Theorem 1   Let $A$ and $B$ be finite sets of the same cardinality. If $f\colon A \to B$ is a surjection then $f$ is a bijection.

Proof. Let $A$ and $B$ be finite sets with $|A| = |B| = n$ Let $C =\{f^{-1}\left(\{b\}\right)\mid b \in B \}$ Then $\bigcup C \subseteq A$ so $|\bigcup C| \le n$ Since $f$ is a surjection, $|f^{-1}\left(\{b\}\right)| \ge 1$ for each $b \in B$ The sets in $C$ are pairwise disjoint because $f$ is a function; therefore, $n \le |\bigcup C|$ and \begin{equation*} \\ \left|\bigcup C \right|=\sum_{b\in B}|f^{-1}\left(\{b\}\right)| .\end{equation*}In the last equation, $n$ has been expressed as the sum of $n$ positive integers; thus $|f^{-1}\left(\{b\}\right)| = 1$ for each $b \in B$ so $f$ is injective.