Subset of Domain is Subset of Preimage of Image

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

Then:
 * $$A \subseteq S \implies A \subseteq \left({f^{-1} \circ f}\right) \left({A}\right)$$
 * $$B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) \subseteq B$$

Proof
As a mapping is by definition also a relation, we apply Preimage of Image directly:
 * $$A \subseteq S \implies A \subseteq \left({\mathcal R^{-1} \circ \mathcal R}\right) \left({A}\right)$$

where $$\mathcal R$$ is a relation.

Hence the first result:
 * $$A \subseteq S \implies A \subseteq \left({f^{-1} \circ f}\right) \left({A}\right)$$

The second result:
 * $$B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) \subseteq B$$

is a weaker version of Image of Preimage of Mapping:
 * $$B \subseteq T \implies \left({f \circ f^{-1}}\right) \left({B}\right) = B \cap f \left({S}\right)$$