Exercise 6 in Chapter 2 of Steve Awodey's "Category theory" asks,
Show that the epis among posets are the surjections (on elements)
The answer at the back is:
First, in the category Pos, an arrow is epi iff it is surjective: suppose that f: A -> B is surjective and let g, h: B-> C with gf = hf. In Pos, this means that g and h agree on the image of f, which by surjectivity is all of B. Hence g = h and f is epi. On the other hand, suppose f is not epi and that g, h: B -> C witness this. Since $g \neq h$, there is some $b \in B$ with $g(b) \neq h(b)$. But from this $b \notin f(A)$, and so A is not surjective.
Isn't this proof only of one direction (and then the contrapositive)? Does it remain to prove that an epimorphism in Pos is surjective?