Suppose that f is a function from A to B. where A and B are finite sets with |A| = |B|. Show that f is one-to-one if and only if it is onto.

Solution:Step 1:The objective of this question is to show that f is one-to-one if and only if it is onto.Step 2:Assume A and B are finite set with = and that : f: A Give us a chance to consider f is injective and f is not onto. Since = each ai Can be put with precisely one bi . Presently since f is injective, if f(ai) = f(aj) = bi. At that point ai = aj: if f is not onto then there is abifor which there is no ai such that f(ai) =bi however this is a disagreement since this would imply that no less than two inequal ai , aj would guide to a similar bi .therefore f must be onto if f is injective.