As a follow-up to my previous question: Preorders as categories vs category of preorders, I’m trying to construct a functor $J \colon \mathrm{Cat} \to \mathrm{Preord}$ such that $J ∘ I = 1$. First problem: does this even work for general categories, or only small ones? The natural choice for the object part of $J_0$ of $J$ would be to map a object part $\mathrm{ob}(\mathcal{C})$, of a category $\mathcal{C}$ to $\mathrm{ob}(\mathcal{C})$, but this doesn’t have to be a set (which is required for being a preorder). Is there any way to ‘shrink down’ a proper class to a set in a suitable way? We ignore this for now to move on.
Then $J_0$ has to do something with the morphism part $\mathrm{Mor}(\mathcal{C}) = (\mathrm{Hom}_{\mathcal{C}}(X, Y))_{X, Y ∈ \mathrm{ob}(\mathcal{C})}$ of the category $\mathcal{C}$. A natural choice seems to me to take the set(??) $$\{(X, Y) ∈ \mathrm{ob}(\mathcal{C})^2 \mid \lvert \mathrm{Hom}_{\mathcal{C}}(X, Y) \rvert \leq 1\}. $$ Apart from $\mathrm{ob}(\mathcal{C})$ not being a set, this at least seems well-defined to me.
Then to the morphism part $J_1$ of $J$. Clearly, it should map, for each pair of objects $(X, Y)$ of $\mathcal{C}$, a morphism (i.e., functor) $F \colon \mathcal{C} \to \mathcal{D}$ to a monotone function $J_1(F) \colon \mathrm{ob}(\mathcal{C}) \to \mathrm{ob}(\mathcal{D})$.
Before I go on, am I at least in the right direction? I’m sure there’s lots to check afterwards (most importantly, how to ensure that $J_1(F)$ is monotone).
Also, I seem to have forgotten that $J_1$ not just has to send ‘the functor $F$’ somewhere, but $F$ of course itself consists of two parts, $F_0$ and $F_1$. That’s where I get really confused.
I hope that when all this is cleared up (i.e., what $J_0(F_0), J_1(F_0), J_1(F_0), J_1(F_1)$ are all doing, exactly), the result that $J ∘ I = 1$ shouldn’t be to hard to see.
$\mathbf{Cat}$ usually denotes the category of small categories. If $\mathcal{C}$ is a small category, define a preorder on its objects by letting $X \leq Y$ if there is a morphism $X \to Y$. If $\mathcal{C} \to \mathcal{D}$ is a functor, the resulting map is monotonic. So we get the desired functor $\mathbf{Cat} \to \mathbf{PreOrd}$.