Preimage of Subset under Composite Mapping

Theorem
Let $S_1, S_2, S_3$ be sets, and let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.

Denote with $g \circ f: S_1 \to S_3$ the composition of $g$ and $f$.

Let $S_3' \subseteq S_3$ be a subset of $S_3$.

Then:


 * $\left({g \circ f}\right)^{-1} \left({S_3'}\right) = f^{-1} \left({ g^{-1} \left({S_3'}\right)}\right)$

where $g^{-1} \left({S_3'}\right)$ denotes the preimage of $S_3'$ under $g$.

Proof
A mapping is a specific kind of relation.

Hence, Inverse of Composite Relation applies, and it follows that:


 * $\left({g \circ f}\right)^{-1} \left({S_3'}\right) = \left({f^{-1} \circ g^{-1}}\right) \left({S_3'}\right)$

From the definition of composition, the latter equals:


 * $\left({f^{-1} \circ g^{-1}}\right) \left({S_3'}\right) = f^{-1} \left({ g^{-1} \left({S_3'}\right)}\right)$

yielding the result.