Cartesian Product of Bijections is Bijection

Theorem
Let $S_1 \times S_2$ be the Cartesian product of two sets $S_1$ and $S_2$.

Let $T_1 \times T_2$ be the Cartesian product of two sets $T_1$ and $T_2$.

Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be bijections.

Let $f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$ be the Cartesian product of $f_1$ and $f_2$ defined as:
 * $\forall \left({s_1, s_2}\right) \in S_1 \times S_2: f_1 \times f_2 \left({s_1, s_2}\right) := \left({f_1 \left({s_1}\right), f_2 \left({s_2}\right)}\right)$

Then $f_1 \times f_2$ is a bijection.

Proof
Because $f_1$ and $f_2$ are both bijections, it follows by definition that they are both surjections.

Let $\left({t_1, t_2}\right) \in T_1 \times T_2$.

Then:
 * as $f_1$ is a surjection, $\exists s_1 \in S_1: f_1 \left({s_1}\right) = t_1$
 * as $f_2$ is a surjection, $\exists s_2 \in S_2: f_2 \left({s_2}\right) = t_2$

Thus:
 * $\exists \left({s_1, s_2}\right) \in S_1 \times S_2: f_1 \times f_2 \left({s_1, s_2}\right) = \left({t_1, t_2}\right)$

So $f_1 \times f_2$ is a surjection.

Because $f_1$ and $f_2$ are both bijections, it follows by definition that they are both injections.

Let:
 * $f_1 \times f_2 \left({a_1, a_2}\right) = \left({c_1, c_2}\right), f_1 \times f_2 \left({b_1, b_2}\right) = \left({d_1, d_2}\right)$

for some $\left({a_1, a_2}\right), \left({b_1, b_2}\right) \in S_1 \times S_2$.

Suppose $\left({c_1, c_2}\right) = \left({d_1, d_2}\right)$.

Then by Equality of Ordered Pairs:
 * $c_1 = d_1$
 * $c_2 = d_2$

By definition of $f_1 \times f_2$:
 * $f_1 \left({a_1}\right) = c_1$
 * $f_1 \left({b_1}\right) = d_1$

and:
 * $f_2 \left({a_2}\right) = c_2$
 * $f_2 \left({b_2}\right) = d_2$

As $f_1$ is an injection:
 * $c_1 = d_1 \implies a_1 = b_1$

and as $f_2$ is an injection:
 * $c_2 = d_2 \implies a_2 = b_2$

Thus it follows that:
 * $\left({c_1, c_2}\right) = \left({d_1, d_2}\right) \implies \left({a_1, a_2}\right) = \left({b_1, b_2}\right)$

and so $f_1 \times f_2$ is an injection.

So $f_1 \times f_2$ is a surjection and also an injection.

Hence by definition, $f_1 \times f_2$ is a bijection.