Let $A$ and $B$ be sets. Show that $f: A \times B \rightarrow B \times A$ defined by $f(a, b) = (b, a)$ is a bijective function.

(N/A) The function is defined as $f: A \times B \rightarrow B \times A$ where $f(a, b) = (b, a)$.
$1.$ To show $f$ is one-one (injective):
Let $(a_1, b_1), (a_2, b_2) \in A \times B$ such that $f(a_1, b_1) = f(a_2, b_2)$.
This implies $(b_1, a_1) = (b_2, a_2)$.
Equating the components,we get $b_1 = b_2$ and $a_1 = a_2$.
Therefore,$(a_1, b_1) = (a_2, b_2)$.
Since $f(a_1, b_1) = f(a_2, b_2) \Rightarrow (a_1, b_1) = (a_2, b_2)$,$f$ is one-one.
$2.$ To show $f$ is onto (surjective):
Let $(b, a) \in B \times A$ be any arbitrary element.
By the definition of the Cartesian product,$b \in B$ and $a \in A$,which implies $(a, b) \in A \times B$.
For this $(a, b) \in A \times B$,we have $f(a, b) = (b, a)$.
Since every element in the codomain $B \times A$ has a pre-image in the domain $A \times B$,$f$ is onto.
Since $f$ is both one-one and onto,$f$ is a bijective function.