a surjection between finite sets of the same cardinality is bijective

Theorem.

Let $A$ and $B$ be finite sets of the same cardinality. If $f\mathrm{:}A\mathrm{\to }B$ 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

$$\left|\bigcup C\right|=\sum _{b\in B}|f^{-1}\left(\{b\}\right)|.$$

In the last equation, $n$ has been expressed asthe sum of $n$ positive integers; thus $|f^{-1}\left(\{b\}\right)|=1$ for each $b\in B$, so $f$ is injective.∎