# Equivalence of Mappings between Sets of Same Cardinality

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## 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.

$\Box$

$(3)$ 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.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 17$: Theorem $17.7$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 22.4$: Injections; bijections; inverse of a bijection - 1996: H. Jerome Keisler and Joel Robbin:
*Mathematical Logic and Computability*... (previous) ... (next): Appendix $\text{A}.10$: Finite Sequences: Proposition $\text{A}.10.1: \ 3$ - 2008: David Joyner:
*Adventures in Group Theory*(2nd ed.) ... (previous) ... (next): Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Ponderable $2.1.6$