Union of Mappings which Agree is Mapping
Theorem
Let $A, B, Y$ be sets.
Let $f: A \to Y$ and $g: B \to Y$ be mappings.
Let $X = A \cup B$.
Let $f$ and $g$ agree on $A \cap B$.
Then $f \cup g: X \to Y$ is a mapping.
Family of Sets
Let $Y$ be a set.
Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $A_i \cap A_j$.
Then:
- $f : X \to Y$ is a mapping
Proof
By definition, $f \cup g$ is a relation whose domain is $X = A \cup B$.
Let $\tuple {x, y_1} \in f \cup g$ and $\tuple {x, y_2} \in f \cup g$.
At least one of the following must be true:
- $(1): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in f$
- $(2): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in g$
- $(3): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in g$
- $(4): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in f$
Because $f$ and $g$ are mappings, $(1)$ and $(2)$ imply that $y_1 = y_2$.
If $(3)$ holds, then $y_1 = \map f x$ and $y_2 = \map g x$.
But then $x \in A \cap B$.
So by hypothesis:
- $y_1 = \map f x = \map g x = y_2$
Similarly if $(4)$ holds.
Thus in all cases:
- $\tuple {x, y_1}, \tuple {x, y_2} \in f \cup g \implies y_1 = y_2$
and so by definition $f \cup g: X \to Y$ is a mapping.
$\blacksquare$
Sources
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.13$: The Restriction of a Function: Theorem $13.2$