Universal Property: do people study terminal objects in $(X\downarrow U)$?

Suppose that $U:D\to C$ is a functor and $X$ is an object of $C$. In defining Universal property, Wikipedia writes about terminal objects in $(U\downarrow X)$ and initial objects in the category $(X\downarrow U)$. The images below appear to illustrate the idea.

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I'm wondering why I've never encountered discussion of terminal objects in $(X\downarrow U)$ or initial objects in $(U\downarrow X)$? Are these objects simply less interesting? Are fewer examples available for study?

• Note that, if D has a terminal object $e_D$ and U preserves terminal objects, then the unique morphism $X \to U(e_D)$ is terminal in $(X\downarrow U)$.
– user314
Nov 5, 2014 at 11:26

The two 'universal properties' Wikipedia talks about are in fact dual: terminal morphisms from $U$ to $X$ are exactly the initial morphisms from $X^\mathrm{op}$ to $U$. Terminal objects in $(X \downarrow U)$ on the other hand are a very different thing, and there is no reason for them to be interesting just because the initial objects are. The initial morphism is what other morphisms extend (dually: lift) through, the terminal morphism is what extends through other morphisms, which is something that can hold for trivial reason (cf. Adeel's comment).
To put it in less vague terms, the crucial property of initial morphisms from $X$ to $U$ is that they are equivalent to the representability of $\hom(X, U-)$ (and initial morphisms in general are in fact equivalent to the very notion of representability: functor $F : C → \mathrm{Set}$ is representable iff there is an initial morphism from the singleton to $F$). For terminal objects of $(X \downarrow U)$ nothing similar is true.
• For a very concrete example of a lifting problem (in which the lift terminology probably predates category theory) take a path $s : I → M$ in a smooth manifold, and a fiber bundle $π : E → M$. A lift of $s$ to $E$ is then a section of that bundle along the path $s$. In particular, if $E$ is the tangent bundle $TM$, then the lifts of $s$ are exactly the vector fields along $s$. Mar 14, 2015 at 1:03