# 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.

$\card S = \card {f \sqbrk S}$

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.

$\Box$

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

Let $f$ be surjective.

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

By Cardinality of Surjection, $f$ is bijective.

$\Box$

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

Let $f$ be a bijection.

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

$\blacksquare$
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