question_answer 24)

Let A an B be sets. Show that  such that f(a, b) = (b, a) is bijective functrion.       

Answer:

Let x1 = (a, b) and x2 = (a?, b?)           f(x1) = f(a, b) = (b, a)       f(x2) = f(a?, b?) = (b? a?)       Now f(x1) = f(x2)  (b, a) = (b?, a?)        b = b? and a = a?                   x1 = (a, b) = (a?, b?) = x­2              f(x1) = f(x2)              is one-one.       Let y = (a?, b?)  (co-domain of f)       Also let x = (a, b)  be its pre-image of f              (b, a) = (a?, b?)        b = a?, a = b?        x = (a, b) = (b?, a?)        f(x) = f(b?, a?) = (a?, b?) = y        Corresponding to every ordered pair (a?, b?) there exists (a, b) such that             f(a, b) = (a?, b?)       is onto.       Hence f is bijective.