Equivalence of Mappings between Finite Sets of Same Cardinality

Theorem
Let $S$ and $T$ be finite sets such that $\card S = \card T$.

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 $(3)$:

Let $f$ be an injection.

Then by Cardinality of Image of Injection:
 * $\card S = \card {f \sqbrk S}$

where $f \sqbrk S$ denotes the image of $S$ under $f$.

Therefore the subset $f \sqbrk S$ of $T$ has the same number of elements as $T$.

Therefore:
 * $f \sqbrk S = T$

and so $f$ is a surjection.

$(3)$ implies $(1)$:

Let $f$ be surjective.

We have that $\card S = \card T$.

By Cardinality of Surjection, $f$ is bijective.

$(1)$ implies $(2)$:

Let $f$ be a bijection.

By definition of bijection, $f$ is also an injection.