There is an exercise in "An introduction to Category Theory" by Harold Simmons says:
Consider any pair of $\mathcal{Pos}$-arrows $f: S \rightarrow T$ and $g: T \rightarrow S$.
Show that when both $id_S \le g \circ f$ and $f \circ g \le id_T$, if $f \dashv g$ then
$$f \circ g \circ f = f \quad g \circ f \circ g = g $$
There is a solution to this exercise ([here](http://www.cambridge.org/us/download_file/150877/, exercise 1.3.6(b) on page 12)), which I don't understand:
To prove $f \circ g \circ f = f$, we know that $\forall a \in S, id_s(a) \le (g \circ f)(a)$ and because $f$ is a monotonic function($\mathcal{Pos}$-arrow), we know that $f(a) \le (f \circ g \circ f)(a)$, so far so good.
On the other hand, it states that for each $b \in T$, we have $$ (f \circ g)(b) \le b $$ and hence $$ (f \circ g \circ f)(a) \le f(a) $$ as a particular case.
But here how can we go from $b$ to $f(a)$? It might look obvious but for $f$ we only know that it is monotonic, to make sure $\forall b \in T, \exists a \in S, f(a) = b$ we need $f$ to be surjective. However this concern isn't seen in the proof.
So even if it doesn't sound good to me, I guess monotonicity implies surjectivity?
EDIT:
I think in general if there is a poset $S$, $\forall a,b \in S$, if $a \le b$, then it is fine to "add any monotone to the left side" (i.e $f(a) \le f(b), (g \circ f)(a) \le (g \circ f)(b),(h \circ g \circ f)(a) \le (h \circ g \circ f)(b), \ldots $, sorry I don't know the formal terminology for this) But I doubt the same holds when you are "adding monotone to the right side".
No, monocity doesn't imply surjectivity, but we don't even need surjectivity.
By hypothesis, we know that for all $b\in T$ we have $(f\circ g)(b)\le b$. In particular, for any $a\in S$, it also holds for the element $f(a)\in T$ written in place of $b$.
By the way, the conditions on the pair $f,g$ (to satisfy $id_S \le g\circ f$ and $f\circ g\le id_T$) make the definition of $f$ being left adjoint to $g$, i.e. $f\dashv g$ (also called a 'Galois connection' between $S$ and $T^{op}$).