Element in Preimage of Image under Mapping

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

Then:
 * $\forall x \in S: x \in f^{-1} \sqbrk {\map f x}$

Proof
A mapping is by definition a left-total relation.

Therefore Preimage of Image under Left-Total Relation is Superset applies:
 * $A \subseteq S \implies A \subseteq f^{-1} \sqbrk {f \sqbrk A}$

Thus:
 * $\set x \subseteq S \implies \set x \subseteq f^{-1} \sqbrk {f \sqbrk A}$

Hence the result.