Answer :

Given A = {1, 2, 3, 4} and B = {a, b, c, d}.

We need to define bijections f_{1}, f_{2}, f_{3} and f_{4} from A to B.

Consider f_{1} = {(1, a), (2, b), (3, c), (4, d)}

(1) f_{1} is one-one because no two elements of the domain are mapped to the same element.

f_{1} is also onto because each element in the co-domain has a pre-image in the domain.

Thus, f_{1} is a bijection from A to B.

We have f_{1}^{-1} = {(a, 1), (b, 2), (c, 3), (d, 4)}

Using similar explanation, we also have the following bijections defined from A to B -

(2) f_{2} = {(1, b), (2, c), (3, d), (4, a)}

We have f_{2}^{-1} = {(b, 1), (c, 2), (d, 3), (a, 4)}

(3) f_{3} = {(1, c), (2, d), (3, a), (4, b)}

We have f_{3}^{-1} = {(c, 1), (d, 2), (a, 3), (b, 4)}

(4) f_{4} = {(1, d), (2, a), (3, b), (4, c)}

We have f_{4}^{-1} = {(d, 1), (a, 2), (b, 3), (c, 4)}

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