$$
\newcommand{op}[1]{\operatorname{#1}}
\newcommand{cat}[1]{\op{#1}}
\newcommand{hom}[1][2]{\cat{Hom}(#1,#2)}
$$
On what at this point is becoming a series, I'm going through an exercise of Emily Riehl's Category Theory in Context,
Exercise 2.4.v. Define a contravariant functor $F: \cat{Set}^{op} \rightarrow \cat{Set}$ that carries a set to the set
of preorders on it. What is its category of elements? Is $F$ representable?
I've defined this functor as
$$
\mathfrak{P} : \cat{Set}^{op} \longrightarrow \cat{Set} \\
X \longmapsto \mathfrak{P}(X) \\
(f:X \to Y) \mapsto \mathfrak{P}(f)
$$
with $\mathfrak{P}(X) = \{\leq \ \subseteq \ X \times X : \leq \text{ is a preorder on } X\}$ and, given a function $f: X \to Y$,
$$
(x,y) \in \mathfrak{P}(f)(\leq) \iff f(x) \leq f(y)
$$
If this is the correct/usual construction of such functor then its category of elements has objects as sets with a fixed preorder $(X, \leq )$ and morphisms as functions that behave as previously described, i.e. $f : (X,\leq) \rightarrow (Y, \leq')$ is a morphism in $\int \mathfrak{P}$ if and only if
$$
x \leq x' \iff f(x) \leq f(x') \quad (\forall x, x' \in X)
$$
Now, as is proven in this exercises' chapter, a contravariant functor is representable if and only if its category of elements has a terminal object. In this case, this means that $\mathfrak{P}$ would be representable if and only if there exists a set $X_F$ equipped with a preorder $\preceq$ so that, for any preordered set $(X,\leq)$, there exists a unique function $f : X \rightarrow X_F$ such that
$$
x \leq x' \iff f(x) \preceq f(x') \quad (\forall x,x' \in X)
$$
My intuition is that this will not happen, although I am not fully convinced about my (apparent) proof, which goes as follows: suppose that such a final object $(X_F, \preceq)$ exists. Therefore, there should exist a unique function $s : \mathbb{N} \longrightarrow X_F$ that defines the usual order on $\mathbb{N}$ and thus $s(\mathbb{N})$ gives a countably infinite chain on $X_F$. In particular, there exist distinct elements $x \preceq x' \preceq x''$ in $X_F$.
However, this contradicts the unicity of morphisms towards $X_F$ in the case of $T = \{1,2\}$ with the usual order, since the functions
$$
f_1: T \rightarrow X_F \\
1 \mapsto x \\
2 \mapsto x'\\
$$
$$
f_2: T \rightarrow X_F \\
1 \mapsto x' \\
2 \mapsto x''\\
$$
both verify $y \leq y' \iff f_i(y) \leq f_i(y') \ (\forall y,y' \in T)$.
Have I missed something along the way? I had a strong intuition about $\mathfrak{P}$ being representable when I started thinking about this, but now I am leaning towards a negative answer.
Best Answer
Yes, that argument looks good. You could just use the order $0<1<2$, instead of going all the way to $\mathbb{N}$.