Prove that the composition of two bijective functions is bijective. Solution A f

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A function is bijective if it is both injective and surjective. A bijective function is a bijection (one-to-one correspondence). A function is bijective if and only if every possible image is mapped to by exactly one argument. This equivalent condition is formally expressed as follows. The function f: A -> B is bijective iff for all b -> B, there is a unique a->A, such that f(a) = b. A function f : A ? B is bijective if and only if it is invertible, that is, there is a function g: B ? A such that g o f = identity function on A and f o g = identity function on B. This function maps each image to its unique preimage. The composition of two bijections is again a bijection, but if g o f is a bijection, then it can only be concluded that f is injective and g is surjective. (See the figure at right and the remarks above regarding injections and surjections.) The bijections from a set to itself form a group under composition, called the symmetric group.

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