# Union of Bijections with Disjoint Domains and Codomains is Bijection

## Theorem

Let $A$, $B$, $C$, and $D$ be sets or classes.

Let $A \cap B = C \cap D = \O$.

Let $f: A \to C$ and $g: B \to D$ be bijections.

Then $f \cup g: A \cup B \to C \cup D$ is also a bijection.

### Corollary

Let $A$, $B$, $C$, and $D$ be sets or classes.

Let $A \cap B = C \cap D = \O$.

Let $f: A \to C$ and $g: D \to B$ be bijections.

Then $f \cup g^{-1}: A \cup B \to C \cup D$ is also a bijection.

## Proof

By the definition of bijection, $f$ and $g$ are many-to-one and one-to-many relations.

$f \cup g$ is many-to-one and one-to-many.

Thus to show $f \cup g$ is a bijection requires us only to demonstrate that it is both left-total and right-total.

We will show that $f \cup g$ is left-total.

Let $x \in A \cup B$.

Then $x \in A$ or $x \in B$.

If $x \in A$ then since $f$ is left-total there is a $y \in C$ such that $\tuple {x, y} \in f$.

By the definition of union, $\tuple {x, y} \in f \cup g$.

If $x \in B$ then since $g$ is left-total there is a $y \in D$ such that $\tuple {x, y} \in g$.

Then by the definition of union, $\tuple {x, y} \in f \cup g$.

As this holds for all $x$, $f \cup g$ is left-total.

The proof that $f \cup g$ is right-total is similar.

Thus it has been demonstrated that:

$f \cup g$ is many-to-one
$f \cup g$ is one-to-many
$f \cup g$ is left-total
$f \cup g$ is right-total

and therefore, by definition, a bijection.

$\blacksquare$