Image of Intersection under One-to-Many Relation/Family of Sets

Theorem
Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation.

Then $\mathcal R$ is a one-to-many relation iff:
 * $\displaystyle \mathcal R \left({\bigcap_{i \mathop \in I} S_i}\right) = \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right)$

where $\left\langle{S_i}\right\rangle_{i \in I}$ is any family of subsets of $S$.

Sufficient Condition
Suppose:
 * $\displaystyle \mathcal R \left({\bigcap_{i \mathop \in I} S_i}\right) = \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right)$

where $\left\langle{S_i}\right\rangle_{i \in I}$ is any family of subsets of $S$.

Then by definition of $\left\langle{S_i}\right\rangle_{i \in I}$:
 * $\forall i, j \in I: \mathcal R \left({S_i \cap S_j}\right) = \mathcal R \left({S_i}\right) \cap \mathcal R \left({S_j}\right)$

and the sufficient condition applies for One-to-Many Image of Intersections.

So $\mathcal R$ is one-to-many.

Necessary Condition
Suppose $\mathcal R$ is one-to-many.

From Image of Intersection/Family of Sets, we already have:
 * $\displaystyle \mathcal R \left({\bigcap_{i \mathop \in I} S_i}\right) \subseteq \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right)$

so we just need to show:
 * $\displaystyle \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right) \subseteq \mathcal R \left({\bigcap_{i \mathop \in I} S_i}\right)$

Let:
 * $\displaystyle t \in \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right)$

Then:

So if $\mathcal R$ is one-to-many, it follows that:
 * $\displaystyle \mathcal R \left({\bigcap_{i \mathop \in I} S_i}\right) = \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right)$

Putting the results together:

$\mathcal R$ is one-to-many iff:
 * $\displaystyle \mathcal R \left({\bigcap_{i \mathop \in I} S_i}\right) = \bigcap_{i \mathop \in I} \mathcal R \left({S_i}\right)$

where $\left\langle{S_i}\right\rangle_{i \in I}$ is any family of subsets of $S$.