Quasiorders and their associated partial orders

Question

Consider a quasiordered set $Q$ (order relation $\lesssim$) and define an equivalence relation in the usual way.

$$x \sim y \quad \Leftrightarrow \quad x \lesssim y \,\mbox{ and }\, y \lesssim x$$

Consider also the partial order $P$ (order relation $\leq)$ obtained by identifying equivalent elements of $Q,$ and let $f : Q \rightarrow P$ denote the natural surjection.

Does every order-preserving map $\phi : Q \rightarrow P'$, where $P'$ is a partially ordered set, factor uniquely through $f$? In the sense that there always exists a unique order-preserving map $g : P \rightarrow P'$ such that $g \circ f = \phi$?

If so, what is the general name for this kind of thing?

Answer 1

Yes, since any order preserving mapping from a qusi-poset to a poset must identify two equivalent elements, thus $f$ factors uniquely through the set of equivalence classes.

This is a special case of what is known as an adjunction. Let $Pos$ be the category of posets and let $QPos$ be the category of quasi-posets. The process you describe above, sending $Q$ to $P$, is actually a functor $F:QPos\to Pos$ (so now I'll write $F(Q)$ for your $P$). Now, there is clearly also a functor $U:Pos\to QPos$ (since any poset is also a quasi-poset). Such a functor is called a forgetful functor (since it forgets some aspects of a poset only to remember that it is a quasi-poset).

This pair of functors is an example of an adjoint pair of functors, with $F$ being left adjoint to $U$. This means that for every poset $P$ and quasi-poset $Q$, there is a natural bijection between all order preserving mappings $F(Q)\to P$ and the set of all order preserving mappings $Q\to U(P)$. In particular, there is a mappings $Q\to UF(Q)$ which corresponds to $id:F(Q)\to F(Q)$. It turns out that this mappings is precisely the projection you describe above. Moreover, in general for adjunctions, the correspondence is obtained by precomposing with this projection. Thus, your question can now be re-cast as follows. Given a mappings $Q\to UP$, does it factor uniquely through $Q\to UFQ$. The answer is yes, since $Q\to UP$ corresponds by adjunction to some unique $FQ\to P$, and the correspondence is given by $Q\to UP$ is equal to $Q\to UFQ\to UP$.

Answer 2

Your definition of the equivalence relation $\sim$ is wrong; it should say $x\lesssim y$ and $y\lesssim x$. With "or" instead of "and" it usually won't define an equivalence relation.

With this correction, it becomes clear that any order-preserving map $\phi$ into a partially ordered set $P'$ will send equivalent elements to the same element (because of the antisymmetry of the order of $P'$), so it is constant on each equivalence class and thus determines a map $g:P\to P'$. It is trivial to check that $g\circ f=\phi$ and that $g$ is order-preserving. Finally, the factorization is unique because $f$ is surjective.