Preimage of Subset under Mapping equals Union of Preimages of Elements

Theorem
Let $S$ and $T$ be sets.

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

Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, defined as:


 * $f^{-1} = \left\{{\left({t, s}\right): f \left({s}\right) = t}\right\}$

Let $Y \subseteq T$ be a subset of $T$.

Then:
 * $\displaystyle f^{-1} \left[{Y}\right] = \bigcup_{y \mathop \in Y} f^{-1} \left({y}\right)$

where:
 * $f^{-1} \left[{Y}\right]$ is the preimage of the subset $Y$ under $f$
 * $f^{-1} \left({y}\right)$ is the preimage of the element $y$ under $f$.

Proof
By definition, $f^{-1} \subseteq T \times S$ is a relation on $T \times S$.

Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly.