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.
$\blacksquare$
Warning
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): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.8$