Let $\mathcal{C}$ be a category with finite products and $A,B$ be objects in $\mathcal{C}$ . An object $E$ in $\mathcal{C}$ is called an exponential object from $A$ to $B$ if it satisfies the following conditions:

• there is a morphism $f:E\times A\to B$ , called an evaluation morphism
• for any morphism $g:C\times A\to B$ , there is a unique morphism $h:C\to E$ such that $f\circ (h\times 1_A)=g$ , where $h\times 1_A:C\times A\to E\times A$ is the product morphism of $h$ and the identity morphism on $A$ .

For example, in the category of sets, ${Set}$ , where products clearly exist (empty set is a set!) between pairs of objects (sets), the exponential from $A$ to $B$ is the set $B^A$ , which is defined as the set of all functions from $A$ to $B$ . The evaluation morphism is the function $ev: B^A\times A\to B$ given by $ev(f,a)=f(a)$ , where $f\in B^A$ and $a\in A$ . If $g:C\times A\to B$ is any function, then we define $h:C\to B^A$ by $h(c)(a)=g(c,a)$ . Then $ev\circ (h\times 1_A)(c,a)=ev(h(c),a)=h(c)(a)=g(c,a)$ , and $ev$ is universal (in the sense of the second condition above).

Since each $h$ is uniquely determined by $g$ in the above definition, and conversely every $h$ determines a $g$ by the formula $g=f\circ (h\times 1_A)$ , we have a bijection $$\hom(C\times A,B)\cong \hom(C,B^A).$$

If an exponential object exists between every pair of objects in category $C$ with finite products, then we say that $C$ has exponentials. According to the bijection above, we see that the functor $\cdot\times A:\mathcal{C}\to \mathcal{C}$ has a right adjoint, namely $\cdot ^A:\mathcal{C}\to\mathcal{C}$ , called the exponential functor.

It can be seen that a category $C$ with finite products has exponentials iff the covariant function $\cdot\times A:\mathcal{C}\to\mathcal{C}$ has a right adjoint for every object $A$ in $\mathcal{C}$ .