up, nearest) are monotonic but not order-embedding.

More brutally, take any constant function from one partially ordered set to another (such as the one-element poset).

Generalizing both examples, any order-preserving function from poset $P$ to a "smaller" poset $p$, such as a quotient of $P$, or a placement of that quotient into a poset with additional elements. And/or additional relations.