Proof of the bijection principle

Proof

We compose bijections with the counting bijections in the definition of cardinality.

1. Let $n = |A|$. By definition, there is a bijection $h : \set{1, \dots, n} \to A$.
2. Let $f : A \to B$ be a bijection. A composite of bijections is a bijection, since injections compose and surjections compose.
3. So $f \circ h : \set{1, \dots, n} \to B$ is a bijection. Thus $B$ has $n$ elements.
4. The number of elements of a finite set is unique, so $|B| = n = |A|$.
5. Conversely, let $|A| = |B| = n$. There are bijections $g : \set{1, \dots, n} \to A$ and $k : \set{1, \dots, n} \to B$.
6. A bijection has a two-sided inverse, so $g^{-1} : A \to \set{1, \dots, n}$ exists and is a bijection.
7. By step 2, $k \circ g^{-1} : A \to B$ is a bijection. This proves the converse.