You are here
Homeexponential object
Primary tabs
exponential object
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 $A$.
The two conditions can be summarized by the following commutative diagram:
$\xymatrix@R=20pt{E\times A\ar[dr]^{f}\\ &B\\ C\times A\ar[ur]_{g}\ar[uu]^{{h\times 1_{A}}}}$
where $h$ is uniquely determined by $g$. It is easy to see that any two exponential objects from $A$ to $B$ are isomorphic, hence the existence of an exponential objects between two objects is a universal property. We may write $B^{A}(\cong E$ above) the exponential object from $A$ to $B$.
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 $\mathcal{C}$.
Mathematics Subject Classification
18D15 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections