Bijection between Specific Elements

Lemma
Let $A$ and $B$ be sets.

Let $f: A \leftrightarrow B$ be a bijection.

Then given $a \in A$ and $b \in B$, there exists another bijection $g: A \leftrightarrow B$ such that $g \left ({a}\right) = b$.

Proof
Let $f \left ({a}\right) = b'$.

Since $f$ is surjective, there is $a' \in A$ such that $f \left({a'}\right) = b$.

We define the mapping $g: A \to B$ as:
 * $g \left({x}\right) = \begin{cases}

b & : x = a \\ b' & : x = a' \\ f \left({x}\right) & : x \in A \setminus \left\{{a, a'}\right\}\end{cases}$

$g$ is Surjective
Let $y \in B \setminus \left\{{b, b'}\right\}$.

Then:
 * $\exists x \in A: f \left({x}\right) = y$

and so by definition of $g$:


 * $\exists x \in A: g \left({x}\right) = y$

Otherwise:
 * $y = b \implies b = g \left({a}\right)$


 * $y = b' \implies b = g \left({a'}\right)$

and so $y$ is seen to be surjective.

$g$ is Injective
Suppose $g \left({x}\right) = g \left({x'}\right)$.

If $g \left({x}\right), g \left({x'}\right) \in B \setminus \left\{{b, b'}\right\}$ then $f \left({x}\right) = f \left({x'}\right)$.

As $f$ is bijective and so injective it follows that $x = x'$.

If $g \left({x}\right), g \left({x'}\right) \in \left\{{b, b'}\right\}$ then either:
 * $g \left({x}\right) = g \left({x'}\right) = b$ and so $x = x' = a$

or
 * $g \left({x}\right) = g \left({x'}\right) = b'$ and so $x = x' = a'$

For the other two cases:
 * $g \left({x}\right) \in \left\{{b, b'}\right\}, g \left({x'}\right) \notin \left\{{b, b'}\right\}$


 * $g \left({x}\right) \notin \left\{{b, b'}\right\}, g \left({x'}\right) \in \left\{{b, b'}\right\}$

it is clear that $g \left({x}\right) \ne g \left({x'}\right)$.

So $g$ has been shown to be injective.

$g$ is Bijective
So $g$ is both injective and surjective, and so by definition a bijection.

Comment
What this means is that given any two equivalent sets, it is always possible to find a bijection that connects any given pair of elements, one in one set and one in the other.