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.

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

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

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

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

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.

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

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

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

and the sufficient condition applies for the main proof.

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

Necessary Condition (General Result)
Suppose $$\mathcal R$$ is one-to-many.

From Image of Intersection, we already have:
 * $$\mathcal R \left({\bigcap \mathbb S}\right) \subseteq \bigcap_{X \in \, \mathbb S} \mathcal R \left({X}\right)$$

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

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

Then:

$$ $$ $$ $$ $$ $$

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

Putting the results together:

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

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