Suppose A and B are finite sets with |A| = |B| and that f: A B is a function. Prove: f is one-to-one iff f is onto.

Proof that f is onto: Suppose f is injective and f is not onto. Since |A| = |B| every can be paired with exactly one . Now since f is injective, if , then . If f is not onto then there is a for which there is no such that . But this is a contradiction since this would mean that at least two unequal would map to the same . Therefore f must be onto if f is injective.

Proof that f is injective: Suppose f is onto and f is not injective. Then there exists and such that . Now again |A| = |B| means that every can be paired with exactly one . If there exists an such that and then there must exist a for which there is no such that which is a contradiction. This means that f is injective if f is onto. QED

Would someone be kind enough to critique this proof? Thanks.