Condition for Agreement of Family of Mappings

Theorem

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

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

$\forall i \in I: f_i \in \mathcal F \left({A_i, B_i}\right)$

We have that:

$\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i }\right)$

if and only if:

$\displaystyle \forall i, j \in I: \operatorname{Dom} f_i \cap \operatorname{Dom} f_j \ne \varnothing \implies \left({\forall a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j}\right), \left({a, b}\right) \in f_i \implies \left({a, b}\right) \in f_j}\right)$

Proof

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

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

$\forall i \in I: f_i \in \mathcal F \left({A_i, B_i}\right)$

Sufficient Condition

Let:

$\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}\right)$

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

$\operatorname{Dom} f_i \cap \operatorname{Dom} f_j \ne \varnothing$

Let $a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j}\right)$

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

$\left({a, b}\right) \in f_i$

Aiming for a contradiction, suppose:

$\left({a, b}\right) \notin f_j$

As $a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j} \right)$:

$\displaystyle \exists c \in \bigcup_{i \mathop \in I} B_i: \left({a, c}\right) \in f_j$

As $\left({a, b}\right) \in f_i$:

$\displaystyle \left({a, b}\right) \in \bigcup_{i \mathop \in I} f_i$

Thus:

$\displaystyle \left({a, b}\right), \left({a, c}\right) \in \bigcup_{i \mathop \in I} f_i$

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

This is a contradiction.

Thus the supposition that the fact $\left({a, b}\right) \notin f_j$ was false.

So:

$\left({a, b}\right) \in f_j$

$\Box$

Necessary Condition

Let:

$\forall i, j \in I: \operatorname{Dom} f_i \cap \operatorname{Dom} f_j \ne \varnothing \implies \left({\forall a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j}\right), \left({a, b}\right) \in f_i \implies \left({a, b}\right) \in f_j}\right)$

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

Hence:

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

Let $k \in I$.

Thus:

$a \in \operatorname{Dom} f_k$

Let $l = f_k \left({a}\right)$

It follows that:

$\left({a, l}\right) \in f_k$

and so:

$\displaystyle \left({a, l}\right) \in \bigcup_{i \mathop \in I} f_i$

Aiming for a contradiction, suppose:

$\displaystyle \exists m \in \bigcup_{i \mathop \in I} B_i: \left({\left({a, m}\right) \in \bigcup_{i \mathop \in I} f_i \land m \ne l}\right)$

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

Let $j \in I$ be such that:

$\left({a, m}\right) \in f_j$

We have:

$a \in \left({\operatorname{Dom} f_k \cap \operatorname{Dom} f_j}\right)$

As $\left({a, l}\right) \in f_k$:

$\left(a,l \right) \in f_j$.

Therefore:

$\left({a, m}\right), \left({a, l}\right) \in f_j$

where $f_j \in \mathcal F \left({A_j, B_j}\right)$ and $m \ne l$.

This contradicts the definition of mapping.

So:

$\displaystyle \nexists m \in \bigcup_{i \mathop \in I} B_i: \left({\left({a, m}\right) \in \bigcup_{i \mathop \in I} f_i \land m \ne l}\right)$

and so:

$\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}\right)$

$\blacksquare$