Union of Mappings with Disjoint Domains is Mapping

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

Let $h = f \cup g$.

If $S_1 \cap S_2 = \varnothing$, 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 $\left({x, y_1}\right), \left({x, y_2}\right) \in h$.

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

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

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

So $\left({x, y_1}\right), \left({x, y_2}\right) \in h \implies y_1 = y_2$.

Now suppose $x \in \operatorname{Dom} \left({h}\right)$.

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

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

In either case, $\exists y \in T_1 \cup T_2: \left({x, y}\right) \in h$.

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

Note
If $S_1 \cap S_2 \ne \varnothing$ then there may exist $\left({x, y_1}\right) \in f$ and $\left({x, y_2}\right) \in g$ such that $y_1 \ne y_2$.

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