Corollary 4

Let E and F be two finite sets with the same number of elements, and let $f$ be a mapping of E into F. Then the following statements are equivalent :

(a) $f$ is an injection; (b) $f$ is a surjection; (c) $f$ is a bijection.

It is enough to prove that (a) and (b) are equivalent. If $f$ is injective, then $\mathrm{Card}(f(\mathrm{E})) = \mathrm{Card}\,(\mathrm{E}) = \mathrm{Card}\,(\mathrm{F})$, whence $f(\mathrm{E}) = \mathrm{F}$ (Corollary 2). If $f$ is not injective, let $x$ and $x'$ be two elements of E such that $x \neq x'$ and $f(x) = f(x')$. Then, putting $\mathrm{E}' = \mathrm{E} - \{x\}$, we have $f(\mathrm{E}') = f(\mathrm{E})$, whence $\mathrm{Card}(f(\mathrm{E})) \leqslant \mathrm{Card}\,(\mathrm{E}') < \mathrm{Card}\,(\mathrm{E})$ by virtue of Corollary 2; but since $\mathrm{Card}\,(\mathrm{F}) = \mathrm{Card}\,(\mathrm{E})$, it follows that $f(\mathrm{E}) \neq \mathrm{F}$.