How is the cartesian product $E^F$ defined?

Question

For arbitrary sets $E, F$, how is $E^F$ defined? It seems to be the set of all maps from $F$ to $E$, i.e. $E^F := \{\phi : \phi\colon F \rightarrow E\}$? Is that right?

Answer 1

Yes; however, this is not the Cartesian product. The Cartesian product of two sets, $A$ and $B$ - denoted "$A\times B$" - is the set of pairs $(a, b)$ with $a\in A$ and $b\in B$.

$E^F$ can be thought of as a (possibly infinitary) Cartesian product - the Cartesian power: $E\times E\times E\times . . . $ with "$F$-many" factors of $E$. This takes some thought to make precise, though.

Answer 2

The short answer is basically "yes", but I consider this to be a construction, not a definition. I would also consider the description of $A\times B$ as the set of ordered pairs $\{(a,b)\mid a\in A,\ b\in B\}$ to be a construction, not a definition. The reason for this is that these constructions do not always give the "correct" notion of product when applied to objects more complicated than sets.

If $\{A_i\}_{i\in I}$ is a collection of sets indexed by some (possibly infinite) set $I$, the Cartesian product $$\prod_{i\in I} A_i$$ is the (necessarily unique up to unique isomorphism) pair $(X,\{\pi_i\}_{i\in I})$ consisting of a set $X$ and a family of functions $\pi_i:X\to A_i$ satisfying the following universal property: for any other pair $(Y, \{\sigma_i\}_{i\in I})$ of the same type, there is a unique function $h:Y\to X$ such that $\pi_i\circ h = \sigma_i$ for all $i\in I$.

You can check that the set of functions $X = \{\phi \mid \phi:F\to E\}$ together with the functions $\{\pi_f\}_{f\in F}$ defined by $\pi_f(\phi) = \phi(f)$ satisfies this universal property for the collection $\{E_f\}_{f\in F}$ where each $E_f$ is a copy of $E$.