Proof that f is onto: Suppose f is injective and f is not onto. Since |A| = |B| every \(\displaystyle a_{i}\in A\) can be paired with exactly one \(\displaystyle b_{i}\in B\). Now since f is injective, if \(\displaystyle f(a_{i})=f(a_{j})=b_{i}\), then \(\displaystyle a_{i}=a_{j}\). If f is not onto then there is a \(\displaystyle b_{i}\in B\) for which there is no \(\displaystyle a_{i}\in A\) such that \(\displaystyle f(a_{i})=b_{i}\). But this is a contradiction since this would mean that at least two unequal \(\displaystyle a_{i},a_{j}\in A\) would map to the same \(\displaystyle b_{i}\in B\). 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 \(\displaystyle a_{i},a_{j}\in A\) and \(\displaystyle b_{i}\in B\) such that \(\displaystyle f(a_{i})=f(a_{j})=b_{i}where a_{i}\neq a_{j}\). Now again |A| = |B| means that every \(\displaystyle a_{i}\) can be paired with exactly one \(\displaystyle b_{i}\). If there exists an \(\displaystyle a_{i}\neq a_{j}\) such that \(\displaystyle f(a_{i})=f(a_{j})=b_{i}\) and \(\displaystyle i\neq j\) then there must exist a \(\displaystyle b_{j}\) for which there is no \(\displaystyle a_{j}\) such that \(\displaystyle f(a_{j})=b_{j}\) 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.