Preorders as categories vs category of preorders I understand how a preorder can be interpreted as a category (reflexivity translating into identity morphisms and transitivity to associativity). But now I want to understand the category of all preorders, Preord. Here, the morphisms are the monotone/order-preserving maps.
Take two preorders, $(X, ≤), (Y, ⪯)$, say, and a monotone function $f : (X, ≤) → (Y, ⪯)$.
I want to understand the inclusion functor $I:$ Preord → Cat. I would guess the particular preorders $(X, ≤)$ are mapped to the categories $I((X, ≤)) := X$ where the elements of $X$ are unchanged (but now considered as object in the category $X$, and the ≤'s are replaced by arrows? (Do I need to check something here?)
Now we need to map the monotone functions $f : (X, ≤) → (Y, ⪯)$ to functors $I(f) : I(X) → I(Y)$. But here I get confused. As a morphism in Cat, it should be a functor. That means $I(f)$ must assign both an object of (the category) $I(X) = X$ to an object of $I(Y) = Y$. So far, no problems here (just set $I(f)(x) = f(x)$). But what to do with the binary relation (≤)...? Should we view it as a subset of $X \times X$? Where to map that?
EDIT: We'd have to check that $x ≤ y$ (which in $I(X)$ I guess means there is an arrow from $x$ to $y$) implies that there is an arrow from $f(x)$ to $f(y)$? Why is it true that $x ≤ y$ implies that there is a unique arrow from $x$ to $y$ in $I(X)$?
Addendum:
Once I understand this, I'd like to construct a left-inverse functor $J$ s.t. $JI = 1$, but I'll first await your anwers here because that might then be easy :)
Cheers, and a happy 12023!
 A: Maybe here it helps to write down the definitions with less words and more formulas:
Let $(X,\leq)$ be a preorder. Define a category $I(X,\leq)$ has follows: $\mathrm{Ob}(I(X,\leq)) := X$ and $$\mathrm{Hom}(x,x') := \begin{cases} \{(x,x')\} & x \leq x'  \\ \emptyset & \text{else} \end{cases}.$$
This might seem awkward at first, but notice that this definition makes the Hom-sets disjoint, which is what we really need to have a well-defined source and target of a morphism.
Alternatively, you could define $\mathrm{Mor}(I(X,\leq))$ to be the set $\leq$. Remember that $\leq$ is a subset of $X \times X$. Then define source and target $s,t : {\leq} \to X$ as the two projections.
Let $f : (X,\leq) \to (Y,\leq)$ be a monotone map. We define a functor $ I(f) : I(X,\leq) \to I(Y,\leq)$. The map on objects is just $f : X \to Y$. The action on morphisms is a family of maps
$$\mathrm{Hom}(x,x') \to \mathrm{Hom}(f(x),f(x')).$$
Since $\emptyset$ is the initial set, there is nothing to define unless $x \leq x'$, in which case also $f(x) \leq f(x')$ (by assumption) and we have to define a map
$$\{(x,x')\} \to \{(f(x),f(x'))\},$$
which is of course unique and exists.
Alternatively, notice that $f \times f$ restricts to a map of sets ${\leq_X} \to {\leq_Y}$, and this is the "global" morphism part of $I(f)$.
It is trivial to check that $I(f)$ is indeed a functor, and that we have $I(\mathrm{id})=\mathrm{id}$ and $I(fg)=I(f)I(g)$. Thus, $I$ is a functor $\mathbf{PreOrd} \to \mathbf{Cat}$. One can show that $I$ is fully faithful and the image consists of all thin small categories.
A: 
Why is it true that $x\le y$ implies that there is a unique arrow from $x$ to $y$ in $I(X)$?

Because that's the very definition of the (canonical) category associated to a preorder $(X,\le)$.
$I(X)$ is defined to have object class the set of all $x\in X$ (that is, each $x\in  X$ is a distinct object of $I(X)$) and the arrows in the category $I(X)$ are defined as follows: $$I(X)(x,y)=\begin{cases}\{!\}&x<y\\\{\mathrm{id}_x\}&x=y\\\emptyset&\text{otherwise}\end{cases}$$
I use $!$ to denote a unique non-identity arrow $x\to y$. Use whichever set theoretic trick of your choice to produce a singleton set for this purpose.
An order preserving function $f:(X,\le_X)\to (Y,\le_Y)$ is assigned to the functor $I((X,\le_X))\to I((Y,\le_Y))$ which takes every object $x$ to the object $f(x)$ and every arrow $x\to y$ (when they exist) (there is only one!) to the unique arrow $f(x)\to f(y)$. I know this arrow exists since the existence of an arrow $x\to y$ means $x\le_X y$, so $f(x)\le_Y f(y)$ since $f$ is order-preserving: by very definition, there is then a unique arrow $f(x)\to f(y)$ in $I((Y,\le_Y))$.
