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 $$\mathcal{R} \left({X}\right) = \mathrm{Im} \left({\mathcal{R}}\right) \subseteq T$$ from Image of Domain is Subset of Range.


 * Finally, suppose $$\varnothing \subset X \subset S$$. From Image is Subset of Range, 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.