Functors amount to order-preserving maps 
Let $A,B$ be ordered sets and let $\mathcal A,\mathcal B$ be the corresponding categories. Show that a functor $\mathcal A\to \mathcal B$ amounts to an order-preserving map $A\to B$.

What is the precise meaning of "amounts to"? It sounds too informal. Does it mean that a functor $\mathcal A\to \mathcal B$ gives rise to an order-preserving map $A\to B$, and conversely?
Here's how I understand the correspondence between ordered sets and categories. Given an ordered set $A$, the corresponding category $\mathcal A$ has as objects the elements of $A$, and if $a,b$ are objects, then $a\to b$ iff $a\le b$ in $A$. There is at most one arrow between any two objects. A function $A\to B$ between ordered sets is a functor between the corresponding categories. I don't have a firm understanding of the latter -- formally speaking, why is it the case? (I assumed it is the case because I have a feeling that it should be the case.)
Now let $F:\mathcal A\to \mathcal B$ be a functor and $a,b$ be objects of $\mathcal A$. Suppose $a\to b$ in $\mathcal A$. Then $F(a)\to F(b)$ in $\mathcal B$ by the definition of a functor. That is, if $a\le b$ in $B$, then $F(a)\le F(b)$ in $B$. So it looks like the claim we want to prove is trivial. Is this indeed so? The only thing which remains murky is the correspondence between functions and functors I asked in the previous paragraph.
Update:
I think here's what I was looking for:
1) Let $F:\mathcal A\to \mathcal B$ be a functor and $a,b$ be objects of $\mathcal A$. Suppose $a\to b$ in $\mathcal A$. Then $F(a)\to F(b)$ in $\mathcal B$ by the definition of a functor. By the definition of $\to$, this means $F(a)\le F(b)$ in $B$. Since $F$ is, in particular, a function $Ob(\mathcal A)=A\to Ob(\mathcal B)=B$, the functor $F$ indeed  gives rise to an order-preserving function $A\to B$.
2) Now suppose we have an order-preserving function $F:A\to B$. So if $a\le b$, then $F(a)\le F(b)$. Because of the this, $F$ can be regarded as a functor $\mathcal A\to \mathcal B$. We already know that it is a function $Ob(\mathcal A)\to Ob(\mathcal B)$, so it suffices to show that it can be also regarded as a function $\mathcal A(a,b)\to\mathcal B(F(a),F(b))$ for all $a,b\in Ob(\mathcal A)$ (that satisfies functoriality). Indeed, if $a\to b$ in $\mathcal A$, then $a\le b$ in $A$, so since $F:A\to B$ is order-preserving, it follows that $F(a)\le F(b)$, hence $F(a)\to F(b)$. This gives a map $\mathcal A(a,b)\to\mathcal B(F(a),F(b))$. I don't quite see how to prove functoriality.
 A: Basically you are right and maybe simply overanalysing a simple situation. The 'amount to' notion means that if you have ordered sets $A,B$ and you view them as categories as you describe, then when you look into what it means to have a functor $A\to B$ you find essentially the same information as when you look into what it means to have an order preserving function $A\to B$. And yes, the argument is very simple. 
At the level indicated by the question I'd say you are expecting something more complicated than it actually is. You can safely move one. On a slightly more precise manner though there is a way to make this business very rigorous. Firstly, notice that a given ordered set $A$ can be viewed as a category in the way you describe in infinitely many ways. This is so because you must actually name the morphism $a\to b$ that must exist whenever $a\le b$. That name is an arbitrary choice and cannot be made in a canonical way. So, you have infinitely many categories that all represent the same given ordered set $A$. In the other direction though, given a category with at most one morphism between any two objects, there is a canonical ordered set associated to it, simply declare $a\le b$ when there exists a morphism $a\to b$ in the category. If we denote by $Pos$ the category of ordered sets and order preserving functions and by $Pos_C$ the category of categories with at most one morphism between any two objects, then this remark says: there is a canonical choice for a functor $Pos_C\to Pos$ but there are infinitely many choices for functors in the other direction. Now, the upshot of all of this is that the two categories in question are equivalent. The canonical functor $Pos_C\to Pos$ is an equivalence of categories, and any of the non-canonical choices in the other direction is also an equivalence, in fact a right inverse of the canonical one. Spelling this out for objects is basically what you've done in the middle paragraph. Spelling it out for morphisms is basically your last paragraph. So, the 'amounts to' phrase can be understood as a functoriality claim for the process you describe on objects. 
