Same Cardinality Bijective Injective Surjective

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 Cardinality of Subset of Finite Set:

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 $(2)$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 17$: Theorem $17.7$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 22.4$
• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.10$: Proposition $\text{A}.10.1: 3$