Image of Subset under Relation is Subset of Image

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

Let $$A, B \subseteq S$$ such that $$A \subseteq B$$.

Then the image of $$A$$ is a subset of the image of $$B$$:
 * $$A \subseteq B \implies \mathcal{R} \left({A}\right) \subseteq \mathcal{R} \left({B}\right)$$

In the language of induced mappings, that would be written:
 * $$A \subseteq B \implies f_{\mathcal{R}} \left({A}\right) \subseteq f_{\mathcal{R}} \left({B}\right)$$

Corollary 1
The same applies to the preimage.

Let $$C, D \subseteq T$$.

Then:
 * $$C \subseteq D \implies f_{\mathcal{R}^{-1}} \left({C}\right) \subseteq f_{\mathcal{R}^{-1}} \left({D}\right)$$

where $$\mathcal{R}^{-1}$$ is the inverse of $$\mathcal{R}$$.

Corollary 2
The same applies for a mapping $$f: S \to T$$ and its inverse $$f^{-1} \subseteq T \times S$$, whether $$f^{-1}$$ is a mapping or not.

Let $$f: S \to T$$ be a mapping.

Let:
 * $$A, B \subseteq S$$
 * $$C, D \subseteq T$$

Then:
 * $$A \subseteq B \implies f \left({A}\right) \subseteq f \left({B}\right)$$
 * $$C \subseteq D \implies f^{-1} \left({C}\right) \subseteq f^{-1} \left({D}\right)$$

Proof
Suppose $$\mathcal{R} \left({A}\right) \not \subseteq \mathcal{R} \left({B}\right)$$.

$$ $$ $$ $$

... and the result follows by the Rule of Transposition.

Proof of Corollary 1
As $$\mathcal{R}^{-1}$$ is itself a relation, by definition of inverse relation, the main result applies directly.

Proof of Corollary 2
As $$f: S \to T$$ is a mapping, it is also a relation, and thus:
 * $$f \subseteq S \times T$$

and so is its inverse:
 * $$f^{-1} \subseteq T \times S$$

Hence, as for Corollary 1, the main result applies directly.