# Union of Mappings with Disjoint Domains is Mapping

## Theorem

Let $S_1, S_2, T_1, T_2$ be sets.

Let $f: S_1 \to T_1$ and $g: S_2 \to T_2$ be mappings.

Let $h = f \cup g$ be their union.

If $S_1 \cap S_2 = \O$, then $h: S_1 \cup S_2 \to T_1 \cup T_2$ is a mapping whose domain is $S_1 \cup S_2$.

## Proof

From the definition of mapping, it is clear that $h$ is a relation.

Suppose $\tuple {x, y_1}, \tuple {x, y_2} \in h$.

Clearly $x \in S_1$ or $x \in S_2$ but as $S_1 \cap S_2 = \O$ it is not in both.

If $x \in S_1$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in f$, and $f$ is a mapping.

Similarly, if $x \in S_2$ then $y_1 = y_2$ as $\tuple {x, y_1}, \tuple {x, y_2} \in g$, and $g$ is a mapping.

So $\tuple {x, y_1}, \tuple {x, y_2} \in h \implies y_1 = y_2$.

Now suppose $x \in \Dom h$.

Either $x \in S_1$ or $x \in S_2$.

As both $f$ and $g$ are mappings it follows that either $\exists y \in T_1: \tuple {x, y} \in f$ or $\exists y \in T_2: \tuple {x, y} \in g$.

In either case, $\exists y \in T_1 \cup T_2: \tuple {x, y} \in h$.

So $h$ is a mapping whose domain is $S_1 \cup S_2$, as we were to show.

$\blacksquare$

## Note

If $S_1 \cap S_2 \ne \O$ then there may exist $\tuple {x, y_1} \in f$ and $\tuple {x, y_2} \in g$ such that $y_1 \ne y_2$.

In such a case $h = f \cup g$ is *not* a mapping.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $3.8$