Building a function $p : \mathcal{D} \rightarrow \mathbf{Ord}$ from a faithful functor $U : \mathcal{C} \rightarrow \mathcal{D}.$

Question

For simplicity, I will ignore size issues in this question.

Let $\mathcal{D}$ and $\mathcal{O}$ denote categories. By a function $\mathcal{D} \rightarrow \mathcal{O},$ let us mean a functor from the core of $\mathcal{D}$ into the core of $\mathcal{O}$. Now let $\mathcal{C}$ and $\mathcal{D}$ denote categories, and suppose we're given a faithful functor $U : \mathcal{C} \rightarrow \mathcal{D}$. Then we get a function $p : \mathcal{D} \rightarrow \mathbf{Ord}$ as follows. (By $\mathbf{Ord}$, I mean the category of preordered sets).

Fix $D,D' \in \mathbf{D}$. Then:

1. As a set, define $p(D) = \{(X,i) \mid X \in \mathrm{Obj}(\mathbf{C}),\; i : \mathrm{Iso}(UX,D)\}.$
2. $(X,i) \leq (Y,j)$ in $p(D)$ iff there exists a morphism $f : X \rightarrow Y$ with $i = j \circ Uf$.
3. Given an isomorphism $f : D \rightarrow D',$ define $p(f)$ to be the mapping $(X,i) \in D \mapsto (X,f \circ i) \in D'.$

Examples.

Let $\mathcal{D}$ denote the category of sets, and consider a set $D$.

1. If $\mathcal{C}$ is the category of graphs, then $p(D)$ is equivalent to the poset of all graphs with underlying set $D$, ordered in the usual way.

2. If $\mathcal{C}$ is the category of topological spaces, then $p(D)$ is equivalent to the poset of all topologies on $D$, ordered in reverse to the usual way.

Questions.

1. Does this construction have a name?
2. Where can I learn more?