# What is an exponential object?

One interpretation for a power of natural numbers, such as $10^6$, is that it is counting a certain collection of functions: in this case, the number of functions from a six-element set to a ten-element set. (One way to think about it is that $10^6$ is the number of six-digit combinations from 000000 to 999999, and each combination corresponds to a function from $\{1,\ldots,6\}$ to $\{0,\dots,9\}$: the first digit is the image of $1$, the second digit is the image of $2$, etc.)

In general, the number $m^n$ is the number of functions from an $n$-element set to a $m$-element set. This is somewhat analogous to standard interpretations of other arithmetical operations:

• $m\times n$ is the number of elements in the Cartesian product of an $m$-element set and an $n$-element set.
• $m + n$ is the number of elements in the disjoint union of an $m$-element set and an $n$-element set.

So $\times$ and $+$ are arithmetical shadows of the categorical operations “Cartesian product” and “disjoint union” on sets. Is exponentiation the shadow of a categorical operation sending a pair of sets $S$ and $T$ to the set of functions from $S$ to $T$? The answer is yes: this set of functions is an example of an exponential object in the category of sets, and is written $T^S$. This is one way of interpreting similarly notated sets like $\mathbb{R}^3$: as the set of functions from a 3-element set (such as $\{1,2,3\}$) to $\mathbb{R}$. We could instead let $S$ be any set, and consider $\mathbb{R}^S$ as the set of functions $S\to \mathbb{R}$, i.e. the set of $S$-indexed families of elements of $\mathbb{R}$.

So what are the categorical properties of this exponential object? Well, $\mathbb{R}^3$ comes with three maps $\mathbb{R}^3\to \mathbb{R}$: projection onto each coordinate. In other words, there’s a map

$\{1,2,3\}\times \mathbb{R}^3 \to \mathbb{R}$

that takes a number $i$ and a point $p\in \mathbb{R}^3$ and sends it to the $i$th coordinate of $p$. More generally, there’s a canonical evaluation map

$\text{eval}: S\times T^S \to T$

that sends an element $s\in S$ and a function $f:S\to T$ to the element $f(s)\in T$. In fact, $T^S$ is universal among sets $A$ equipped with a function $S\times A\to T$: given such a function, each element $a\in A$ gives us a function $S\to T$ obtained from $S\times A\to T$ by holding the second input constant at $a$, so we get a function $A\to T^S$.

This universal property is the categorical definition of an exponential object: if $\mathcal{C}$ is any category with finite products (“$\times$“), then for any two objects $S,T$ in $\mathcal{C}$, their exponential object $T^S$ is an object with the property that morphisms $A\to T^S$ are in natural one-to-one correspondence with morphisms $S\times A\to T$.

Here are some examples of exponential objects:

1. The ordered set $\{0,1\}$, regarded as a category with the $\leq$ relation as its morphisms, is a very simple category with nonetheless interesting exponential objects.

The product operation in this category, as in any total order, is the “min” operation denoted $\wedge$, so that $0\wedge 0 = 0$, $0\wedge 1 = 0$, and $1\wedge 1 = 1$. (If we interpret $0$ and $1$ as the truth values False and True, then this categorical product is the “and” operation. “Max” is the categorical coproduct, denoted by $\vee$.)

Then given two truth values $a$ and $b$, the exponential $b^a$ is the largest element such that $a\wedge b^a \leq b$. You can check that this works out to: $b^a = 0$ if $a=1$ and $b=0$, and $b^a = 1$ otherwise. In other words, the exponential object $b^a$ is the implication $a\Rightarrow b$!

2. More generally, if $\mathcal{B}$ is any Boolean algebra, regarded as a category via its poset structure, the exponential object $b^a$ for two elements $a,b\in B$ is the implication $\neg a \vee b$.
3. If $G$ is a group, the category of $G$-sets has exponential objects: if $X$ and $Y$ are two $G$-sets, then the exponential object $Y^X$ is the set of all functions $X\to Y$ (whether or not they respect the $G$-action!) and the $G$-action on $Y^X$ is given by conjugation: the action of $g\in G$ on $f:X\to Y$ is the function $X\to Y: x\to g\cdot f(g^{-1}\cdot x)$.
4. If $\mathcal{C}$ and $\mathcal{D}$ are categories, i.e. objects in the category of categories, then the exponential object $\mathcal{D}^\mathcal{C}$ is the category whose objects are functors $\mathcal{C}\to \mathcal{D}$ and whose morphisms are natural transformations between them.

Do exponential objects always exist?

No: in fact, it’s rather common for exponential objects not to exist. If a category has an exponential object $T^S$ for every pair of objects $S$ and $T$, then one way of interpreting the correspondence

“morphisms $S\times A\to T$ correspond to morphisms $A\to T^S$

is as an adjunction between the functors $S\times (-)$ and $(-)^S$; as such, the functor $S\times (-)$ automatically must preserve all colimits. In the category of sets, this is just a souped-up version of the statement that cartesian products distribute over disjoint unions:

$S\times(A\coprod B) \cong (S\times A) \coprod (S\times B)$,

In other categories, however, this can fail. For example, in the category of abelian groups, the categorical product and coproduct are both direct sums, and we definitely don’t have a general isomorphism

$S\oplus (A\oplus B) \cong (S\oplus A) \oplus (S\oplus B)$.

But in some ways this is the only problem: often if a category has enough colimits and a functor (such as $S\times (-)$) preserves them, that functor automatically has a right adjoint (which we could call $(-)^S$). Such a category would not only have binary products $\times$ and a terminal object $1$, it would also have binary coproducts $+$ and an initial object $0$, and if $S\times (-)$ preserves colimits we must have

$S\times (A+B) = (S\times A) + (S\times B)$
and $S\times 0 = 0$.

In such a category, it’s possible to prove (with the Yoneda lemma, for example) that for all objects $A$, $B$, and $C$:

1. $A^B\times A^C \cong A^{B+C}$ and $A^0 \cong 1$.
2. $(A^B)^C \cong A^{B\times C}$ and $A^1 \cong A$.
3. $A^C\times B^C \cong (A\times B)^C$ and $1^C \cong 1$.

These are exactly the laws of exponentiation for natural numbers, promoted to categorical isomorphisms, which we deduced in the first place from the interpretation of $m^n$ as the number of functions from an $n$-element set to an $m$-element set. But now we know they hold in a general category with finite products, finite coproducts, and exponential objects—for example, if $P$, $Q$, and $R$ are abstract propositions in the Boolean algebra of propositional logic, where product is “and”, coproduct is “or”, and exponentiation is “implies”, they read:

1. $P$ implies $R$ and $Q$ implies $R$ if and only if ($P$ or $Q$) implies $R$. Also, a false statement implies anything.
2. $P$ implies ($Q$ implies $R$) if and only if ($P$ and $Q$) implies $R$. Also, a proposition holds if and only if it is implied by a true statement.
3. ($P$ implies $Q$) and ($P$ implies $R$) if and only if $P$ implies ($Q$ and $R$). Also, everything implies a true statement.

Amazing, huh?