Monotonic function which is not an order-embedding 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).
 A: 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.
A: 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.
A: Category-Theoretic Interpretation
The problem also has a nice category-theoretic interpretation as follows:
The preordered sets are precisely the thin (small) categories and the monotone functions are precisely the functors between such categories.
Moreover, functors between thin categories are trivially faithful on morphisms,
and an order embedding corresponds precisely to being full on morphisms,
while injective/surjective corresponds precisely to injective/surjective on objects.
Now, as in most of category theory, there's little to almost no relation between the behavior of a functor on objects to its behavior on morphisms and this remains the case also for functors between thin categories:

Between thin categories, a functor may easily be injective but still fail to be full,
  and conversely, a functor may easily be full but still fail to be injective!

The situation improves just a little if antisymmetry enters the game, namely:

Between partially ordered sets; if a functor is full then it is injective.

Proof: $$F(x)=F(x')\implies F(x)\leq F(x')\leq F(x)\stackrel{F:\mathrm{full}}{\implies} x\leq x'\leq x\stackrel{X:\mathrm{POSET}}{\implies} x=x'$$
