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) $$f$$ is bijective;
 * 2) $$f$$ is injective;
 * 3) $$f$$ is surjective.

Proof

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

If $$f$$ is injective, then $$\left|{S}\right| = \left|{f \left({S}\right)}\right|$$.

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 $$2$$.

Source

 * : $$\S 22.4$$