Set of All Mappings of Cartesian Product

Theorem
Let $R$, $S$, and $T$ be sets.

Then:
 * $R^{S \times T} \sim \left({ R^S }\right)^T$

Proof
Define the mapping $F : \left({ R^S }\right)^T \to R^{S \times T}$ as follows:


 * $F\left({ f }\right) \left({ x,y }\right) = \left({f\left({x}\right)}\right) \left({y}\right)$ for all $x \in S, y \in T$.

Suppose $F\left({f_1}\right) = F\left({f_2}\right)$.

Then $\left({f_1\left({x}\right)}\right)\left({y}\right) = \left({f_2\left({x}\right)}\right)\left({y}\right)$ for all $x \in S, y \in T$ by the definition of $F$.

Therefore, $f_1\left({x}\right) = f_2\left({x}\right)$ and $f_1 = f_2$ by Equality of Mappings.

It follows that $F$ is an injection.

Take any $g \in R^{S \times T}$.

Define a function $f$ as $\left({f\left({x}\right)}\right)\left({y}\right) = g\left({x,y}\right)$.

It follows that:

Therefore, $F$ is a surjection.

Thus, $F$ is a bijection.

It follows that $ R^{S \times T} \sim \left({ R^S }\right)^T$ by the definition of set equivalence.