Condition for Agreement of Family of Mappings

Theorem

Let $\family {A_i}_{i \mathop \in I}, \family {B_i}_{i \mathop \in I}$ be families of non empty sets.

Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings such that:

$\forall i \in I: f_i \in \map \FF {A_i, B_i}$

We have that:

$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$

if and only if:

$\ds \forall i, j \in I: \Dom {f_i} \cap \Dom {f_j} \ne \O \implies \paren {\forall a \in \paren {\Dom {f_i} \cap \Dom {f_j} }, \tuple {a, b} \in f_i \implies \tuple {a, b} \in f_j}$

Proof

Let $\family {A_i}_{i \mathop \in I}, \family {B_i}_{i \mathop \in I}$ be families of non empty sets.

Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings such that:

$\forall i \in I: f_i \in \map \FF {A_i, B_i}$

Sufficient Condition

Let:

$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$

Let $i, j \in I$ be such that:

$\Dom {f_i} \cap \Dom {f_j} \ne \O$

Let $a \in \paren {\Dom {f_i} \cap \Dom {f_j} }$

Let $\ds b \in \bigcup_{i \mathop \in I} B_i$ be such that:

$\tuple {a, b} \in f_i$

Aiming for a contradiction, suppose:

$\tuple {a, b} \notin f_j$

As $a \in \paren {\Dom {f_i} \cap \Dom {f_j} }$:

$\ds \exists c \in \bigcup_{i \mathop \in I} B_i: \tuple {a, c} \in f_j$

As $\tuple {a, b} \in f_i$:

$\ds \tuple {a, b} \in \bigcup_{i \mathop \in I} f_i$

Thus:

$\ds \tuple {a, b}, \tuple {a, c} \in \bigcup_{i \mathop \in I} f_i$

such that $b \ne c$ and $\ds \bigcup_{i \mathop \in I} f_i$ is a mapping.

This is a contradiction.

Thus the supposition that the fact $\tuple {a, b} \notin f_j$ was false.

So:

$\tuple {a, b} \in f_j$

$\Box$

Necessary Condition

Let:

$\forall i, j \in I: \Dom {f_i} \cap \Dom {f_j} \ne \O \implies \paren {\forall a \in \paren {\Dom {f_i} \cap \Dom {f_j} }, \tuple {a, b} \in f_i \implies \tuple {a, b} \in f_j}$

Let $\ds a \in \bigcup_{i \mathop \in I} A_i$.

Hence:

$\exists k \in I: a \in A_k$

Let $k \in I$.

Thus:

$a \in \operatorname{Dom} f_k$

Let $l = \map {f_k} a$.

It follows that:

$\tuple {a, l} \in f_k$

and so:

$\ds \tuple {a, l} \in \bigcup_{i \mathop \in I} f_i$

Aiming for a contradiction, suppose:

$\ds \exists m \in \bigcup_{i \mathop \in I} B_i: \paren {\tuple {a, m} \in \bigcup_{i \mathop \in I} f_i \land m \ne l}$

Let $\ds m \in \bigcup_{i \mathop \in I} B_i$.

Let $j \in I$ be such that:

$\tuple {a, m} \in f_j$

We have:

$a \in \paren {\Dom {f_k} \cap \Dom {f_j} }$

As $\tuple {a, l} \in f_k$:

$\tuple {a, l} \in f_j$.

Therefore:

$\tuple {a, m}, \tuple {a, l} \in f_j$

where $f_j \in \map \FF {A_j, B_j}$ and $m \ne l$.

This contradicts the definition of mapping.

So:

$\ds \nexists m \in \bigcup_{i \mathop \in I} B_i: \paren {\tuple {a, m} \in \bigcup_{i \mathop \in I} f_i \land m \ne l}$

and so:

$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$

$\blacksquare$