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Let , where and are finite sets with
Let us show that is one-to-one if and only if it is onto.
If is one-to-one, then for any and hence the set has the same number of different elements as . It follows that Since and is finite, we conclude that and hence is onto.
If is onto, then the preimage for any Let us prove using the method by contradiction. Suppose that is not one-to-one. Then for some Then Since and we conclude that for some We get that and so we have the contradiction with This contraction proves that is one-to-one.