Image of Intersection under One-to-Many Relation/General Result

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

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Then:
 * $\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): \mathcal R \left({\bigcap \mathbb S}\right) = \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right)$

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

Sufficient Condition
Suppose:
 * $\displaystyle \mathcal R \left({\bigcap \mathbb S}\right) = \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right)$

where $\mathbb S$ is any subset of $\mathcal P \left({S}\right)$.

Then by definition of $\mathbb S$:
 * $\forall S_1, S_2 \in \mathbb S: \mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$

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/General Result, we already have:
 * $\displaystyle \mathcal R \left({\bigcap \mathbb S}\right) \subseteq \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right)$

so we just need to show:
 * $\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right) \subseteq \mathcal R \left({\bigcap \mathbb S}\right)$

Let:
 * $\displaystyle t \in \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right)$.

Then:

So if $\mathcal R$ is one-to-many, it follows that:
 * $\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): \mathcal R \left({\bigcap \mathbb S}\right) = \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right)$

Putting the results together:

$\mathcal R$ is one-to-many iff:
 * $\displaystyle \mathcal R \left({\bigcap \mathbb S}\right) = \bigcap_{X \mathop \in \mathbb S} \mathcal R \left({X}\right)$

where $\mathbb S$ is any subset of $\mathcal P \left({S}\right)$.