Solution:
Stage 1: Check for Injectivity:
Let (a1, b1) and (a2, b2) ∈ A x B with the end goal that f (a1, b1) = (a2, b2)
This suggests, (b1 , a1) and (b2, a2 ) b1 = b2 and a1 = a2
(a1, b1) = (a2, b2) for all (a1, b1) and (a2, b2) ∈ A x B Therefore, f is injective.
Stage 2: Check for Surjectivity:
Let (b, a) be any component of B x A. Then, at that point a ∈ An and b ∈ B This infers (a, b) ∈ A x B
For all (b, a) ∈ B x A, their exists (a, b) ∈ A x B Therefore, f: A x B – > B x An is bijective capacity.