Condition for Agreement of Family of Mappings

Theorem
Let $\displaystyle{\displaystyle\left(A_i\right)}_{i \in I},\displaystyle{\displaystyle\left(B_i\right)}_{i \in I}$ be a family of non empty sets and $\displaystyle{\left(f_i\right)}_{i \in I}$ a family of functions such that $\forall i \in I, f_i \in \mathscr{F} \left( A_i, B_i\right)$. We have that $\bigcup_{i \in I} f_i \in \mathscr{F} \left( \bigcup_{i \in I}A_i, \bigcup_{i \in I}B_i\right)$ if, and only if,

$ \displaystyle\forall i,j \in I, \displaystyle\text{dom} f_i \cap \displaystyle\text{dom} f_j \neq \emptyset \implies \displaystyle\left(\forall a \in \left(\text{dom} f_i \cap \text{dom} f_j\right), \left(a,b \right) \in f_i \implies \left(a,b \right) \in f_j \right) $

Proof
Let ${\left(A_i\right)}_{i \in I},{\left(B_i\right)}_{i \in I}$ be a family of non empty sets and ${\left(f_i\right)}_{i \in I}$ a family of functions such that $\forall i \in I, f_i \in \mathscr{F} \left( A_i, B_i\right)$.

Let's suppose, in first place, that

$\bigcup_{i \in I} f_i \in \mathscr{F} \left( \bigcup_{i \in I}A_i, \bigcup_{i \in I}B_i\right)$.

Let $i,j \in I$ be such that $\text{dom} f_i \cap \text{dom} f_j \neq \emptyset$. Let $a \in \left(\text{dom} f_i \cap \text{dom} f_j\right)$ and let $b \in \bigcup_{i \in I}B_i$ such that $\left(a,b\right) \in f_i$. Let's suppose overlooking the absurdum, that $\left(a,b\right) \notin f_j$. As $a \in \left(\text{dom} f_i \cap \text{dom} f_j \right)$, then

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

As $\left(a,b\right) \in f_i$, we can conclude that $\left(a,b\right) \in \bigcup_{i \in I} f_i$. In these conditions we have that $\left(a,b\right), \left(a,c\right) \in \bigcup_{i \in I} f_i$ with $b \neq c$ and $\bigcup_{i \in I} f_i$ is a function, which is an absurdum.

So we have that $\left(a,b \right) \in f_j$.

Let's finally suppose that

$ \forall i,j \in I, \text{dom} f_i \cap \text{dom} f_j \neq \emptyset \implies \left(\forall a \in \left(\text{dom} f_i \cap \text{dom} f_j\right), \left(a,b \right) \in f_i \implies \left(a,b \right) \in f_j \right) $.

Let $a \in \bigcup_{i \in I}A_i$. We have, in these conditions, that

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

Let $k \in I$ be in these conditions. So $ a\in \text{dom} f_k$. Let $l=f_k(a)$. We can conclude that $\left(a,l \right) \in f_k$ and so $\left(a,l\right) \in \bigcup_{i \in I} f_i$.

Let's suppose overlooking the absurdum, that $\exists m \in \bigcup_{i \in I}B_i: \left( \left(a,m \right) \in \bigcup_{i \in I} f_i \land m\neq l \right)$

Let $m \in \bigcup_{i \in I}B_i$ be in these conditions and let $j \in I$ be such that $\left(a,m \right) \in f_j$. Having in mind our hypothesis, we have that

$a \in \left( \text{dom} f_k \cap\text{dom} f_j \right)$

and, as $\left(a,l \right) \in f_k$, we have that $\left(a,l \right) \in f_j$. Therefore, $\left( a,m \right),\left(a,l\right) \in f_j$, where $f_j \in \mathscr{F} \left( A_j, B_j\right)$, and $m \neq l$, which is an absurdum.

So we have that

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

So, $ \bigcup_{i \in I} f_i \in \mathscr{F} \left( \bigcup_{i \in I}A_i, \bigcup_{i \in I}B_i\right) $

Source
Translated and adapted by the author of the article: