What do we call an exponential object in a concrete category that basically equals the one in Set So suppose you've got a concrete category $\mathcal C$ and two objects $X, Y$ of $\mathcal C$. For instance, $\mathcal C$ could be the category of topological spaces and we could also suppose that $Y$ is locally compact, so that $X^Y$, if endowed with the compact-open topology, is an exponential object of $X$ to the power of $Y$.
Now in this scenario, the space $X^Y$ is nothing but a subset of the set $X^Y$ (that is, the continuous functions from $Y$ to $X$) with a topology on it, so that the underlying set happens to be a subset of the exponential object in the category $\mathbf{Set}$. Moreover, all the functions associated to this exponential object are merely restrictions of those given in the category $\mathbf{Set}$.
Now my question is:
Is there a special name for an exponential object of $X$ to the power of $Y$ which, when projected to $\mathbf{Set}$, is nothing but a subset of the set $X^Y$, together with the canonical evaluation map and the canonical assignment of the $\lambda$ function as explained in the Wikipedia article "Exponential object"?
 A: The example of $\text{Top}$ has the additional property that the underlying set functor is represented by the terminal object $1$; in other words, it is the functor $\text{Hom}(1, -)$ of global points. This is not true of many concrete categories, but when it is true, and assuming as usual that "concrete category" means the underlying set functor is faithful, applying $\text{Hom}(1, -)$ to an evaluation homomorphism
$$X \times Y^X \to Y$$
produces the concrete evaluation homomorphism
$$\text{Hom}(1, X) \times \text{Hom}(X, Y) \to \text{Hom}(1, Y)$$
given by the action of $\text{Hom}(X, Y)$ on global points. Currying this evaluation homomorphism gives a map $\text{Hom}(X, Y) \to \text{Hom}(1, Y)^{\text{Hom}(1, X)}$ and you want to know when this map is an injection. This is guaranteed by the assumption that $\text{Hom}(1, -)$ is faithful, which says exactly that a morphism is determined by its action on global points.
So the key condition is that the underlying set functor is (faithful and) the global points functor; we might call such a category globally concrete (I don't know if it has a standard name).
A: Yes. If the underlying set of your exponential object is just an exponential in $\mathsf{Set}$ and the evaluation maps get sent to evaluation maps in $\mathsf{Set}$, then we call the category Concretely Cartesian Closed (over $\mathsf{Set}$, if we want to be precise).
More generally, from The Joy of Cats (Chapter 27, Definition 11):

A concrete category $(\mathbb{A}, U)$ over $\mathbb{X}$ is called Concretely Cartesian Closed provided that the following hold:

*

*$\mathbb{A}$ and $\mathbb{X}$ are cartesian closed.

*$U$ preserves finite products, power objects, and evaluation; in particular, whenever $A \times B^A \overset{\text{ev}}{\longrightarrow} B$ is an evaluation in $\mathbb{A}$, then
$$U \left ( A \times B^A \overset{\text{ev}}{\longrightarrow} B \right ) = UA \times UB^{UA} \overset{\text{ev}}{\longrightarrow} UB$$
is an evaluation in $\mathbb{X}$.



Edit:
As commenters have noted, this definition is too restrictive. In fact, it excludes the example that you gave.
The authors of Joy of Cats actually mention that every concretely cartesian closed category over $\mathsf{Set}$ with a discrete terminal object is, up to concrete isomorphism, a full subcategory of $\mathsf{Set}$. So this is a very restrictive definition.
They give another definition (Chapter 27, Definition 17), which tries to
be more accommodating:

A construct $(\mathbb{A}, U)$ is said to have function spaces provided the following hold:

*

*$(\mathbb{A}, U)$ has finite concrete products

*$\mathbb{A}$ is cartesian closed and the evaluation morphisms $A \times B^A \overset{\text{ev}}{\longrightarrow} B$ can be chosen in such a way that $U \left ( B^A \right ) = \text{hom}_\mathbb{A}(A,B)$ and $\text{ev}$ is the restriction of the canonical evaluation map in $\mathsf{Set}$

This is still too restrictive for your example, though, since it requires all exponentials to exist. However, if we restrict attention to a "nice" subcategory of $\mathsf{Top}$ (for instance, compact hausdorff spaces) then we do have function spaces, and they agree with your example.
I haven't seen a definition which allows for only some exponential objects, but based on the language of "has function spaces", I think calling such an exponential a function space when it exists would be reasonable.

I hope this helps ^_^
