Let, A and B be sets. Show that $f:A \times B \to B \times A$ such that $f(a,\;b) = (b,\;a)$ is bijective function.

Injectivity

Let $({a_1},\;{b_1})\;\;and\;\;({a_2},\;{b_2}) \in A \times B$ such that,
$f({a_1},\;{b_1}) = f({a_2},\;{b_2})$
$\Rightarrow$ $({b_1},\;{a_1}) = ({b_2},\;{a_2}) \Rightarrow {b_1} = {b_2}\;\;and\;\;{a_1} = {a_2}$
$\Rightarrow$ $({a_1},\;{b_1}) = ({a_2},\;{b_2})$

Thus, $f({a_1},\;{b_1}) = f({a_2},\;{b_2}) \Rightarrow ({a_1},\;{b_1}) = ({a_2},\;{b_2})$ [For all $({a_1},\;{b_1}),\;({a_2},\;{b_2}) \in A \times B]$
So, f is an injective function.

Surjectivity

Let (b, a) be an arbitrary element of $B \times A$, where, $b \in B$ and a$\in$A $\Rightarrow$ (a, b)$\in$ $A \times B$
Thus, for all (b, a)$\in$B x A, their exists (a, b) $\in$ (A x B) such that, $f(a,\;b) = (b,\;a)$
So, $f:A \times B \to B \times A$ is an onto function.

Hence, f is a bijective function.