Equivalence of Mappings between Finite Sets of Same Cardinality

Theorem
Let $S$ and $T$ be finite sets such that $\left|{S}\right| = \left|{T}\right|$.

Let $f: S \to T$ be a mapping.

Then the following statements are equivalent:


 * $(1): \quad f$ is bijective
 * $(2): \quad f$ is injective
 * $(3): \quad f$ is surjective.

Proof

 * $(2)$ implies $(3)$ by Proper Subset has Fewer Elements:

If $f$ is injective, then $\left|{S}\right| = \left|{f \left({S}\right)}\right|$ from Cardinality of Image of Injection.

Therefore the subset $f \left({S}\right)$ of $T$ has the same number of elements as $T$ and so therefore is $f \left({S}\right) = T$.


 * By Cardinality of Surjection, $(3)$ implies $(1)$.


 * By definition of bijection, $(1)$ implies $(2)$.