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:
- $\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i }\right)$
- $\displaystyle \forall i, j \in I: \operatorname{Dom} f_i \cap \operatorname{Dom} f_j \ne \varnothing \implies \left({\forall a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j}\right), \left({a, b}\right) \in f_i \implies \left({a, b}\right) \in f_j}\right)$
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:
- $\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}\right)$
$\blacksquare$