Let A and B be sets then the function f: A×B →B×A such that (a, b) = (b, a)  is-

one-one and onto both

It is given f: A×B →B×A such that (a, b) = (b, a) 

Let (a₁, b₁), (a₂, b₂) ∈ A×B such that f(a₁, b₁) = f(a₂, b₂)

⇒ (b₁, a₁) =  (b₂, a₂)

⇒  b₁ = b₂ and a₁ =a₂

⇒  (a₁, b₁) = (a₂, b₂)

so, f is one-one.

Now let (b, a) ∈ B×A be any element.

Then, there exist (a, b) ∈ A×B such that f(a, b) = (b, a).

so f is onto.

Hence f is one- one and onto both.