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Exercise 1 Let A and B be finite sets. Let f : A - B be an injective (also called 1-to-1) function from the elements of A to the elements of B-recall that this means each element a E A is mapped by f to a unique element b E B, so that a, * a, implies f(a,) # f(a2). Similarly, let g : B - A be an injective function from the elements of B to the elements of A. Prove that f and g must also be bijections. A bijection f : A - B is injective, and additionally surjective (also called onto): for each b E B there is a unique a E A where f(a) = b. (A similar statement holds for bijective g : B - A-work it out. ..) Thus f and g "pair off" the elements of A and B where each of the elements in A and B appears in exactly one pair; as a consequence, the two finite sets have the same size. Prove that f and g are both bijections without using "baby facts" about integers such as I S y and y