Direct Image Mapping of Left-Total Relation is Empty iff Argument is Empty

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

Let $\mathcal R: S \to T$ be a left-total relation on $S \times T$.

Let $\mathcal R^\to$ be the direct image mapping of $\mathcal R$:


 * $\mathcal R^\to: \powerset S \to \powerset T: \map {\mathcal R^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R}$

Then:
 * $\map {\mathcal R^\to} X = \O \iff X = \O$

Sufficient Condition
Let $\map {\mathcal R^\to} X = \O$.

By definition of direct image mapping:
 * $\set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R} = \O$

That is:
 * $\neg \exists s \in X: \tuple {s, t} \in \mathcal R$

But as $\mathcal R$ is a left-total relation:
 * $\forall s \in X: \exists t \in T: \tuple {s, t} \in \mathcal R$

Thus:
 * $\neg \exists s \in X$

and so:
 * $X = \O$

Sufficient Condition
Let $X = \O$.

Then by definition of direct image mapping:
 * $\map {\mathcal R^\to} X = \O$