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.