Define exponential object without reference to cartesian product?

Question

In category theory, the exponential object $X^Y$ is defined, roughly, by saying that if we take some other object $Z$ and the product between $Z$ and $X$, then any morphism from that product to $Y$ has an equivalent morphism that “goes through” $X^Y$.

However, what I find weird about this construction (note that category theory in general is weird to me), is that we need a third object $Z$ and the assumption that a cartesian product exists between them, to define the exponential object $X^Y$.

Is it not possible to define it more directly somehow? Intuitely it seems like we don’t need to talk about $Z$ or about cartesian products, to talk about functions from $X$ to $Y$ (I know that category theory is not only about functions, but this is how I intuitively think about it).

Answer 1

First note that the idea of a universal property in category theory is virtually always defined in terms of some other objects, so the idea of adding $Z$ in the definition isn't special for exponentials:

• $T$ is a terminal object in a category if, for any other object $X$, there is a unique morphism $X \to T$.

• $C$ is the product of $A$ and $B$ if there are maps $C \to A$ and $C \to B$ and for any other object $D$ with maps $f: D \to A$ and $g: D \to B$, there is a unique morphism $D \to C$ which $f,g$ factor through.

The point is that to define an object, you can only refer to other objects in the category - so this is really the most direct way also to define the exponential object.

In terms of requiring binary products to exist, this may seem odd at first glance, but seemingly turns out to be the most sensible definition. Personally, I think of the exponential object $X^Y$ as the most general object (in some loose sense) for which giving a $Y$ combines with it to give an $X$.

Answer 2

The closest thing you can do is to define a closed category. This is one for which there exist functorial choices of exponentials together with a unit object $I$ equipped with a map to every endomorphism object $X^X$ which is neutral for a given associative family of composition maps. Everything has to be said in curried form: for instance the composition map goes $X^Y\to (Z^X)^{Y^Z}$. If you ask your exponentiation functors to be right adjoints, then you have a monoidal closed category, not in general a Cartesian closed one. There is a proposal on the nLab page for closed categories that there should be a characterization of the internal Hom structures whose left adjoints, if they exist, are Cartesian monoidal in terms of the existence of constant-morphism maps $X\to X^Y$ and another family corresponding to diagonals.

Closed categories are used relatively infrequently, in part for psychological reasons: contravariant homs create extra natural transformations, and people find these annoying. There is also the practical reason that closed categories which aren't monoidal are rather rare, and the theoretical reason that one can embed a closed category in a closed monoidal one.