Image of Intersection under One-to-Many Relation

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

Then:
 * $\forall S_1, S_2 \subseteq S: \mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$

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

Sufficient Condition
Suppose that:
 * $\forall S_1, S_2 \subseteq S: \mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$

If $S$ is singleton, the result follows immediately as $\mathcal R$ would have to be one-to-many.

So, assume $S$ is not singleton, and suppose $\mathcal R$ is specifically not one-to-many.

So:
 * $\exists x, y \in S: \exists z \in T: \left({x, z}\right) \in T, \left({y, z}\right) \in T, x \ne y$.

and of course $\left\{{x}\right\} \subseteq S, \left\{{y}\right\} \subseteq S$.

So:
 * $z \in \mathcal R \left({\left\{{x}\right\}}\right)$
 * $z \in \mathcal R \left({\left\{{y}\right\}}\right)$

and so by definition of intersection:
 * $z \in \mathcal R \left({\left\{{x}\right\}}\right) \cap \mathcal R \left({\left\{{y}\right\}}\right)$

But $\left\{{x}\right\} \cap \left\{{y}\right\} = \varnothing$.

Thus from Image of Null is Null:
 * $\mathcal R \left({\left\{{x}\right\} \cap \left\{{y}\right\}}\right) = \varnothing$

and so:
 * $\mathcal R \left({\left\{{x}\right\} \cap \left\{{y}\right\}}\right) \ne \mathcal R \left({\left\{{x}\right\}}\right) \cap \mathcal R \left({\left\{{y}\right\}}\right)$

Necessary Condition
Let $\mathcal R$ be one-to-many.

From Image of Intersection, we already have:


 * $\mathcal R \left({S_1 \cap S_2}\right) \subseteq \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$.

So we just need to show:


 * $\mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right) \subseteq \mathcal R \left({S_1 \cap S_2}\right)$.

Let $t \in \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$.

Then:

So if $\mathcal R$ is one-to-many, it follows that:
 * $\mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$

Putting the results together, we see that:
 * $\mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$ iff $\mathcal R$ is one-to-many.