Direct Image Mapping of Relation is Mapping

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

Let $f_\mathcal R$ be the mapping induced by $\mathcal R$:


 * $f_\mathcal R: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right): f_\mathcal R \left({X}\right) = \mathcal R \left({X}\right)$

Then $f_\mathcal R$ is indeed a mapping.

Proof
Take the general relation $\mathcal R \subseteq S \times T$.

Let $X \subseteq S$, i.e. $X \in \mathcal P \left({S}\right)$.


 * Suppose $X = \varnothing$. Then $\mathcal R \left({X}\right) = \varnothing \subseteq T$, from Image of Null is Null.


 * Suppose $X = S$. Then $\mathcal R \left({X}\right) = \operatorname{Im} \left({\mathcal R}\right) \subseteq T$ from Image of Domain is Subset of Codomain.


 * Finally, suppose $\varnothing \subset X \subset S$. From Image is Subset of Codomain, again we see that $\mathcal R \left({X}\right) \subseteq T$.

Now, from the definition of the power set, we have that $Y \subseteq T \iff Y \in \mathcal P \left({T}\right)$.

We defined $f_\mathcal R \subseteq \mathcal P \left({S}\right) \times \mathcal P \left({T}\right)$ such that:


 * $f_\mathcal R : \mathcal P \left({S}\right) \to \mathcal P \left({T}\right): f_\mathcal R \left({X}\right) = \mathcal R \left({X}\right)$

Clearly, by definition, there is only one $\mathcal R \left({X}\right)$ for any given $X$, and so $f_\mathcal R$ is functional.

We have shown that $\forall X \subseteq S: \mathcal R \left({X}\right) \in \mathcal P \left({T}\right)$.

So:
 * $\forall X \in \mathcal P \left({S}\right): \exists_1 Y \in \mathcal P \left({T}\right): \mathcal R \left({X}\right) = Y$

and thus:
 * $\forall X \in \mathcal P \left({S}\right): \exists_1 Y \in \mathcal P \left({T}\right): f_\mathcal R \left({X}\right) = Y$.

So:
 * $f_\mathcal R$ is defined for all $X \in \mathcal P \left({S}\right)$

and therefore $f_\mathcal R$ is a mapping.

Also see

 * Mapping Induced on Power Set by Inverse Relation